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NCERT Solutions · Class 12 Maths · NCERT Mathematics Part I and II Part II · Chapter 7

Chapter 7: Integrals (Calculus)

Step-by-step solutions to all exercises of Chapter 7, Integrals (NCERT Class 12 Maths Part II, 2026-27): Exercise 7.1 (anti-derivatives), 7.2 (substitution), 7.3 (trigonometric identities), 7.4 (special integrals), 7.5 (partial fractions), 7.6 (integration by parts), 7.7 (square-root forms), 7.8 and 7.9 (definite integrals), 7.10 (properties of definite integrals) and the Miscellaneous Exercise. All 261 questions are answered, with the key answer highlighted.

NCERT textbook Chapter 7 (PDF)

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Free NCERT solutions by Notes Bazar · www.notesbazar.in/ncert-solutions/class-12-maths/chapter-7-integrals

C denotes the constant of integration throughout. Standard forms: $\int \frac{d x}{x ^{2} + a ^{2}} = \frac{1}{a} tan^{- 1} \frac{x}{a}$ ; $\int \frac{d x}{x ^{2} - a ^{2}} = \frac{1}{2 a} lo g \frac{x - a}{x + a}$ ; $\int \frac{d x}{a ^{2} - x ^{2}} = \frac{1}{2 a} lo g \frac{a + x}{a - x}$ ; $\int \frac{d x}{a ^{2} - x ^{2}} = sin^{- 1} \frac{x}{a}$ ; $\int \frac{d x}{x ^{2} \pm a ^{2}} = lo g x + x^{2} \pm a^{2}$ . By parts: $\int uv d x = u \int v d x - \int (u^{'} \int v d x) d x$ ; $\int e^{x} [f (x) + f^{'} (x)] d x = e^{x} f (x)$ .

Exercise 7.1

Find an anti derivative (or integral) of sin 2x by the method of inspection.

Solution

We look for a function whose derivative is sin 2x. Since $\frac{d}{d x} (cos 2 x) = - 2 sin 2 x$ , we get $\frac{d}{d x} (- \frac{1}{2} cos 2 x) = sin 2 x$ .

$- \frac{1}{2} cos 2 x$

Find an anti derivative of cos 3x by the method of inspection.

Solution

$\frac{d}{d x} (sin 3 x) = 3 cos 3 x$ , so $\frac{d}{d x} (\frac{1}{3} sin 3 x) = cos 3 x$ .

$\frac{1}{3} sin 3 x$

Find an anti derivative of e²ˣ by the method of inspection.

Solution

$\frac{d}{d x} (e^{2 x}) = 2 e^{2 x}$ , so $\frac{d}{d x} (\frac{1}{2} e^{2 x}) = e^{2 x}$ .

$\frac{1}{2} e^{2 x}$

Find an anti derivative of (ax + b)² by the method of inspection.

Solution

$\frac{d}{d x} (a x + b)^{3} = 3 a (a x + b)^{2}$ , so $\frac{d}{d x} \frac{( a x + b ) ^{3}}{3 a} = (a x + b)^{2}$ .

$\frac{1}{3 a} (a x + b)^{3}$

Find an anti derivative of sin 2x – 4e³ˣ by the method of inspection.

Solution

$\frac{d}{d x} (- \frac{1}{2} cos 2 x - \frac{4}{3} e^{3 x}) = sin 2 x - 4 e^{3 x}$ .

$- \frac{1}{2} cos 2 x - \frac{4}{3} e^{3 x}$

Find

\int (4 e^{3 x} + 1) d x

Solution

$= 4 \cdot \frac{e ^{3 x}}{3} + x + C$

$\frac{4}{3} e^{3 x} + x + C$

Find

\int x^{2} (1 - \frac{1}{x ^{2}}) d x

Solution

$= \int (x^{2} - 1) d x = \frac{x ^{3}}{3} - x + C$

$\frac{x ^{3}}{3} - x + C$

Find

\int (a x^{2} + b x + c) d x

Solution

$\frac{a x ^{3}}{3} + \frac{b x ^{2}}{2} + c x + C$

Find

\int (2 x^{2} + e^{x}) d x

Solution

$\frac{2}{3} x^{3} + e^{x} + C$

Find

\int (x - \frac{1}{x})^{2} d x

Solution

$(x - \frac{1}{x})^{2} = x - 2 + \frac{1}{x}$ , so the integral is $\frac{x ^{2}}{2} - 2 x + lo g ∣ x ∣ + C$ .

$\frac{x ^{2}}{2} + lo g ∣ x ∣ - 2 x + C$

Find

\int \frac{x ^{3} + 5 x ^{2} - 4}{x ^{2}} d x

Solution

$= \int (x + 5 - 4 x^{- 2}) d x = \frac{x ^{2}}{2} + 5 x - 4 \cdot \frac{x ^{- 1}}{- 1} + C$

$\frac{x ^{2}}{2} + 5 x + \frac{4}{x} + C$

Find

\int \frac{x ^{3} + 3 x + 4}{x} d x

Solution

$= \int (x^{5/2} + 3 x^{1/2} + 4 x^{- 1/2}) d x = \frac{x ^{7/2}}{7/2} + 3 \cdot \frac{x ^{3/2}}{3/2} + 4 \cdot \frac{x ^{1/2}}{1/2} + C$

$\frac{2}{7} x^{7/2} + 2 x^{3/2} + 8 x + C$

Find

\int \frac{x ^{3} - x ^{2} + x - 1}{x - 1} d x

Solution

$x^{3} - x^{2} + x - 1 = x^{2} (x - 1) + (x - 1) = (x - 1) (x^{2} + 1)$ , so the integrand is $x^{2} + 1$ .

$\frac{x ^{3}}{3} + x + C$

Find

\int (1 - x) x d x

Solution

$= \int (x^{1/2} - x^{3/2}) d x$

$\frac{2}{3} x^{3/2} - \frac{2}{5} x^{5/2} + C$

Find

\int x (3 x^{2} + 2 x + 3) d x

Solution

$= \int (3 x^{5/2} + 2 x^{3/2} + 3 x^{1/2}) d x = 3 \cdot \frac{2}{7} x^{7/2} + 2 \cdot \frac{2}{5} x^{5/2} + 3 \cdot \frac{2}{3} x^{3/2} + C$

$\frac{6}{7} x^{7/2} + \frac{4}{5} x^{5/2} + 2 x^{3/2} + C$

Find

\int (2 x - 3 cos x + e^{x}) d x

Solution

$x^{2} - 3 sin x + e^{x} + C$

Find

\int (2 x^{2} - 3 sin x + 5 x) d x

Solution

$= \frac{2}{3} x^{3} + 3 cos x + 5 \cdot \frac{2}{3} x^{3/2} + C$

$\frac{2}{3} x^{3} + 3 cos x + \frac{10}{3} x^{3/2} + C$

Find

\int sec x (sec x + tan x) d x

Solution

$= \int (sec^{2} x + sec x tan x) d x$

$tan x + sec x + C$

Find

\int \frac{sec ^{2} x}{cosec ^{2} x} d x

Solution

$\frac{sec ^{2} x}{cosec ^{2} x} = \frac{sin ^{2} x}{cos ^{2} x} = tan^{2} x = sec^{2} x - 1$

$tan x - x + C$

Find

\int \frac{2 - 3 sin x}{cos ^{2} x} d x

Solution

$= \int (2 sec^{2} x - 3 sec x tan x) d x$

$2 tan x - 3 sec x + C$

The anti derivative of

(x + \frac{1}{x})

equals (A)

\frac{1}{3} x^{1/3} + 2 x^{1/2} + C

(B)

\frac{2}{3} x^{2/3} + \frac{1}{2} x^{2} + C

(C)

\frac{2}{3} x^{3/2} + 2 x^{1/2} + C

(D)

\frac{3}{2} x^{3/2} + \frac{1}{2} x^{1/2} + C

Solution

$\int (x^{1/2} + x^{- 1/2}) d x = \frac{2}{3} x^{3/2} + 2 x^{1/2} + C$

(C)

\frac{d}{d x} f (x) = 4 x^{3} - \frac{3}{x ^{4}}

such that f (2) = 0. Then f (x) is (A)

x^{4} + \frac{1}{x ^{3}} - \frac{129}{8}

(B)

x^{3} + \frac{1}{x ^{4}} + \frac{129}{8}

(C)

x^{4} + \frac{1}{x ^{3}} + \frac{129}{8}

(D)

x^{3} + \frac{1}{x ^{4}} - \frac{129}{8}

Solution

$f (x) = \int (4 x^{3} - 3 x^{- 4}) d x = x^{4} + x^{- 3} + C$ .

$f (2) = 16 + \frac{1}{8} + C = 0 \Rightarrow C = - \frac{129}{8}$ .

(A)

Exercise 7.2

Integrate

\frac{2 x}{1 + x ^{2}}

Solution

Put $t = 1 + x^{2}$ , $d t = 2 x d x$ : $\int \frac{d t}{t} = lo g ∣ t ∣$ .

$lo g (1 + x^{2}) + C$

Integrate

\frac{( lo g x ) ^{2}}{x}

Solution

Put $t = lo g ∣ x ∣$ , $d t = \frac{d x}{x}$ : $\int t^{2} d t = \frac{t ^{3}}{3}$ .

$\frac{1}{3} (lo g ∣ x ∣)^{3} + C$

Integrate

\frac{1}{x + x lo g x}

Solution

$= \frac{1}{x ( 1 + lo g x )}$ . Put $t = 1 + lo g x$ , $d t = \frac{d x}{x}$ : $\int \frac{d t}{t}$ .

$lo g ∣1 + lo g x ∣ + C$

Integrate sin x · sin (cos x).

Solution

Put $t = cos x$ , $d t = - sin x d x$ : $- \int sin t d t = cos t$ .

$cos (cos x) + C$

Integrate sin (ax + b) cos (ax + b).

Solution

$= \frac{1}{2} sin 2 (a x + b)$ . $\int \frac{1}{2} sin (2 a x + 2 b) d x = - \frac{cos ( 2 a x + 2 b )}{4 a}$ .

$- \frac{1}{4 a} cos 2 (a x + b) + C$

Integrate

a x + b

Solution

Put $t = a x + b$ , $d t = a d x$ : $\frac{1}{a} \int t^{1/2} d t = \frac{1}{a} \cdot \frac{2}{3} t^{3/2}$ .

$\frac{2}{3 a} (a x + b)^{3/2} + C$

Integrate

x x + 2

Solution

Put $t = x + 2$ ( $x = t - 2$ , $d x = d t$ ): $\int (t - 2) t^{1/2} d t = \int (t^{3/2} - 2 t^{1/2}) d t = \frac{2}{5} t^{5/2} - \frac{4}{3} t^{3/2}$ .

$\frac{2}{5} (x + 2)^{5/2} - \frac{4}{3} (x + 2)^{3/2} + C$

Integrate

x 1 + 2 x^{2}

Solution

Put $t = 1 + 2 x^{2}$ , $d t = 4 x d x$ : $\frac{1}{4} \int t^{1/2} d t = \frac{1}{4} \cdot \frac{2}{3} t^{3/2}$ .

$\frac{1}{6} (1 + 2 x^{2})^{3/2} + C$

Integrate

(4 x + 2) x^{2} + x + 1

Solution

Put $t = x^{2} + x + 1$ , $d t = (2 x + 1) d x$ , so $(4 x + 2) d x = 2 d t$ : $2 \int t^{1/2} d t = \frac{4}{3} t^{3/2}$ .

$\frac{4}{3} (x^{2} + x + 1)^{3/2} + C$

Integrate

\frac{1}{x - x}

Solution

$\frac{1}{x - x} = \frac{1}{x ( x - 1 )}$ . Put $t = x - 1$ , $d t = \frac{d x}{2 x}$ : $2 \int \frac{d t}{t}$ .

$2 lo g ∣ x - 1∣ + C$

Integrate

\frac{x}{x + 4}

, x > 0.

Solution

Put $t = x + 4$ : $\int \frac{t - 4}{t} d t = \int (t^{1/2} - 4 t^{- 1/2}) d t = \frac{2}{3} t^{3/2} - 8 t^{1/2} = \frac{2}{3} t (t - 12)$ .

$\frac{2}{3} x + 4 (x - 8) + C$

Integrate

(x^{3} - 1)^{1/3} x^{5}

Solution

Put $t = x^{3} - 1$ , $d t = 3 x^{2} d x$ , $x^{3} = t + 1$ : $\int t^{1/3} (t + 1) \frac{d t}{3} = \frac{1}{3} [\frac{3}{7} t^{7/3} + \frac{3}{4} t^{4/3}]$ .

$\frac{1}{7} (x^{3} - 1)^{7/3} + \frac{1}{4} (x^{3} - 1)^{4/3} + C$

Integrate

\frac{x ^{2}}{( 2 + 3 x ^{3} ) ^{3}}

Solution

Put $t = 2 + 3 x^{3}$ , $d t = 9 x^{2} d x$ : $\frac{1}{9} \int t^{- 3} d t = \frac{1}{9} \cdot \frac{t ^{- 2}}{- 2}$ .

$- \frac{1}{18 ( 2 + 3 x ^{3} ) ^{2}} + C$

Integrate

\frac{1}{x ( lo g x ) ^{m}}

, x > 0, m ≠ 1.

Solution

Put $t = lo g x$ : $\int t^{- m} d t = \frac{t ^{1 - m}}{1 - m}$ .

$\frac{( lo g x ) ^{1 - m}}{1 - m} + C$

Integrate

\frac{x}{9 - 4 x ^{2}}

Solution

Put $t = 9 - 4 x^{2}$ , $d t = - 8 x d x$ : $- \frac{1}{8} \int \frac{d t}{t}$ .

$- \frac{1}{8} lo g ∣9 - 4 x^{2} ∣ + C$

Integrate e^(2x + 3).

Solution

Put $t = 2 x + 3$ : $\frac{1}{2} \int e^{t} d t$ .

$\frac{1}{2} e^{2 x + 3} + C$

Integrate

\frac{x}{e ^{x^{2}}}

Solution

Put $t = x^{2}$ , $d t = 2 x d x$ : $\frac{1}{2} \int e^{- t} d t = - \frac{1}{2} e^{- t}$ .

$- \frac{1}{2 e ^{x^{2}}} + C$

Integrate

\frac{e ^{t a n^{- 1} x}}{1 + x ^{2}}

Solution

Put $t = tan^{- 1} x$ , $d t = \frac{d x}{1 + x ^{2}}$ : $\int e^{t} d t$ .

$e^{t a n^{- 1} x} + C$

Integrate

\frac{e ^{2 x} - 1}{e ^{2 x} + 1}

Solution

Divide numerator and denominator by $e^{x}$ : $\frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}}$ . The numerator is the derivative of the denominator.

$lo g (e^{x} + e^{- x}) + C$

Integrate

\frac{e ^{2 x} - e ^{- 2 x}}{e ^{2 x} + e ^{- 2 x}}

Solution

Put $t = e^{2 x} + e^{- 2 x}$ , $d t = 2 (e^{2 x} - e^{- 2 x}) d x$ : $\frac{1}{2} \int \frac{d t}{t}$ .

$\frac{1}{2} lo g (e^{2 x} + e^{- 2 x}) + C$

Integrate tan² (2x – 3).

Solution

$tan^{2} (2 x - 3) = sec^{2} (2 x - 3) - 1$ .

$\frac{1}{2} tan (2 x - 3) - x + C$

Integrate sec² (7 – 4x).

Solution

$\int sec^{2} (7 - 4 x) d x = \frac{tan ( 7 - 4 x )}{- 4}$ .

$- \frac{1}{4} tan (7 - 4 x) + C$

Integrate

\frac{sin ^{- 1} x}{1 - x ^{2}}

Solution

Put $t = sin^{- 1} x$ , $d t = \frac{d x}{1 - x ^{2}}$ : $\int t d t = \frac{t ^{2}}{2}$ .

$\frac{1}{2} (sin^{- 1} x)^{2} + C$

Integrate

\frac{2 cos x - 3 sin x}{6 cos x + 4 sin x}

Solution

$= \frac{1}{2} \cdot \frac{2 cos x - 3 sin x}{3 cos x + 2 sin x}$ , and $\frac{d}{d x} (3 cos x + 2 sin x) = 2 cos x - 3 sin x$ .

$\frac{1}{2} lo g ∣2 sin x + 3 cos x ∣ + C$

Integrate

\frac{1}{cos ^{2} x ( 1 - tan x ) ^{2}}

Solution

Put $t = 1 - tan x$ , $d t = - sec^{2} x d x$ : $- \int t^{- 2} d t = \frac{1}{t}$ .

$\frac{1}{1 - tan x} + C$

Integrate

\frac{cos x}{x}

Solution

Put $t = x$ , $d t = \frac{d x}{2 x}$ : $2 \int cos t d t$ .

$2 sin x + C$

Integrate

sin 2 x cos 2 x

Solution

Put $t = sin 2 x$ , $d t = 2 cos 2 x d x$ : $\frac{1}{2} \int t^{1/2} d t = \frac{1}{3} t^{3/2}$ .

$\frac{1}{3} (sin 2 x)^{3/2} + C$

Integrate

\frac{cos x}{1 + sin x}

Solution

Put $t = 1 + sin x$ : $\int t^{- 1/2} d t = 2 t$ .

$2 1 + sin x + C$

Integrate cot x log sin x.

Solution

Put $t = lo g sin x$ , $d t = cot x d x$ : $\int t d t$ .

$\frac{1}{2} (lo g sin x)^{2} + C$

Integrate

\frac{sin x}{1 + cos x}

Solution

Put $t = 1 + cos x$ , $d t = - sin x d x$ : $- \int \frac{d t}{t}$ .

$- lo g ∣1 + cos x ∣ + C$

Integrate

\frac{sin x}{( 1 + cos x ) ^{2}}

Solution

Put $t = 1 + cos x$ : $- \int t^{- 2} d t = \frac{1}{t}$ .

$\frac{1}{1 + cos x} + C$

Integrate

\frac{1}{1 + cot x}

Solution

$\frac{1}{1 + cot x} = \frac{sin x}{sin x + cos x} = \frac{1}{2} \cdot \frac{( sin x + cos x ) - ( cos x - sin x )}{sin x + cos x} = \frac{1}{2} [1 - \frac{cos x - sin x}{sin x + cos x}]$

The second fraction has numerator = derivative of denominator.

$\frac{x}{2} - \frac{1}{2} lo g ∣ cos x + sin x ∣ + C$

Integrate

\frac{1}{1 - tan x}

Solution

$\frac{cos x}{cos x - sin x} = \frac{1}{2} [1 + \frac{cos x + sin x}{cos x - sin x}]$ , and $\frac{d}{d x} (cos x - sin x) = - (sin x + cos x)$ .

$\int = \frac{x}{2} - \frac{1}{2} lo g ∣ cos x - sin x ∣$

$\frac{x}{2} - \frac{1}{2} lo g ∣ cos x - sin x ∣ + C$

Integrate

\frac{tan x}{sin x cos x}

Solution

$\frac{tan x}{sin x cos x} = \frac{tan x sec ^{2} x}{tan x} = \frac{sec ^{2} x}{tan x}$ . Put $t = tan x$ : $\int t^{- 1/2} d t = 2 t$ .

$2 tan x + C$

Integrate

\frac{( 1 + lo g x ) ^{2}}{x}

Solution

Put $t = 1 + lo g x$ : $\int t^{2} d t$ .

$\frac{1}{3} (1 + lo g x)^{3} + C$

Integrate

\frac{( x + 1 ) ( x + lo g x ) ^{2}}{x}

Solution

Put $t = x + lo g x$ , $d t = (1 + \frac{1}{x}) d x = \frac{x + 1}{x} d x$ : $\int t^{2} d t$ .

$\frac{1}{3} (x + lo g x)^{3} + C$

Integrate

\frac{x ^{3} sin ( tan ^{- 1} x ^{4} )}{1 + x ^{8}}

Solution

Put $t = tan^{- 1} x^{4}$ , $d t = \frac{4 x ^{3}}{1 + x ^{8}} d x$ : $\frac{1}{4} \int sin t d t = - \frac{1}{4} cos t$ .

$- \frac{1}{4} cos (tan^{- 1} x^{4}) + C$

\int \frac{10 x ^{9} + 1 0 ^{x} lo g _{e} 10}{x ^{10} + 1 0 ^{x}} d x

equals (A) 10ˣ – x¹⁰ + C (B) 10ˣ + x¹⁰ + C (C) (10ˣ – x¹⁰)⁻¹ + C (D) log (10ˣ + x¹⁰) + C

Solution

The numerator is the derivative of $x^{10} + 1 0^{x}$ .

(D)

\int \frac{d x}{sin ^{2} x cos ^{2} x}

equals (A) tan x + cot x + C (B) tan x – cot x + C (C) tan x cot x + C (D) tan x – cot 2x + C

Solution

$\frac{sin ^{2} x + cos ^{2} x}{sin ^{2} x cos ^{2} x} = sec^{2} x + cosec^{2} x$ , so the integral is $tan x - cot x + C$ .

(B)

Exercise 7.3

Integrate sin² (2x + 5).

Solution

$sin^{2} (2 x + 5) = \frac{1 - cos ( 4 x + 10 )}{2}$

$\frac{x}{2} - \frac{1}{8} sin (4 x + 10) + C$

Integrate sin 3x cos 4x.

Solution

$sin 3 x cos 4 x = \frac{1}{2} [sin 7 x + sin (- x)] = \frac{1}{2} (sin 7 x - sin x)$

$- \frac{1}{14} cos 7 x + \frac{1}{2} cos x + C$

Integrate cos 2x cos 4x cos 6x.

Solution

$cos 2 x cos 4 x = \frac{1}{2} (cos 6 x + cos 2 x)$ . Multiply by $cos 6 x$ :

$\frac{1}{2} (cos^{2} 6 x + cos 2 x cos 6 x) = \frac{1}{2} [\frac{1 + cos 12 x}{2} + \frac{cos 8 x + cos 4 x}{2}] = \frac{1}{4} (1 + cos 12 x + cos 8 x + cos 4 x)$

$\frac{1}{4} [x + \frac{sin 12 x}{12} + \frac{sin 8 x}{8} + \frac{sin 4 x}{4}] + C$

Integrate sin³ (2x + 1).

Solution

$sin^{3} (2 x + 1) = [1 - cos^{2} (2 x + 1)] sin (2 x + 1)$ . Put $t = cos (2 x + 1)$ , $d t = - 2 sin (2 x + 1) d x$ :

$- \frac{1}{2} \int (1 - t^{2}) d t = - \frac{1}{2} (t - \frac{t ^{3}}{3})$

$- \frac{1}{2} cos (2 x + 1) + \frac{1}{6} cos^{3} (2 x + 1) + C$

Integrate sin³ x cos³ x.

Solution

$= sin x (1 - cos^{2} x) cos^{3} x$ . Put $t = cos x$ , $d t = - sin x d x$ : $- \int (t^{3} - t^{5}) d t = - \frac{t ^{4}}{4} + \frac{t ^{6}}{6}$ .

$\frac{1}{6} cos^{6} x - \frac{1}{4} cos^{4} x + C$

Integrate sin x sin 2x sin 3x.

Solution

$sin x sin 3 x = \frac{1}{2} (cos 2 x - cos 4 x)$ . Multiply by $sin 2 x$ :

$\frac{1}{2} [\frac{1}{2} sin 4 x - \frac{1}{2} (sin 6 x - sin 2 x)] = \frac{1}{4} (sin 2 x + sin 4 x - sin 6 x)$

$\frac{1}{4} [\frac{cos 6 x}{6} - \frac{cos 4 x}{4} - \frac{cos 2 x}{2}] + C$

Integrate sin 4x sin 8x.

Solution

$sin 4 x sin 8 x = \frac{1}{2} (cos 4 x - cos 12 x)$

$\frac{1}{2} [\frac{sin 4 x}{4} - \frac{sin 12 x}{12}] + C$

Integrate

\frac{1 - cos x}{1 + cos x}

Solution

$\frac{2 sin ^{2} \frac{x}{2}}{2 cos ^{2} \frac{x}{2}} = tan^{2} \frac{x}{2} = sec^{2} \frac{x}{2} - 1$

$2 tan \frac{x}{2} - x + C$

Integrate

\frac{cos x}{1 + cos x}

Solution

$= 1 - \frac{1}{1 + cos x} = 1 - \frac{1}{2} sec^{2} \frac{x}{2}$

$x - tan \frac{x}{2} + C$

Integrate sin⁴ x.

Solution

$sin^{4} x = (\frac{1 - cos 2 x}{2})^{2} = \frac{1}{4} (1 - 2 cos 2 x + \frac{1 + cos 4 x}{2}) = \frac{3}{8} - \frac{1}{2} cos 2 x + \frac{1}{8} cos 4 x$

$\frac{3 x}{8} - \frac{1}{4} sin 2 x + \frac{1}{32} sin 4 x + C$

Integrate cos⁴ 2x.

Solution

$cos^{4} 2 x = \frac{1}{4} (1 + cos 4 x)^{2} = \frac{1}{4} (1 + 2 cos 4 x + \frac{1 + cos 8 x}{2}) = \frac{3}{8} + \frac{1}{2} cos 4 x + \frac{1}{8} cos 8 x$

$\frac{3 x}{8} + \frac{1}{8} sin 4 x + \frac{1}{64} sin 8 x + C$

Integrate

\frac{sin ^{2} x}{1 + cos x}

Solution

$\frac{( 1 - cos x ) ( 1 + cos x )}{1 + cos x} = 1 - cos x$

$x - sin x + C$

Integrate

\frac{cos 2 x - cos 2 α}{cos x - cos α}

Solution

$cos 2 x - cos 2 α = 2 cos^{2} x - 2 cos^{2} α = 2 (cos x - cos α) (cos x + cos α)$ , so the integrand is $2 (cos x + cos α)$ .

$2 (sin x + x cos α) + C$

Integrate

\frac{cos x - sin x}{1 + sin 2 x}

Solution

$1 + sin 2 x = (cos x + sin x)^{2}$ . Put $t = cos x + sin x$ , $d t = (cos x - sin x) d x$ : $\int t^{- 2} d t = - \frac{1}{t}$ .

$- \frac{1}{cos x + sin x} + C$

Integrate tan³ 2x sec 2x.

Solution

$= (sec^{2} 2 x - 1) sec 2 x tan 2 x$ . Put $t = sec 2 x$ , $d t = 2 sec 2 x tan 2 x d x$ : $\frac{1}{2} \int (t^{2} - 1) d t$ .

$\frac{1}{6} sec^{3} 2 x - \frac{1}{2} sec 2 x + C$

Integrate tan⁴ x.

Solution

$tan^{4} x = tan^{2} x (sec^{2} x - 1) = tan^{2} x sec^{2} x - (sec^{2} x - 1)$

$\int tan^{2} x sec^{2} x d x = \frac{tan ^{3} x}{3}$ (put $t = tan x$ ).

$\frac{1}{3} tan^{3} x - tan x + x + C$

Integrate

\frac{sin ^{3} x + cos ^{3} x}{sin ^{2} x cos ^{2} x}

Solution

$= \frac{sin x}{cos ^{2} x} + \frac{cos x}{sin ^{2} x} = sec x tan x + cosec x cot x$

$sec x - cosec x + C$

Integrate

\frac{cos 2 x + 2 sin ^{2} x}{cos ^{2} x}

Solution

$cos 2 x + 2 sin^{2} x = (1 - 2 sin^{2} x) + 2 sin^{2} x = 1$ , so the integrand is $sec^{2} x$ .

$tan x + C$

Integrate

\frac{1}{sin x cos ^{3} x}

Solution

$\frac{sin ^{2} x + cos ^{2} x}{sin x cos ^{3} x} = \frac{sin x}{cos ^{3} x} + \frac{1}{sin x cos x} = tan x sec^{2} x + \frac{sec ^{2} x}{tan x}$

Both parts use $t = tan x$ : $\int t d t + \int \frac{d t}{t}$ .

$lo g ∣ tan x ∣ + \frac{1}{2} tan^{2} x + C$

Integrate

\frac{cos 2 x}{( cos x + sin x ) ^{2}}

Solution

$\frac{( cos x - sin x ) ( cos x + sin x )}{( cos x + sin x ) ^{2}} = \frac{cos x - sin x}{cos x + sin x}$ (numerator = derivative of denominator).

$lo g ∣ cos x + sin x ∣ + C$

Integrate sin⁻¹ (cos x).

Solution

$sin^{- 1} (cos x) = sin^{- 1} (sin (\frac{π}{2} - x)) = \frac{π}{2} - x$ (for $0 \leq x \leq π$ ).

$\frac{π x}{2} - \frac{x ^{2}}{2} + C$

Integrate

\frac{1}{cos ( x - a ) cos ( x - b )}

Solution

Multiply and divide by $sin (a - b) = sin [(x - b) - (x - a)]$ :

$\frac{1}{sin ( a - b )} \cdot \frac{sin ( x - b ) cos ( x - a ) - cos ( x - b ) sin ( x - a )}{cos ( x - a ) cos ( x - b )} = \frac{1}{sin ( a - b )} [tan (x - b) - tan (x - a)]$

$\int = \frac{1}{sin ( a - b )} [- lo g ∣ cos (x - b) ∣ + lo g ∣ cos (x - a) ∣]$

$\frac{1}{sin ( a - b )} lo g \frac{cos ( x - a )}{cos ( x - b )} + C$

\int \frac{sin ^{2} x - cos ^{2} x}{sin ^{2} x cos ^{2} x} d x

is equal to (A) tan x + cot x + C (B) tan x + cosec x + C (C) – tan x + cot x + C (D) tan x + sec x + C

Solution

$= sec^{2} x - cosec^{2} x$ , whose integral is $tan x + cot x + C$ .

(A)

\int \frac{e ^{x} ( 1 + x )}{cos ^{2} ( e ^{x} x )} d x

equals (A) – cot (exˣ) + C (B) tan (xeˣ) + C (C) tan (eˣ) + C (D) cot (eˣ) + C

Solution

Put $t = x e^{x}$ , $d t = e^{x} (1 + x) d x$ : $\int sec^{2} t d t = tan t$ .

(B)

Exercise 7.4

Integrate

\frac{3 x ^{2}}{x ^{6} + 1}

Solution

Put $t = x^{3}$ , $d t = 3 x^{2} d x$ : $\int \frac{d t}{t ^{2} + 1} = tan^{- 1} t$ .

$tan^{- 1} x^{3} + C$

Integrate

\frac{1}{1 + 4 x ^{2}}

Solution

Put $t = 2 x$ , $d x = \frac{d t}{2}$ : $\frac{1}{2} \int \frac{d t}{1 + t ^{2}} = \frac{1}{2} lo g t + 1 + t^{2}$ .

$\frac{1}{2} lo g 2 x + 1 + 4 x^{2} + C$

Integrate

\frac{1}{( 2 - x ) ^{2} + 1}

Solution

Put $t = 2 - x$ , $d t = - d x$ : $- \int \frac{d t}{t ^{2} + 1} = - lo g t + t^{2} + 1$ , and $(2 - x)^{2} + 1 = x^{2} - 4 x + 5$ .

$- lo g 2 - x + x^{2} - 4 x + 5 + C$

Integrate

\frac{1}{9 - 25 x ^{2}}

Solution

Put $t = 5 x$ : $\frac{1}{5} \int \frac{d t}{3 ^{2} - t ^{2}} = \frac{1}{5} sin^{- 1} \frac{t}{3}$ .

$\frac{1}{5} sin^{- 1} \frac{5 x}{3} + C$

Integrate

\frac{3 x}{1 + 2 x ^{4}}

Solution

Put $t = 2 x^{2}$ , $d t = 22 x d x$ : $\frac{3}{2 2} \int \frac{d t}{1 + t ^{2}}$ .

$\frac{3}{2 2} tan^{- 1} (2 x^{2}) + C$

Integrate

\frac{x ^{2}}{1 - x ^{6}}

Solution

Put $t = x^{3}$ , $d t = 3 x^{2} d x$ : $\frac{1}{3} \int \frac{d t}{1 - t ^{2}} = \frac{1}{3} \cdot \frac{1}{2} lo g \frac{1 + t}{1 - t}$ .

$\frac{1}{6} lo g \frac{1 + x ^{3}}{1 - x ^{3}} + C$

Integrate

\frac{x - 1}{x ^{2} - 1}

Solution

Split: $\int \frac{x}{x ^{2} - 1} d x - \int \frac{d x}{x ^{2} - 1} = x^{2} - 1 - lo g x + x^{2} - 1$ (first part by $t = x^{2} - 1$ ).

$x^{2} - 1 - lo g x + x^{2} - 1 + C$

Integrate

\frac{x ^{2}}{x ^{6} + a ^{6}}

Solution

Put $t = x^{3}$ , $d t = 3 x^{2} d x$ : $\frac{1}{3} \int \frac{d t}{t ^{2} + ( a ^{3} ) ^{2}} = \frac{1}{3} lo g t + t^{2} + a^{6}$ .

$\frac{1}{3} lo g x^{3} + x^{6} + a^{6} + C$

Integrate

\frac{sec ^{2} x}{tan ^{2} x + 4}

Solution

Put $t = tan x$ , $d t = sec^{2} x d x$ : $\int \frac{d t}{t ^{2} + 2 ^{2}}$ .

$lo g tan x + tan^{2} x + 4 + C$

Integrate

\frac{1}{x ^{2} + 2 x + 2}

Solution

$x^{2} + 2 x + 2 = (x + 1)^{2} + 1$ . So $\int \frac{d x}{( x + 1 ) ^{2} + 1} = lo g (x + 1) + (x + 1)^{2} + 1$ .

$lo g x + 1 + x^{2} + 2 x + 2 + C$

Integrate

\frac{1}{9 x ^{2} + 6 x + 5}

Solution

$9 x^{2} + 6 x + 5 = (3 x + 1)^{2} + 4$ . Put $t = 3 x + 1$ : $\frac{1}{3} \int \frac{d t}{t ^{2} + 2 ^{2}} = \frac{1}{3} \cdot \frac{1}{2} tan^{- 1} \frac{t}{2}$ .

$\frac{1}{6} tan^{- 1} (\frac{3 x + 1}{2}) + C$

Integrate

\frac{1}{7 - 6 x - x ^{2}}

Solution

$7 - 6 x - x^{2} = 16 - (x + 3)^{2}$ . So $\int \frac{d x}{4 ^{2} - ( x + 3 ) ^{2}} = sin^{- 1} \frac{x + 3}{4}$ .

$sin^{- 1} (\frac{x + 3}{4}) + C$

Integrate

\frac{1}{( x - 1 ) ( x - 2 )}

Solution

$(x - 1) (x - 2) = x^{2} - 3 x + 2 = (x - \frac{3}{2})^{2} - \frac{1}{4}$ . So the integral is $lo g x - \frac{3}{2} + (x - \frac{3}{2})^{2} - \frac{1}{4}$ .

$lo g x - \frac{3}{2} + x^{2} - 3 x + 2 + C$

Integrate

\frac{1}{8 + 3 x - x ^{2}}

Solution

$8 + 3 x - x^{2} = \frac{41}{4} - (x - \frac{3}{2})^{2}$ . So $sin^{- 1} \frac{x - \frac{3}{2}}{\frac{41}{2}}$ .

$sin^{- 1} (\frac{2 x - 3}{41}) + C$

Integrate

\frac{1}{( x - a ) ( x - b )}

Solution

$(x - a) (x - b) = x^{2} - (a + b) x + ab = (x - \frac{a + b}{2})^{2} - (\frac{a - b}{2})^{2}$ .

$lo g x - \frac{a + b}{2} + (x - a) (x - b) + C$

Integrate

\frac{4 x + 1}{2 x ^{2} + x - 3}

Solution

$\frac{d}{d x} (2 x^{2} + x - 3) = 4 x + 1$ . Put $t = 2 x^{2} + x - 3$ : $\int t^{- 1/2} d t = 2 t$ .

$2 2 x^{2} + x - 3 + C$

Integrate

\frac{x + 2}{x ^{2} - 1}

Solution

$\int \frac{x}{x ^{2} - 1} d x + 2 \int \frac{d x}{x ^{2} - 1} = x^{2} - 1 + 2 lo g x + x^{2} - 1$ .

$x^{2} - 1 + 2 lo g x + x^{2} - 1 + C$

Integrate

\frac{5 x - 2}{1 + 2 x + 3 x ^{2}}

Solution

Write $5 x - 2 = A \frac{d}{d x} (3 x^{2} + 2 x + 1) + B = A (6 x + 2) + B$ : $A = \frac{5}{6}$ , $B = - 2 - \frac{5}{3} = - \frac{11}{3}$ .

$\int \frac{6 x + 2}{3 x ^{2} + 2 x + 1} d x = lo g ∣3 x^{2} + 2 x + 1∣$ .

$\int \frac{d x}{3 x ^{2} + 2 x + 1} = \frac{1}{3} \int \frac{d x}{( x + \frac{1}{3} ) ^{2} + \frac{2}{9}} = \frac{1}{3} \cdot \frac{3}{2} tan^{- 1} \frac{x + \frac{1}{3}}{\frac{2}{3}} = \frac{1}{2} tan^{- 1} \frac{3 x + 1}{2}$ .

$\frac{5}{6} lo g ∣3 x^{2} + 2 x + 1∣ - \frac{11}{3 2} tan^{- 1} (\frac{3 x + 1}{2}) + C$

Integrate

\frac{6 x + 7}{( x - 5 ) ( x - 4 )}

Solution

$(x - 5) (x - 4) = x^{2} - 9 x + 20$ ; $6 x + 7 = 3 (2 x - 9) + 34$ .

$3 \int \frac{2 x - 9}{x ^{2} - 9 x + 20} d x = 6 x^{2} - 9 x + 20$ .

$34 \int \frac{d x}{( x - \frac{9}{2} ) ^{2} - \frac{1}{4}} = 34 lo g x - \frac{9}{2} + x^{2} - 9 x + 20$ .

$6 x^{2} - 9 x + 20 + 34 lo g x - \frac{9}{2} + x^{2} - 9 x + 20 + C$

Integrate

\frac{x + 2}{4 x - x ^{2}}

Solution

$x + 2 = - \frac{1}{2} (4 - 2 x) + 4$ .

$- \frac{1}{2} \int \frac{4 - 2 x}{4 x - x ^{2}} d x = - 4 x - x^{2}$ ; $4 \int \frac{d x}{4 - ( x - 2 ) ^{2}} = 4 sin^{- 1} \frac{x - 2}{2}$ .

$- 4 x - x^{2} + 4 sin^{- 1} (\frac{x - 2}{2}) + C$

Integrate

\frac{x + 2}{x ^{2} + 2 x + 3}

Solution

$x + 2 = \frac{1}{2} (2 x + 2) + 1$ .

$\frac{1}{2} \int \frac{2 x + 2}{x ^{2} + 2 x + 3} d x = x^{2} + 2 x + 3$ ; $\int \frac{d x}{( x + 1 ) ^{2} + 2} = lo g x + 1 + x^{2} + 2 x + 3$ .

$x^{2} + 2 x + 3 + lo g x + 1 + x^{2} + 2 x + 3 + C$

Integrate

\frac{x + 3}{x ^{2} - 2 x - 5}

Solution

$x + 3 = \frac{1}{2} (2 x - 2) + 4$ .

$\frac{1}{2} \int \frac{2 x - 2}{x ^{2} - 2 x - 5} d x = \frac{1}{2} lo g ∣ x^{2} - 2 x - 5∣$ .

$4 \int \frac{d x}{( x - 1 ) ^{2} - ( 6 ) ^{2}} = 4 \cdot \frac{1}{2 6} lo g \frac{x - 1 - 6}{x - 1 + 6}$ .

$\frac{1}{2} lo g ∣ x^{2} - 2 x - 5∣ + \frac{2}{6} lo g \frac{x - 1 - 6}{x - 1 + 6} + C$

Integrate

\frac{5 x + 3}{x ^{2} + 4 x + 10}

Solution

$5 x + 3 = \frac{5}{2} (2 x + 4) - 7$ .

$\frac{5}{2} \int \frac{2 x + 4}{x ^{2} + 4 x + 10} d x = 5 x^{2} + 4 x + 10$ ; $- 7 \int \frac{d x}{( x + 2 ) ^{2} + 6} = - 7 lo g x + 2 + x^{2} + 4 x + 10$ .

$5 x^{2} + 4 x + 10 - 7 lo g x + 2 + x^{2} + 4 x + 10 + C$

\int \frac{d x}{x ^{2} + 2 x + 2}

equals (A) x tan⁻¹ (x + 1) + C (B) tan⁻¹ (x + 1) + C (C) (x + 1) tan⁻¹ x + C (D) tan⁻¹ x + C

Solution

$x^{2} + 2 x + 2 = (x + 1)^{2} + 1$ , so the integral is $tan^{- 1} (x + 1) + C$ .

(B)

\int \frac{d x}{9 x - 4 x ^{2}}

equals (A)

\frac{1}{9} sin^{- 1} (\frac{9 x - 8}{8}) + C

(B)

\frac{1}{2} sin^{- 1} (\frac{8 x - 9}{9}) + C

(C)

\frac{1}{3} sin^{- 1} (\frac{9 x - 8}{8}) + C

(D)

\frac{1}{2} sin^{- 1} (\frac{9 x - 8}{9}) + C

Solution

$9 x - 4 x^{2} = 4 [\frac{81}{64} - (x - \frac{9}{8})^{2}]$ .

$\int = \frac{1}{2} \int \frac{d x}{( \frac{9}{8} ) ^{2} - ( x - \frac{9}{8} ) ^{2}} = \frac{1}{2} sin^{- 1} \frac{x - \frac{9}{8}}{\frac{9}{8}} = \frac{1}{2} sin^{- 1} \frac{8 x - 9}{9}$

(B)

Exercise 7.5

Integrate

\frac{x}{( x + 1 ) ( x + 2 )}

Solution

$\frac{x}{( x + 1 ) ( x + 2 )} = \frac{A}{x + 1} + \frac{B}{x + 2}$ , $x = A (x + 2) + B (x + 1)$ .

$x = - 1$ : $A = - 1$ . $x = - 2$ : $- 2 = - B \Rightarrow B = 2$ .

$\int = - lo g ∣ x + 1∣ + 2 lo g ∣ x + 2∣$

$lo g \frac{( x + 2 ) ^{2}}{∣ x + 1∣} + C$

Integrate

\frac{1}{x ^{2} - 9}

Solution

$\frac{1}{( x - 3 ) ( x + 3 )} = \frac{1}{6} [\frac{1}{x - 3} - \frac{1}{x + 3}]$

$\frac{1}{6} lo g \frac{x - 3}{x + 3} + C$

Integrate

\frac{3 x - 1}{( x - 1 ) ( x - 2 ) ( x - 3 )}

Solution

$3 x - 1 = A (x - 2) (x - 3) + B (x - 1) (x - 3) + C (x - 1) (x - 2)$ .

$x = 1$ : $2 = 2 A \Rightarrow A = 1$ . $x = 2$ : $5 = - B \Rightarrow B = - 5$ . $x = 3$ : $8 = 2 C \Rightarrow C = 4$ .

$lo g ∣ x - 1∣ - 5 lo g ∣ x - 2∣ + 4 lo g ∣ x - 3∣ + C$

Integrate

\frac{x}{( x - 1 ) ( x - 2 ) ( x - 3 )}

Solution

$x = A (x - 2) (x - 3) + B (x - 1) (x - 3) + C (x - 1) (x - 2)$ .

$x = 1$ : $1 = 2 A \Rightarrow A = \frac{1}{2}$ . $x = 2$ : $2 = - B \Rightarrow B = - 2$ . $x = 3$ : $3 = 2 C \Rightarrow C = \frac{3}{2}$ .

$\frac{1}{2} lo g ∣ x - 1∣ - 2 lo g ∣ x - 2∣ + \frac{3}{2} lo g ∣ x - 3∣ + C$

Integrate

\frac{2 x}{x ^{2} + 3 x + 2}

Solution

$\frac{2 x}{( x + 1 ) ( x + 2 )} = \frac{A}{x + 1} + \frac{B}{x + 2}$ : $x = - 1$ : $- 2 = A$ ; $x = - 2$ : $- 4 = - B \Rightarrow B = 4$ .

$4 lo g ∣ x + 2∣ - 2 lo g ∣ x + 1∣ + C$

Integrate

\frac{1 - x ^{2}}{x ( 1 - 2 x )}

Solution

Improper fraction. Dividing: $\frac{1 - x ^{2}}{x - 2 x ^{2}} = \frac{1}{2} + \frac{1 - \frac{x}{2}}{x ( 1 - 2 x )}$ .

$1 - \frac{x}{2} = A (1 - 2 x) + B x$ : $x = 0$ : $A = 1$ ; $x = \frac{1}{2}$ : $\frac{3}{4} = \frac{B}{2} \Rightarrow B = \frac{3}{2}$ .

$\int = \frac{x}{2} + lo g ∣ x ∣ + \frac{3}{2} \cdot \frac{lo g ∣1 - 2 x ∣}{- 2}$

$\frac{x}{2} + lo g ∣ x ∣ - \frac{3}{4} lo g ∣1 - 2 x ∣ + C$

Integrate

\frac{x}{( x ^{2} + 1 ) ( x - 1 )}

Solution

$\frac{x}{( x ^{2} + 1 ) ( x - 1 )} = \frac{A}{x - 1} + \frac{B x + C}{x ^{2} + 1}$ ; $x = A (x^{2} + 1) + (B x + C) (x - 1)$ .

$x = 1$ : $A = \frac{1}{2}$ . Coefficient of $x^{2}$ : $0 = A + B \Rightarrow B = - \frac{1}{2}$ . Constant: $0 = A - C \Rightarrow C = \frac{1}{2}$ .

$\int = \frac{1}{2} lo g ∣ x - 1∣ - \frac{1}{2} \int \frac{x}{x ^{2} + 1} d x + \frac{1}{2} \int \frac{d x}{x ^{2} + 1}$

$\frac{1}{2} lo g ∣ x - 1∣ - \frac{1}{4} lo g (x^{2} + 1) + \frac{1}{2} tan^{- 1} x + C$

Integrate

\frac{x}{( x - 1 ) ^{2} ( x + 2 )}

Solution

$x = A (x - 1) (x + 2) + B (x + 2) + C (x - 1)^{2}$ .

$x = 1$ : $1 = 3 B \Rightarrow B = \frac{1}{3}$ . $x = - 2$ : $- 2 = 9 C \Rightarrow C = - \frac{2}{9}$ . Coefficient of $x^{2}$ : $0 = A + C \Rightarrow A = \frac{2}{9}$ .

$\int = \frac{2}{9} lo g ∣ x - 1∣ - \frac{1}{3 ( x - 1 )} - \frac{2}{9} lo g ∣ x + 2∣$

$\frac{2}{9} lo g \frac{x - 1}{x + 2} - \frac{1}{3 ( x - 1 )} + C$

Integrate

\frac{3 x + 5}{x ^{3} - x ^{2} - x + 1}

Solution

$x^{3} - x^{2} - x + 1 = (x - 1)^{2} (x + 1)$ . $3 x + 5 = A (x - 1) (x + 1) + B (x + 1) + C (x - 1)^{2}$ .

$x = 1$ : $8 = 2 B \Rightarrow B = 4$ . $x = - 1$ : $2 = 4 C \Rightarrow C = \frac{1}{2}$ . $x^{2}$ : $0 = A + C \Rightarrow A = - \frac{1}{2}$ .

$\int = - \frac{1}{2} lo g ∣ x - 1∣ - \frac{4}{x - 1} + \frac{1}{2} lo g ∣ x + 1∣$

$\frac{1}{2} lo g \frac{x + 1}{x - 1} - \frac{4}{x - 1} + C$

Integrate

\frac{2 x - 3}{( x ^{2} - 1 ) ( 2 x + 3 )}

Solution

$\frac{2 x - 3}{( x + 1 ) ( x - 1 ) ( 2 x + 3 )} = \frac{A}{x + 1} + \frac{B}{x - 1} + \frac{C}{2 x + 3}$ .

$x = - 1$ : $- 5 = A (- 2) (1) \Rightarrow A = \frac{5}{2}$ . $x = 1$ : $- 1 = B (2) (5) \Rightarrow B = - \frac{1}{10}$ . $x = - \frac{3}{2}$ : $- 6 = C (- \frac{1}{2}) (- \frac{5}{2}) = \frac{5}{4} C \Rightarrow C = - \frac{24}{5}$ .

$\int = \frac{5}{2} lo g ∣ x + 1∣ - \frac{1}{10} lo g ∣ x - 1∣ - \frac{24}{5} \cdot \frac{1}{2} lo g ∣2 x + 3∣$

$\frac{5}{2} lo g ∣ x + 1∣ - \frac{1}{10} lo g ∣ x - 1∣ - \frac{12}{5} lo g ∣2 x + 3∣ + C$

Integrate

\frac{5 x}{( x + 1 ) ( x ^{2} - 4 )}

Solution

$\frac{5 x}{( x + 1 ) ( x + 2 ) ( x - 2 )} = \frac{A}{x + 1} + \frac{B}{x + 2} + \frac{C}{x - 2}$ .

$x = - 1$ : $- 5 = A (1) (- 3) \Rightarrow A = \frac{5}{3}$ . $x = - 2$ : $- 10 = B (- 1) (- 4) \Rightarrow B = - \frac{5}{2}$ . $x = 2$ : $10 = C (3) (4) \Rightarrow C = \frac{5}{6}$ .

$\frac{5}{3} lo g ∣ x + 1∣ - \frac{5}{2} lo g ∣ x + 2∣ + \frac{5}{6} lo g ∣ x - 2∣ + C$

Integrate

\frac{x ^{3} + x + 1}{x ^{2} - 1}

Solution

Dividing: $\frac{x ^{3} + x + 1}{x ^{2} - 1} = x + \frac{2 x + 1}{x ^{2} - 1}$ .

$\frac{2 x + 1}{( x - 1 ) ( x + 1 )} = \frac{A}{x - 1} + \frac{B}{x + 1}$ : $x = 1$ : $3 = 2 A$ ; $x = - 1$ : $- 1 = - 2 B$ . So $A = \frac{3}{2}$ , $B = \frac{1}{2}$ .

$\frac{x ^{2}}{2} + \frac{3}{2} lo g ∣ x - 1∣ + \frac{1}{2} lo g ∣ x + 1∣ + C$

Integrate

\frac{2}{( 1 - x ) ( 1 + x ^{2} )}

Solution

$2 = A (1 + x^{2}) + (B x + C) (1 - x)$ . $x = 1$ : $A = 1$ . $x^{2}$ : $0 = A - B \Rightarrow B = 1$ . Constant: $2 = A + C \Rightarrow C = 1$ .

$\int [\frac{1}{1 - x} + \frac{x}{1 + x ^{2}} + \frac{1}{1 + x ^{2}}] d x$

$- lo g ∣1 - x ∣ + \frac{1}{2} lo g (1 + x^{2}) + tan^{- 1} x + C$

Integrate

\frac{3 x - 1}{( x + 2 ) ^{2}}

Solution

$3 x - 1 = 3 (x + 2) - 7$ , so $\frac{3 x - 1}{( x + 2 ) ^{2}} = \frac{3}{x + 2} - \frac{7}{( x + 2 ) ^{2}}$ .

$3 lo g ∣ x + 2∣ + \frac{7}{x + 2} + C$

Integrate

\frac{1}{x ^{4} - 1}

Solution

$\frac{1}{( x ^{2} - 1 ) ( x ^{2} + 1 )} = \frac{1}{2} [\frac{1}{x ^{2} - 1} - \frac{1}{x ^{2} + 1}]$

$\int = \frac{1}{2} [\frac{1}{2} lo g \frac{x - 1}{x + 1} - tan^{- 1} x]$

$\frac{1}{4} lo g \frac{x - 1}{x + 1} - \frac{1}{2} tan^{- 1} x + C$

Integrate

\frac{1}{x ( x ^{n} + 1 )}

. [Hint: multiply numerator and denominator by xⁿ⁻¹ and put xⁿ = t]

Solution

$\frac{x ^{n - 1}}{x ^{n} ( x ^{n} + 1 )}$ . Put $t = x^{n}$ , $d t = n x^{n - 1} d x$ : $\frac{1}{n} \int \frac{d t}{t ( t + 1 )} = \frac{1}{n} \int (\frac{1}{t} - \frac{1}{t + 1}) d t$ .

$\frac{1}{n} lo g \frac{x ^{n}}{x ^{n} + 1} + C$

Integrate

\frac{cos x}{( 1 - sin x ) ( 2 - sin x )}

. [Hint: Put sin x = t]

Solution

$t = sin x$ : $\int \frac{d t}{( 1 - t ) ( 2 - t )}$ . $1 = A (2 - t) + B (1 - t)$ : $t = 1$ : $A = 1$ ; $t = 2$ : $B = - 1$ .

$\int [\frac{1}{1 - t} - \frac{1}{2 - t}] d t = - lo g ∣1 - t ∣ + lo g ∣2 - t ∣$

$lo g \frac{2 - sin x}{1 - sin x} + C$

Integrate

\frac{( x ^{2} + 1 ) ( x ^{2} + 2 )}{( x ^{2} + 3 ) ( x ^{2} + 4 )}

Solution

Put $y = x^{2}$ (only for splitting): $\frac{( y + 1 ) ( y + 2 )}{( y + 3 ) ( y + 4 )} = 1 - \frac{4 y + 10}{( y + 3 ) ( y + 4 )}$ .

$\frac{4 y + 10}{( y + 3 ) ( y + 4 )} = \frac{A}{y + 3} + \frac{B}{y + 4}$ : $y = - 3$ : $A = - 2$ ; $y = - 4$ : $- 6 = - B \Rightarrow B = 6$ .

So the integrand is $1 + \frac{2}{x ^{2} + 3} - \frac{6}{x ^{2} + 4}$ .

$x + \frac{2}{3} tan^{- 1} \frac{x}{3} - 3 tan^{- 1} \frac{x}{2} + C$

Integrate

\frac{2 x}{( x ^{2} + 1 ) ( x ^{2} + 3 )}

Solution

Put $t = x^{2}$ , $d t = 2 x d x$ : $\int \frac{d t}{( t + 1 ) ( t + 3 )} = \frac{1}{2} \int [\frac{1}{t + 1} - \frac{1}{t + 3}] d t$ .

$\frac{1}{2} lo g (\frac{x ^{2} + 1}{x ^{2} + 3}) + C$

Integrate

\frac{1}{x ( x ^{4} - 1 )}

Solution

$\frac{x ^{3}}{x ^{4} ( x ^{4} - 1 )}$ . Put $t = x^{4}$ , $d t = 4 x^{3} d x$ : $\frac{1}{4} \int \frac{d t}{t ( t - 1 )} = \frac{1}{4} \int [\frac{1}{t - 1} - \frac{1}{t}] d t$ .

$\frac{1}{4} lo g \frac{x ^{4} - 1}{x ^{4}} + C$

Integrate

\frac{1}{e ^{x} - 1}

. [Hint: Put eˣ = t]

Solution

$t = e^{x}$ , $d x = \frac{d t}{t}$ : $\int \frac{d t}{t ( t - 1 )} = \int [\frac{1}{t - 1} - \frac{1}{t}] d t = lo g \frac{t - 1}{t}$ .

$lo g \frac{e ^{x} - 1}{e ^{x}} + C$

\int \frac{x d x}{( x - 1 ) ( x - 2 )}

equals (A)

lo g \frac{( x - 1 ) ^{2}}{x - 2} + C

(B)

lo g \frac{( x - 2 ) ^{2}}{x - 1} + C

(C)

lo g (\frac{x - 1}{x - 2})^{2} + C

(D) log |(x – 1)(x – 2)| + C

Solution

$x = A (x - 2) + B (x - 1)$ : $A = - 1$ , $B = 2$ . $\int = - lo g ∣ x - 1∣ + 2 lo g ∣ x - 2∣ = lo g \frac{( x - 2 ) ^{2}}{x - 1}$ .

(B)

\int \frac{d x}{x ( x ^{2} + 1 )}

equals (A) log |x| – ½ log (x² + 1) + C (B) log |x| + ½ log (x² + 1) + C (C) – log |x| + ½ log (x² + 1) + C (D) ½ log |x| + log (x² + 1) + C

Solution

$\frac{1}{x ( x ^{2} + 1 )} = \frac{1}{x} - \frac{x}{x ^{2} + 1}$ , giving $lo g ∣ x ∣ - \frac{1}{2} lo g (x^{2} + 1) + C$ .

(A)

Exercise 7.6

Integrate x sin x.

Solution

By parts with $u = x$ : $x (- cos x) - \int (- cos x) d x$ .

$- x cos x + sin x + C$

Integrate x sin 3x.

Solution

$x (- \frac{cos 3 x}{3}) + \int \frac{cos 3 x}{3} d x$

$- \frac{x}{3} cos 3 x + \frac{1}{9} sin 3 x + C$

Integrate x² eˣ.

Solution

$x^{2} e^{x} - \int 2 x e^{x} d x = x^{2} e^{x} - 2 (x e^{x} - e^{x})$

$e^{x} (x^{2} - 2 x + 2) + C$

Integrate x log x.

Solution

$u = lo g x$ : $lo g x \cdot \frac{x ^{2}}{2} - \int \frac{1}{x} \cdot \frac{x ^{2}}{2} d x$

$\frac{x ^{2}}{2} lo g x - \frac{x ^{2}}{4} + C$

Integrate x log 2x.

Solution

$lo g 2 x \cdot \frac{x ^{2}}{2} - \int \frac{1}{x} \cdot \frac{x ^{2}}{2} d x$ (since $\frac{d}{d x} lo g 2 x = \frac{1}{x}$ )

$\frac{x ^{2}}{2} lo g 2 x - \frac{x ^{2}}{4} + C$

Integrate x² log x.

Solution

$lo g x \cdot \frac{x ^{3}}{3} - \int \frac{1}{x} \cdot \frac{x ^{3}}{3} d x$

$\frac{x ^{3}}{3} lo g x - \frac{x ^{3}}{9} + C$

Integrate x sin⁻¹ x.

Solution

$sin^{- 1} x \cdot \frac{x ^{2}}{2} - \frac{1}{2} \int \frac{x ^{2}}{1 - x ^{2}} d x$ .

$\int \frac{x ^{2}}{1 - x ^{2}} d x = \int [\frac{1}{1 - x ^{2}} - 1 - x^{2}] d x = sin^{- 1} x - \frac{x}{2} 1 - x^{2} - \frac{1}{2} sin^{- 1} x = \frac{1}{2} sin^{- 1} x - \frac{x}{2} 1 - x^{2}$

So $I = \frac{x ^{2}}{2} sin^{- 1} x - \frac{1}{4} sin^{- 1} x + \frac{x}{4} 1 - x^{2}$ .

$\frac{2 x ^{2} - 1}{4} sin^{- 1} x + \frac{x}{4} 1 - x^{2} + C$

Integrate x tan⁻¹ x.

Solution

$tan^{- 1} x \cdot \frac{x ^{2}}{2} - \frac{1}{2} \int \frac{x ^{2}}{1 + x ^{2}} d x = \frac{x ^{2}}{2} tan^{- 1} x - \frac{1}{2} \int (1 - \frac{1}{1 + x ^{2}}) d x$

$= \frac{x ^{2}}{2} tan^{- 1} x - \frac{x}{2} + \frac{1}{2} tan^{- 1} x$

$\frac{x ^{2} + 1}{2} tan^{- 1} x - \frac{x}{2} + C$

Integrate x cos⁻¹ x.

Solution

$cos^{- 1} x \cdot \frac{x ^{2}}{2} + \frac{1}{2} \int \frac{x ^{2}}{1 - x ^{2}} d x = \frac{x ^{2}}{2} cos^{- 1} x + \frac{1}{2} [\frac{1}{2} sin^{- 1} x - \frac{x}{2} 1 - x^{2}]$ (from Q7).

Using $sin^{- 1} x = \frac{π}{2} - cos^{- 1} x$ (the constant goes into C):

$\frac{2 x ^{2} - 1}{4} cos^{- 1} x - \frac{x}{4} 1 - x^{2} + C$

Integrate (sin⁻¹ x)².

Solution

Take 1 as the second function: $x (sin^{- 1} x)^{2} - \int 2 sin^{- 1} x \cdot \frac{x}{1 - x ^{2}} d x$ .

For the last integral, $\int \frac{x}{1 - x ^{2}} d x = - 1 - x^{2}$ , so $\int sin^{- 1} x \frac{x}{1 - x ^{2}} d x = - 1 - x^{2} sin^{- 1} x + \int d x = - 1 - x^{2} sin^{- 1} x + x$ .

$x (sin^{- 1} x)^{2} + 2 1 - x^{2} sin^{- 1} x - 2 x + C$

Integrate

\frac{x cos ^{- 1} x}{1 - x ^{2}}

Solution

$u = cos^{- 1} x$ , $\int \frac{x}{1 - x ^{2}} d x = - 1 - x^{2}$ :

$- 1 - x^{2} cos^{- 1} x - \int (- 1 - x^{2}) (\frac{- 1}{1 - x ^{2}}) d x = - 1 - x^{2} cos^{- 1} x - x$

$- [1 - x^{2} cos^{- 1} x + x] + C$

Integrate x sec² x.

Solution

$x tan x - \int tan x d x = x tan x - lo g ∣ sec x ∣$

$x tan x + lo g ∣ cos x ∣ + C$

Integrate tan⁻¹ x.

Solution

$x tan^{- 1} x - \int \frac{x}{1 + x ^{2}} d x$

$x tan^{- 1} x - \frac{1}{2} lo g (1 + x^{2}) + C$

Integrate x (log x)².

Solution

$(lo g x)^{2} \cdot \frac{x ^{2}}{2} - \int 2 lo g x \cdot \frac{1}{x} \cdot \frac{x ^{2}}{2} d x = \frac{x ^{2}}{2} (lo g x)^{2} - \int x lo g x d x$

$= \frac{x ^{2}}{2} (lo g x)^{2} - [\frac{x ^{2}}{2} lo g x - \frac{x ^{2}}{4}]$

$\frac{x ^{2}}{2} (lo g x)^{2} - \frac{x ^{2}}{2} lo g x + \frac{x ^{2}}{4} + C$

Integrate (x² + 1) log x.

Solution

$lo g x (\frac{x ^{3}}{3} + x) - \int \frac{1}{x} (\frac{x ^{3}}{3} + x) d x = (\frac{x ^{3}}{3} + x) lo g x - \int (\frac{x ^{2}}{3} + 1) d x$

$\frac{1}{3} (x^{3} + 3 x) lo g x - \frac{x ^{3}}{9} - x + C$

Integrate eˣ (sin x + cos x).

Solution

Form $e^{x} [f (x) + f^{'} (x)]$ with $f (x) = sin x$ .

$e^{x} sin x + C$

Integrate

\frac{x e ^{x}}{( 1 + x ) ^{2}}

Solution

$\frac{x}{( 1 + x ) ^{2}} = \frac{( 1 + x ) - 1}{( 1 + x ) ^{2}} = \frac{1}{1 + x} - \frac{1}{( 1 + x ) ^{2}}$ : $f = \frac{1}{1 + x}$ , $f^{'} = - \frac{1}{( 1 + x ) ^{2}}$ .

$\frac{e ^{x}}{1 + x} + C$

Integrate

e^{x} (\frac{1 + sin x}{1 + cos x})

Solution

$\frac{1 + 2 sin \frac{x}{2} cos \frac{x}{2}}{2 cos ^{2} \frac{x}{2}} = \frac{1}{2} sec^{2} \frac{x}{2} + tan \frac{x}{2}$ : $f = tan \frac{x}{2}$ , $f^{'} = \frac{1}{2} sec^{2} \frac{x}{2}$ .

$e^{x} tan \frac{x}{2} + C$

Integrate

e^{x} (\frac{1}{x} - \frac{1}{x ^{2}})

Solution

$f = \frac{1}{x}$ , $f^{'} = - \frac{1}{x ^{2}}$ .

$\frac{e ^{x}}{x} + C$

Integrate

\frac{( x - 3 ) e ^{x}}{( x - 1 ) ^{3}}

Solution

$\frac{x - 3}{( x - 1 ) ^{3}} = \frac{( x - 1 ) - 2}{( x - 1 ) ^{3}} = \frac{1}{( x - 1 ) ^{2}} - \frac{2}{( x - 1 ) ^{3}}$ : $f = \frac{1}{( x - 1 ) ^{2}}$ , $f^{'} = - \frac{2}{( x - 1 ) ^{3}}$ .

$\frac{e ^{x}}{( x - 1 ) ^{2}} + C$

Integrate e²ˣ sin x.

Solution

$I = sin x \cdot \frac{e ^{2 x}}{2} - \int cos x \cdot \frac{e ^{2 x}}{2} d x = \frac{e ^{2 x} sin x}{2} - \frac{1}{2} [cos x \cdot \frac{e ^{2 x}}{2} + \int sin x \cdot \frac{e ^{2 x}}{2} d x]$

$I = \frac{e ^{2 x} sin x}{2} - \frac{e ^{2 x} cos x}{4} - \frac{1}{4} I \Rightarrow \frac{5}{4} I = \frac{e ^{2 x}}{4} (2 sin x - cos x)$

$\frac{e ^{2 x}}{5} (2 sin x - cos x) + C$

Integrate

sin^{- 1} (\frac{2 x}{1 + x ^{2}})

Solution

Put $x = tan θ$ : $sin^{- 1} (sin 2 θ) = 2 θ = 2 tan^{- 1} x$ . So $I = 2 \int tan^{- 1} x d x = 2 [x tan^{- 1} x - \frac{1}{2} lo g (1 + x^{2})]$ (Q13).

$2 x tan^{- 1} x - lo g (1 + x^{2}) + C$

\int x^{2} e^{x^{3}} d x

equals (A) ⅓ e^(x³) + C (B) ⅓ e^(x²) + C (C) ½ e^(x³) + C (D) ½ e^(x²) + C

Solution

Put $t = x^{3}$ , $d t = 3 x^{2} d x$ : $\frac{1}{3} \int e^{t} d t$ .

(A)

\int e^{x} sec x (1 + tan x) d x

equals (A) eˣ cos x + C (B) eˣ sec x + C (C) eˣ sin x + C (D) eˣ tan x + C

Solution

$e^{x} (sec x + sec x tan x)$ : $f = sec x$ , $f^{'} = sec x tan x$ .

(B)

Exercise 7.7

Integrate

4 - x^{2}

Solution

$\int a^{2} - x^{2} d x = \frac{x}{2} a^{2} - x^{2} + \frac{a ^{2}}{2} sin^{- 1} \frac{x}{a}$ with $a = 2$ .

$\frac{x}{2} 4 - x^{2} + 2 sin^{- 1} \frac{x}{2} + C$

Integrate

1 - 4 x^{2}

Solution

$1 - 4 x^{2} = 2 (\frac{1}{2})^{2} - x^{2}$ : $2 [\frac{x}{2} \frac{1}{4} - x^{2} + \frac{1}{8} sin^{- 1} 2 x]$ .

$\frac{x}{2} 1 - 4 x^{2} + \frac{1}{4} sin^{- 1} 2 x + C$

Integrate

x^{2} + 4 x + 6

Solution

$x^{2} + 4 x + 6 = (x + 2)^{2} + (2)^{2}$ . Using $\int t^{2} + a^{2} = \frac{t}{2} t^{2} + a^{2} + \frac{a ^{2}}{2} lo g t + t^{2} + a^{2}$ :

$\frac{x + 2}{2} x^{2} + 4 x + 6 + lo g x + 2 + x^{2} + 4 x + 6 + C$

Integrate

x^{2} + 4 x + 1

Solution

$x^{2} + 4 x + 1 = (x + 2)^{2} - (3)^{2}$ . Using $\int t^{2} - a^{2} = \frac{t}{2} t^{2} - a^{2} - \frac{a ^{2}}{2} lo g t + t^{2} - a^{2}$ :

$\frac{x + 2}{2} x^{2} + 4 x + 1 - \frac{3}{2} lo g x + 2 + x^{2} + 4 x + 1 + C$

Integrate

1 - 4 x - x^{2}

Solution

$1 - 4 x - x^{2} = 5 - (x + 2)^{2} = (5)^{2} - (x + 2)^{2}$ .

$\frac{x + 2}{2} 1 - 4 x - x^{2} + \frac{5}{2} sin^{- 1} (\frac{x + 2}{5}) + C$

Integrate

x^{2} + 4 x - 5

Solution

$x^{2} + 4 x - 5 = (x + 2)^{2} - 3^{2}$ .

$\frac{x + 2}{2} x^{2} + 4 x - 5 - \frac{9}{2} lo g x + 2 + x^{2} + 4 x - 5 + C$

Integrate

1 + 3 x - x^{2}

Solution

$1 + 3 x - x^{2} = \frac{13}{4} - (x - \frac{3}{2})^{2}$ , with $a = \frac{13}{2}$ :

$\frac{x - \frac{3}{2}}{2} 1 + 3 x - x^{2} + \frac{13}{8} sin^{- 1} \frac{x - \frac{3}{2}}{\frac{13}{2}}$

$\frac{2 x - 3}{4} 1 + 3 x - x^{2} + \frac{13}{8} sin^{- 1} (\frac{2 x - 3}{13}) + C$

Integrate

x^{2} + 3 x

Solution

$x^{2} + 3 x = (x + \frac{3}{2})^{2} - (\frac{3}{2})^{2}$ :

$\frac{2 x + 3}{4} x^{2} + 3 x - \frac{9}{8} lo g x + \frac{3}{2} + x^{2} + 3 x + C$

Integrate

1 + \frac{x ^{2}}{9}

Solution

$= \frac{1}{3} x^{2} + 9$ : $\frac{1}{3} [\frac{x}{2} x^{2} + 9 + \frac{9}{2} lo g x + x^{2} + 9]$ .

$\frac{x}{6} x^{2} + 9 + \frac{3}{2} lo g x + x^{2} + 9 + C$

\int 1 + x^{2} d x

is equal to (A)

\frac{x}{2} 1 + x^{2} + \frac{1}{2} lo g x + 1 + x^{2} + C

(B)

\frac{2}{3} (1 + x^{2})^{3/2} + C

(C)

\frac{2}{3} x (1 + x^{2})^{3/2} + C

(D)

\frac{x ^{2}}{2} 1 + x^{2} + \frac{1}{2} x^{2} lo g x + 1 + x^{2} + C

Solution

Standard form with $a = 1$ .

(A)

\int x^{2} - 8 x + 7 d x

is equal to (A)

\frac{1}{2} (x - 4) x^{2} - 8 x + 7 + 9 lo g x - 4 + x^{2} - 8 x + 7 + C

(B)

\frac{1}{2} (x + 4) x^{2} - 8 x + 7 + 9 lo g x + 4 + x^{2} - 8 x + 7 + C

(C)

\frac{1}{2} (x - 4) x^{2} - 8 x + 7 - 32 lo g x - 4 + x^{2} - 8 x + 7 + C

(D)

\frac{1}{2} (x - 4) x^{2} - 8 x + 7 - \frac{9}{2} lo g x - 4 + x^{2} - 8 x + 7 + C

Solution

$x^{2} - 8 x + 7 = (x - 4)^{2} - 3^{2}$ , so the integral is $\frac{x - 4}{2} x^{2} - 8 x + 7 - \frac{9}{2} lo g x - 4 + x^{2} - 8 x + 7 + C$ .

(D)

Exercise 7.8

Evaluate

\int_{- 1}^{1} (x + 1) d x

Solution

$[\frac{x ^{2}}{2} + x]_{- 1}^{1} = (\frac{1}{2} + 1) - (\frac{1}{2} - 1) = 2$

Evaluate

\int_{2}^{3} \frac{1}{x} d x

Solution

$[lo g x]_{2}^{3} = lo g 3 - lo g 2$

$lo g \frac{3}{2}$

Evaluate

\int_{1}^{2} (4 x^{3} - 5 x^{2} + 6 x + 9) d x

Solution

$[x^{4} - \frac{5 x ^{3}}{3} + 3 x^{2} + 9 x]_{1}^{2} = (16 - \frac{40}{3} + 12 + 18) - (1 - \frac{5}{3} + 3 + 9) = 33 - \frac{35}{3}$

$\frac{64}{3}$

Evaluate

\int_{0}^{π /4} sin 2 x d x

Solution

$[- \frac{cos 2 x}{2}]_{0}^{π /4} = - \frac{cos \frac{π}{2}}{2} + \frac{cos 0}{2} = 0 + \frac{1}{2}$

$\frac{1}{2}$

Evaluate

\int_{0}^{π /2} cos 2 x d x

Solution

$[\frac{sin 2 x}{2}]_{0}^{π /2} = \frac{sin π - sin 0}{2}$

Evaluate

\int_{4}^{5} e^{x} d x

Solution

$e^{5} - e^{4}$

$e^{4} (e - 1)$

Evaluate

\int_{0}^{π /4} tan x d x

Solution

$[lo g ∣ sec x ∣]_{0}^{π /4} = lo g 2 - lo g 1$

$\frac{1}{2} lo g 2$

Evaluate

\int_{π /6}^{π /4} cosec x d x

Solution

$[lo g ∣ cosec x - cot x ∣]_{π /6}^{π /4} = lo g (2 - 1) - lo g (2 - 3)$

$lo g (\frac{2 - 1}{2 - 3})$

Evaluate

\int_{0}^{1} \frac{d x}{1 - x ^{2}}

Solution

$[sin^{- 1} x]_{0}^{1} = \frac{π}{2} - 0$

$\frac{π}{2}$

Evaluate

\int_{0}^{1} \frac{d x}{1 + x ^{2}}

Solution

$[tan^{- 1} x]_{0}^{1} = \frac{π}{4}$

$\frac{π}{4}$

Evaluate

\int_{2}^{3} \frac{d x}{x ^{2} - 1}

Solution

$[\frac{1}{2} lo g \frac{x - 1}{x + 1}]_{2}^{3} = \frac{1}{2} [lo g \frac{2}{4} - lo g \frac{1}{3}] = \frac{1}{2} lo g (\frac{1}{2} \times 3)$

$\frac{1}{2} lo g \frac{3}{2}$

Evaluate

\int_{0}^{π /2} cos^{2} x d x

Solution

$\int_{0}^{π /2} \frac{1 + cos 2 x}{2} d x = [\frac{x}{2} + \frac{sin 2 x}{4}]_{0}^{π /2} = \frac{π}{4}$

$\frac{π}{4}$

Evaluate

\int_{2}^{3} \frac{x d x}{x ^{2} + 1}

Solution

$[\frac{1}{2} lo g (x^{2} + 1)]_{2}^{3} = \frac{1}{2} (lo g 10 - lo g 5)$

$\frac{1}{2} lo g 2$

Evaluate

\int_{0}^{1} \frac{2 x + 3}{5 x ^{2} + 1} d x

Solution

$= \frac{1}{5} \int_{0}^{1} \frac{10 x}{5 x ^{2} + 1} d x + \frac{3}{5} \int_{0}^{1} \frac{d x}{x ^{2} + \frac{1}{5}}$

$= \frac{1}{5} [lo g (5 x^{2} + 1)]_{0}^{1} + \frac{3}{5} \cdot 5 [tan^{- 1} (5 x)]_{0}^{1} = \frac{1}{5} lo g 6 + \frac{3}{5} tan^{- 1} 5$

$\frac{1}{5} lo g 6 + \frac{3}{5} tan^{- 1} 5$

Evaluate

\int_{0}^{1} x e^{x^{2}} d x

Solution

Put $t = x^{2}$ : $\frac{1}{2} \int_{0}^{1} e^{t} d t = \frac{1}{2} (e - 1)$ .

$\frac{1}{2} (e - 1)$

Evaluate

\int_{1}^{2} \frac{5 x ^{2}}{x ^{2} + 4 x + 3} d x

Solution

$\frac{5 x ^{2}}{x ^{2} + 4 x + 3} = 5 - \frac{20 x + 15}{( x + 1 ) ( x + 3 )}$ , and $\frac{20 x + 15}{( x + 1 ) ( x + 3 )} = - \frac{5/2}{x + 1} + \frac{45/2}{x + 3}$ .

$I = [5 x]_{1}^{2} + \frac{5}{2} [lo g (x + 1)]_{1}^{2} - \frac{45}{2} [lo g (x + 3)]_{1}^{2} = 5 + \frac{5}{2} lo g \frac{3}{2} - \frac{45}{2} lo g \frac{5}{4}$

$5 - \frac{5}{2} (9 lo g \frac{5}{4} - lo g \frac{3}{2})$

Evaluate

\int_{0}^{π /4} (2 sec^{2} x + x^{3} + 2) d x

Solution

$[2 tan x + \frac{x ^{4}}{4} + 2 x]_{0}^{π /4} = 2 + \frac{π ^{4}}{1024} + \frac{π}{2}$

$\frac{π ^{4}}{1024} + \frac{π}{2} + 2$

Evaluate

\int_{0}^{π} (sin^{2} \frac{x}{2} - cos^{2} \frac{x}{2}) d x

Solution

$sin^{2} \frac{x}{2} - cos^{2} \frac{x}{2} = - cos x$ , so $I = - [sin x]_{0}^{π} = 0$ .

Evaluate

\int_{0}^{2} \frac{6 x + 3}{x ^{2} + 4} d x

Solution

$= 3 \int_{0}^{2} \frac{2 x}{x ^{2} + 4} d x + 3 \int_{0}^{2} \frac{d x}{x ^{2} + 4} = 3 [lo g (x^{2} + 4)]_{0}^{2} + \frac{3}{2} [tan^{- 1} \frac{x}{2}]_{0}^{2} = 3 lo g 2 + \frac{3}{2} \cdot \frac{π}{4}$

$3 lo g 2 + \frac{3 π}{8}$

Evaluate

\int_{0}^{1} (x e^{x} + sin \frac{π x}{4}) d x

Solution

$\int x e^{x} = e^{x} (x - 1)$ ; $\int sin \frac{π x}{4} = - \frac{4}{π} cos \frac{π x}{4}$ .

$[e^{x} (x - 1)]_{0}^{1} + [- \frac{4}{π} cos \frac{π x}{4}]_{0}^{1} = (0 + 1) + (- \frac{4}{π 2} + \frac{4}{π})$

$1 + \frac{4}{π} - \frac{2 2}{π}$

\int_{1}^{3} \frac{d x}{1 + x ^{2}}

equals (A) π/3 (B) 2π/3 (C) π/6 (D) π/12

Solution

$tan^{- 1} 3 - tan^{- 1} 1 = \frac{π}{3} - \frac{π}{4} = \frac{π}{12}$

(D)

\int_{0}^{2/3} \frac{d x}{4 + 9 x ^{2}}

equals (A) π/6 (B) π/12 (C) π/24 (D) π/4

Solution

$\frac{1}{9} \int \frac{d x}{x ^{2} + \frac{4}{9}} = \frac{1}{9} \cdot \frac{3}{2} [tan^{- 1} \frac{3 x}{2}]_{0}^{2/3} = \frac{1}{6} \cdot \frac{π}{4} = \frac{π}{24}$

(C)

Exercise 7.9

Evaluate

\int_{0}^{1} \frac{x}{x ^{2} + 1} d x

Solution

Put $t = x^{2} + 1$ (x: 0→1 gives t: 1→2): $\frac{1}{2} \int_{1}^{2} \frac{d t}{t} = \frac{1}{2} lo g 2$ .

$\frac{1}{2} lo g 2$

Evaluate

\int_{0}^{π /2} sin ϕ cos^{5} ϕ d ϕ

Solution

Put $t = sin ϕ$ , $d t = cos ϕ d ϕ$ ; $cos^{4} ϕ = (1 - t^{2})^{2}$ ; limits 0 → 1:

$\int_{0}^{1} t^{1/2} (1 - 2 t^{2} + t^{4}) d t = [\frac{2}{3} t^{3/2} - \frac{4}{7} t^{7/2} + \frac{2}{11} t^{11/2}]_{0}^{1} = \frac{2}{3} - \frac{4}{7} + \frac{2}{11} = \frac{154 - 132 + 42}{231}$

$\frac{64}{231}$

Evaluate

\int_{0}^{1} sin^{- 1} (\frac{2 x}{1 + x ^{2}}) d x

Solution

For $0 \leq x \leq 1$ , $sin^{- 1} \frac{2 x}{1 + x ^{2}} = 2 tan^{- 1} x$ .

$2 \int_{0}^{1} tan^{- 1} x d x = 2 [x tan^{- 1} x - \frac{1}{2} lo g (1 + x^{2})]_{0}^{1} = 2 [\frac{π}{4} - \frac{1}{2} lo g 2]$

$\frac{π}{2} - lo g 2$

Evaluate

\int_{0}^{2} x x + 2 d x

(Put x + 2 = t²).

Solution

$x = t^{2} - 2$ , $d x = 2 t d t$ ; x: 0→2 gives t: √2→2.

$\int_{2}^{2} (t^{2} - 2) t \cdot 2 t d t = [\frac{2 t ^{5}}{5} - \frac{4 t ^{3}}{3}]_{2}^{2} = (\frac{64}{5} - \frac{32}{3}) - (\frac{8 2}{5} - \frac{8 2}{3}) = \frac{32}{15} + \frac{16 2}{15}$

$\frac{16 2}{15} (2 + 1)$

Evaluate

\int_{0}^{π /2} \frac{sin x}{1 + cos ^{2} x} d x

Solution

Put $t = cos x$ , $d t = - sin x d x$ ; x: 0→π/2 gives t: 1→0: $- \int_{1}^{0} \frac{d t}{1 + t ^{2}} = [tan^{- 1} t]_{0}^{1}$ .

$\frac{π}{4}$

Evaluate

\int_{0}^{2} \frac{d x}{x + 4 - x ^{2}}

Solution

$4 + x - x^{2} = \frac{17}{4} - (x - \frac{1}{2})^{2}$ . With $a = \frac{17}{2}$ :

$\frac{1}{17} [lo g \frac{17 - 1 + 2 x}{17 + 1 - 2 x}]_{0}^{2} = \frac{1}{17} lo g [\frac{17 + 3}{17 - 3} \cdot \frac{17 + 1}{17 - 1}] = \frac{1}{17} lo g \frac{20 + 4 17}{20 - 4 17}$

$\frac{5 + 17}{5 - 17} = \frac{( 5 + 17 ) ^{2}}{8} = \frac{21 + 5 17}{4}$ .

$\frac{1}{17} lo g (\frac{21 + 5 17}{4})$

Evaluate

\int_{- 1}^{1} \frac{d x}{x ^{2} + 2 x + 5}

Solution

$x^{2} + 2 x + 5 = (x + 1)^{2} + 4$ : $\frac{1}{2} [tan^{- 1} \frac{x + 1}{2}]_{- 1}^{1} = \frac{1}{2} (tan^{- 1} 1 - 0)$ .

$\frac{π}{8}$

Evaluate

\int_{1}^{2} (\frac{1}{x} - \frac{1}{2 x ^{2}}) e^{2 x} d x

Solution

Put $t = 2 x$ ( $d x = \frac{d t}{2}$ ; t: 2→4): $\frac{1}{x} = \frac{2}{t}$ , $\frac{1}{2 x ^{2}} = \frac{2}{t ^{2}}$ .

$\int_{2}^{4} (\frac{2}{t} - \frac{2}{t ^{2}}) e^{t} \frac{d t}{2} = \int_{2}^{4} e^{t} (\frac{1}{t} - \frac{1}{t ^{2}}) d t = [\frac{e ^{t}}{t}]_{2}^{4} = \frac{e ^{4}}{4} - \frac{e ^{2}}{2}$

$\frac{e ^{2} ( e ^{2} - 2 )}{4}$

The value of the integral

\int_{1/3}^{1} \frac{( x - x ^{3} ) ^{1/3}}{x ^{4}} d x

is (A) 6 (B) 0 (C) 3 (D) 4

Solution

$(x - x^{3})^{1/3} = x (\frac{1}{x ^{2}} - 1)^{1/3}$ , so the integrand is $\frac{( x ^{- 2} - 1 ) ^{1/3}}{x ^{3}}$ .

Put $t = x^{- 2} - 1$ , $d t = - 2 x^{- 3} d x$ ; x: 1/3→1 gives t: 8→0.

$- \frac{1}{2} \int_{8}^{0} t^{1/3} d t = \frac{1}{2} [\frac{3}{4} t^{4/3}]_{0}^{8} = \frac{3}{8} \times 16 = 6$

(A)

f (x) = \int_{0}^{x} t sin t d t

, then f′(x) is (A) cos x + x sin x (B) x sin x (C) x cos x (D) sin x + x cos x

Solution

By the first fundamental theorem, $f^{'} (x) = x sin x$ .

(B)

Exercise 7.10

Evaluate

\int_{0}^{π /2} cos^{2} x d x

using properties.

Solution

$I = \int_{0}^{π /2} cos^{2} x d x = \int_{0}^{π /2} sin^{2} x d x$ (by $\int_{0}^{a} f (x) = \int_{0}^{a} f (a - x)$ ). Adding: $2 I = \int_{0}^{π /2} 1 d x = \frac{π}{2}$ .

$\frac{π}{4}$

Evaluate

\int_{0}^{π /2} \frac{sin x}{sin x + cos x} d x

Solution

Replacing x by $\frac{π}{2} - x$ : $I = \int_{0}^{π /2} \frac{cos x}{cos x + sin x} d x$ . Adding: $2 I = \int_{0}^{π /2} d x = \frac{π}{2}$ .

$\frac{π}{4}$

Evaluate

\int_{0}^{π /2} \frac{sin ^{3/2} x d x}{sin ^{3/2} x + cos ^{3/2} x}

Solution

Same method: $I = \int_{0}^{π /2} \frac{cos ^{3/2} x}{cos ^{3/2} x + sin ^{3/2} x} d x$ ; $2 I = \frac{π}{2}$ .

$\frac{π}{4}$

Evaluate

\int_{0}^{π /2} \frac{cos ^{5} x d x}{sin ^{5} x + cos ^{5} x}

Solution

Same method: $2 I = \int_{0}^{π /2} d x = \frac{π}{2}$ .

$\frac{π}{4}$

Evaluate

\int_{- 5}^{5} ∣ x + 2∣ d x

Solution

$\int_{- 5}^{- 2} - (x + 2) d x + \int_{- 2}^{5} (x + 2) d x = [- \frac{( x + 2 ) ^{2}}{2}]_{- 5}^{- 2} + [\frac{( x + 2 ) ^{2}}{2}]_{- 2}^{5} = \frac{9}{2} + \frac{49}{2}$

Evaluate

\int_{2}^{8} ∣ x - 5∣ d x

Solution

$\int_{2}^{5} (5 - x) d x + \int_{5}^{8} (x - 5) d x = \frac{9}{2} + \frac{9}{2}$

Evaluate

\int_{0}^{1} x (1 - x)^{n} d x

Solution

Replace x by $1 - x$ : $I = \int_{0}^{1} (1 - x) x^{n} d x = \int_{0}^{1} (x^{n} - x^{n + 1}) d x = \frac{1}{n + 1} - \frac{1}{n + 2}$ .

$\frac{1}{( n + 1 ) ( n + 2 )}$

Evaluate

\int_{0}^{π /4} lo g (1 + tan x) d x

Solution

Replace x by $\frac{π}{4} - x$ : $1 + tan (\frac{π}{4} - x) = 1 + \frac{1 - tan x}{1 + tan x} = \frac{2}{1 + tan x}$ .

$I = \int_{0}^{π /4} [lo g 2 - lo g (1 + tan x)] d x = \frac{π}{4} lo g 2 - I \Rightarrow 2 I = \frac{π}{4} lo g 2$ .

$\frac{π}{8} lo g 2$

Evaluate

\int_{0}^{2} x 2 - x d x

Solution

Replace x by $2 - x$ : $I = \int_{0}^{2} (2 - x) x d x = [\frac{4}{3} x^{3/2} - \frac{2}{5} x^{5/2}]_{0}^{2} = \frac{8 2}{3} - \frac{8 2}{5}$ .

$\frac{16 2}{15}$

Evaluate

\int_{0}^{π /2} (2 lo g sin x - lo g sin 2 x) d x

Solution

$2 lo g sin x - lo g (2 sin x cos x) = lo g sin x - lo g cos x - lo g 2$ .

$\int_{0}^{π /2} lo g sin x d x = \int_{0}^{π /2} lo g cos x d x$ (replace x by $\frac{π}{2} - x$ ), so those two cancel. $I = - \frac{π}{2} lo g 2$ .

$\frac{π}{2} lo g \frac{1}{2}$

Evaluate

\int_{- π /2}^{π /2} sin^{2} x d x

Solution

$sin^{2} x$ is even: $I = 2 \int_{0}^{π /2} sin^{2} x d x = 2 \cdot \frac{π}{4}$ (as in Q1).

$\frac{π}{2}$

Evaluate

\int_{0}^{π} \frac{x d x}{1 + sin x}

Solution

Replace x by $π - x$ : $I = \int_{0}^{π} \frac{π - x}{1 + sin x} d x$ . Adding: $2 I = π \int_{0}^{π} \frac{d x}{1 + sin x}$ .

Since $sin (π - x) = sin x$ , $\int_{0}^{π} \frac{d x}{1 + sin x} = 2 \int_{0}^{π /2} \frac{d x}{1 + sin x} = 2 \int_{0}^{π /2} \frac{d x}{1 + cos x} = 2 \int_{0}^{π /2} \frac{1}{2} sec^{2} \frac{x}{2} d x = 2 [tan \frac{x}{2}]_{0}^{π /2} = 2$ .

So $2 I = 2 π$ .

Evaluate

\int_{- π /2}^{π /2} sin^{7} x d x

Solution

$sin^{7} (- x) = - sin^{7} x$ : an odd function on a symmetric interval.

Evaluate

\int_{0}^{2 π} cos^{5} x d x

Solution

$cos^{5} (2 π - x) = cos^{5} x$ , so $I = 2 \int_{0}^{π} cos^{5} x d x$ . And $cos^{5} (π - x) = - cos^{5} x$ , so $\int_{0}^{π} cos^{5} x d x = 0$ .

Evaluate

\int_{0}^{π /2} \frac{sin x - cos x}{1 + sin x cos x} d x

Solution

Replace x by $\frac{π}{2} - x$ : sin and cos swap, so $I = \int_{0}^{π /2} \frac{cos x - sin x}{1 + sin x cos x} d x = - I$ . Hence $2 I = 0$ .

Evaluate

\int_{0}^{π} lo g (1 + cos x) d x

Solution

Replace x by $π - x$ : $I = \int_{0}^{π} lo g (1 - cos x) d x$ . Adding: $2 I = \int_{0}^{π} lo g sin^{2} x d x = 2 \int_{0}^{π} lo g sin x d x = 4 \int_{0}^{π /2} lo g sin x d x$ .

Let $J = \int_{0}^{π /2} lo g sin x d x = \int_{0}^{π /2} lo g cos x d x$ . Then $2 J = \int_{0}^{π /2} lo g \frac{sin 2 x}{2} d x = \int_{0}^{π /2} lo g sin 2 x d x - \frac{π}{2} lo g 2$ . Putting $t = 2 x$ , $\int_{0}^{π /2} lo g sin 2 x d x = \frac{1}{2} \int_{0}^{π} lo g sin t d t = J$ . So $J = - \frac{π}{2} lo g 2$ .

$2 I = 4 J = - 2 π lo g 2$ .

$- π lo g 2$

Evaluate

\int_{0}^{a} \frac{x}{x + a - x} d x

Solution

Replace x by $a - x$ : $I = \int_{0}^{a} \frac{a - x}{a - x + x} d x$ . Adding: $2 I = \int_{0}^{a} d x = a$ .

$\frac{a}{2}$

Evaluate

\int_{0}^{4} ∣ x - 1∣ d x

Solution

$\int_{0}^{1} (1 - x) d x + \int_{1}^{4} (x - 1) d x = \frac{1}{2} + \frac{9}{2}$

Show that

\int_{0}^{a} f (x) g (x) d x = 2 \int_{0}^{a} f (x) d x

, if f and g are defined as f(x) = f(a – x) and g(x) + g(a – x) = 4.

Solution

$I = \int_{0}^{a} f (a - x) g (a - x) d x = \int_{0}^{a} f (x) [4 - g (x)] d x = 4 \int_{0}^{a} f (x) d x - I$ .

So $2 I = 4 \int_{0}^{a} f (x) d x$ .

$I = 2 \int_{0}^{a} f (x) d x$

The value of

\int_{- π /2}^{π /2} (x^{3} + x cos x + tan^{5} x + 1) d x

is (A) 0 (B) 2 (C) π (D) 1

Solution

$x^{3}$ , $x cos x$ , $tan^{5} x$ are odd, so their integrals are 0. $\int_{- π /2}^{π /2} 1 d x = π$ .

(C)

The value of

\int_{0}^{π /2} lo g (\frac{4 + 3 sin x}{4 + 3 cos x}) d x

is (A) 2 (B) 3/4 (C) 0 (D) –2

Solution

Replacing x by $\frac{π}{2} - x$ turns the integrand into $lo g \frac{4 + 3 cos x}{4 + 3 sin x}$ , its negative. So $I = - I$ , $I = 0$ .

(C)

Miscellaneous Exercise on Chapter 7

Integrate

\frac{1}{x - x ^{3}}

Solution

$\frac{1}{x ( 1 - x ) ( 1 + x )} = \frac{A}{x} + \frac{B}{1 - x} + \frac{C}{1 + x}$ : $x = 0$ : $A = 1$ ; $x = 1$ : $B = \frac{1}{2}$ ; $x = - 1$ : $C = - \frac{1}{2}$ .

$\int = lo g ∣ x ∣ - \frac{1}{2} lo g ∣1 - x ∣ - \frac{1}{2} lo g ∣1 + x ∣ = lo g ∣ x ∣ - \frac{1}{2} lo g ∣1 - x^{2} ∣$

$\frac{1}{2} lo g \frac{x ^{2}}{1 - x ^{2}} + C$

Integrate

\frac{1}{x + a + x + b}

Solution

Rationalise: $\frac{x + a - x + b}{( x + a ) - ( x + b )} = \frac{x + a - x + b}{a - b}$ .

$\frac{2}{3 ( a - b )} [(x + a)^{3/2} - (x + b)^{3/2}] + C$

Integrate

\frac{1}{x a x - x ^{2}}

[Hint: Put x = a/t]

Solution

$x = \frac{a}{t}$ , $d x = - \frac{a}{t ^{2}} d t$ ; $a x - x^{2} = \frac{a ^{2} ( t - 1 )}{t ^{2}}$ , so $x a x - x^{2} = \frac{a ^{2} t - 1}{t ^{2}}$ .

$\int = \int \frac{t ^{2}}{a ^{2} t - 1} \cdot (- \frac{a}{t ^{2}}) d t = - \frac{1}{a} \int \frac{d t}{t - 1} = - \frac{2}{a} t - 1$ , and $t - 1 = \frac{a - x}{x}$ .

$- \frac{2}{a} \frac{a - x}{x} + C$

Integrate

\frac{1}{x ^{2} ( x ^{4} + 1 ) ^{3/4}}

Solution

$x^{2} (x^{4} + 1)^{3/4} = x^{5} (1 + x^{- 4})^{3/4}$ . Put $t = 1 + x^{- 4}$ , $d t = - 4 x^{- 5} d x$ : $- \frac{1}{4} \int t^{- 3/4} d t = - t^{1/4}$ .

$- (1 + \frac{1}{x ^{4}})^{1/4} + C$

Integrate

\frac{1}{x ^{1/2} + x ^{1/3}}

[Hint: Put x = t⁶]

Solution

$d x = 6 t^{5} d t$ : $\int \frac{6 t ^{5}}{t ^{3} + t ^{2}} d t = 6 \int \frac{t ^{3}}{t + 1} d t = 6 \int (t^{2} - t + 1 - \frac{1}{t + 1}) d t = 2 t^{3} - 3 t^{2} + 6 t - 6 lo g ∣1 + t ∣$ .

$2 x - 3 x^{1/3} + 6 x^{1/6} - 6 lo g (1 + x^{1/6}) + C$

Integrate

\frac{5 x}{( x + 1 ) ( x ^{2} + 9 )}

Solution

$5 x = A (x^{2} + 9) + (B x + C) (x + 1)$ : $x = - 1$ : $A = - \frac{1}{2}$ ; $x^{2}$ : $B = \frac{1}{2}$ ; constant: $0 = 9 A + C \Rightarrow C = \frac{9}{2}$ .

$\int = - \frac{1}{2} lo g ∣ x + 1∣ + \frac{1}{4} lo g (x^{2} + 9) + \frac{9}{2} \cdot \frac{1}{3} tan^{- 1} \frac{x}{3}$

$- \frac{1}{2} lo g ∣ x + 1∣ + \frac{1}{4} lo g (x^{2} + 9) + \frac{3}{2} tan^{- 1} \frac{x}{3} + C$

Integrate

\frac{sin x}{sin ( x - a )}

Solution

$sin x = sin [(x - a) + a] = sin (x - a) cos a + cos (x - a) sin a$ , so the integrand is $cos a + sin a cot (x - a)$ .

$x cos a + sin a lo g ∣ sin (x - a) ∣ + C$

Integrate

\frac{e ^{5 l o g x} - e ^{4 l o g x}}{e ^{3 l o g x} - e ^{2 l o g x}}

Solution

$e^{k l o g x} = x^{k}$ : $\frac{x ^{5} - x ^{4}}{x ^{3} - x ^{2}} = \frac{x ^{4} ( x - 1 )}{x ^{2} ( x - 1 )} = x^{2}$ .

$\frac{x ^{3}}{3} + C$

Integrate

\frac{cos x}{4 - sin ^{2} x}

Solution

Put $t = sin x$ : $\int \frac{d t}{4 - t ^{2}} = sin^{- 1} \frac{t}{2}$ .

$sin^{- 1} (\frac{sin x}{2}) + C$

Integrate

\frac{sin ^{8} x - cos ^{8} x}{1 - 2 sin ^{2} x cos ^{2} x}

Solution

$sin^{8} x - cos^{8} x = (sin^{4} x - cos^{4} x) (sin^{4} x + cos^{4} x) = (sin^{2} x - cos^{2} x) (1 - 2 sin^{2} x cos^{2} x)$ .

So the integrand is $sin^{2} x - cos^{2} x = - cos 2 x$ .

$- \frac{1}{2} sin 2 x + C$

Integrate

\frac{1}{cos ( x + a ) cos ( x + b )}

Solution

Multiply and divide by $sin (a - b) = sin [(x + a) - (x + b)]$ : integrand $= \frac{tan ( x + a ) - tan ( x + b )}{sin ( a - b )}$ .

$\int = \frac{1}{sin ( a - b )} [- lo g ∣ cos (x + a) ∣ + lo g ∣ cos (x + b) ∣]$

$\frac{1}{sin ( a - b )} lo g \frac{cos ( x + b )}{cos ( x + a )} + C$

Integrate

\frac{x ^{3}}{1 - x ^{8}}

Solution

Put $t = x^{4}$ , $d t = 4 x^{3} d x$ : $\frac{1}{4} \int \frac{d t}{1 - t ^{2}}$ .

$\frac{1}{4} sin^{- 1} (x^{4}) + C$

Integrate

\frac{e ^{x}}{( 1 + e ^{x} ) ( 2 + e ^{x} )}

Solution

Put $t = e^{x}$ : $\int \frac{d t}{( 1 + t ) ( 2 + t )} = \int [\frac{1}{1 + t} - \frac{1}{2 + t}] d t$ .

$lo g \frac{1 + e ^{x}}{2 + e ^{x}} + C$

Integrate

\frac{1}{( x ^{2} + 1 ) ( x ^{2} + 4 )}

Solution

$= \frac{1}{3} [\frac{1}{x ^{2} + 1} - \frac{1}{x ^{2} + 4}]$

$\frac{1}{3} tan^{- 1} x - \frac{1}{6} tan^{- 1} \frac{x}{2} + C$

Integrate

cos^{3} x e^{l o g s i n x}

Solution

$e^{l o g s i n x} = sin x$ . Put $t = cos x$ : $- \int t^{3} d t$ .

$- \frac{1}{4} cos^{4} x + C$

Integrate

e^{3 l o g x} (x^{4} + 1)^{- 1}

Solution

$= \frac{x ^{3}}{x ^{4} + 1}$ . Put $t = x^{4} + 1$ : $\frac{1}{4} \int \frac{d t}{t}$ .

$\frac{1}{4} lo g (x^{4} + 1) + C$

Integrate f′(ax + b) [f (ax + b)]ⁿ.

Solution

Put $t = f (a x + b)$ , $d t = a f^{'} (a x + b) d x$ : $\frac{1}{a} \int t^{n} d t = \frac{t ^{n + 1}}{a ( n + 1 )}$ .

$\frac{[ f ( a x + b ) ] ^{n + 1}}{a ( n + 1 )} + C$

Integrate

\frac{1}{sin ^{3} x sin ( x + α )}

Solution

$sin^{3} x sin (x + α) = sin^{4} x (cos α + cot x sin α)$ , so the integrand is $\frac{cosec ^{2} x}{cos α + sin α cot x}$ .

Put $t = cos α + sin α cot x$ , $d t = - sin α cosec^{2} x d x$ : $- \frac{1}{sin α} \int t^{- 1/2} d t = - \frac{2}{sin α} t$ , and $t = \frac{sin ( x + α )}{sin x}$ .

$- \frac{2}{sin α} \frac{sin ( x + α )}{sin x} + C$

Integrate

\frac{1 - x}{1 + x}

Solution

Put $x = cos^{2} θ$ , $d x = - sin 2 θ d θ$ ; $\frac{1 - cos θ}{1 + cos θ} = tan \frac{θ}{2}$ .

$tan \frac{θ}{2} sin 2 θ = 4 sin^{2} \frac{θ}{2} cos θ = 2 (1 - cos θ) cos θ = 2 cos θ - 1 - cos 2 θ$ .

$\int = - \int (2 cos θ - 1 - cos 2 θ) d θ = - 2 sin θ + θ + sin θ cos θ$ , with $sin θ = 1 - x$ , $θ = cos^{- 1} x$ , $sin θ cos θ = x - x^{2}$ .

$- 2 1 - x + cos^{- 1} x + x - x^{2} + C$

Integrate

\frac{2 + sin 2 x}{1 + cos 2 x} e^{x}

Solution

$\frac{2 + 2 sin x cos x}{2 cos ^{2} x} = sec^{2} x + tan x$ : form $e^{x} [f + f^{'}]$ with $f = tan x$ .

$e^{x} tan x + C$

Integrate

\frac{x ^{2} + x + 1}{( x + 1 ) ^{2} ( x + 2 )}

Solution

$x^{2} + x + 1 = A (x + 1) (x + 2) + B (x + 2) + C (x + 1)^{2}$ : $x = - 1$ : $B = 1$ ; $x = - 2$ : $C = 3$ ; $x^{2}$ : $1 = A + C \Rightarrow A = - 2$ .

$- 2 lo g ∣ x + 1∣ - \frac{1}{x + 1} + 3 lo g ∣ x + 2∣ + C$

Integrate

tan^{- 1} \frac{1 - x}{1 + x}

Solution

Put $x = cos 2 θ$ : $\frac{1 - cos 2 θ}{1 + cos 2 θ} = tan θ$ , so the integrand is $θ = \frac{1}{2} cos^{- 1} x$ .

$\frac{1}{2} \int cos^{- 1} x d x = \frac{1}{2} [x cos^{- 1} x - 1 - x^{2}]$ (by parts).

$\frac{1}{2} (x cos^{- 1} x - 1 - x^{2}) + C$

Integrate

\frac{x ^{2} + 1 [ lo g ( x ^{2} + 1 ) - 2 lo g x ]}{x ^{4}}

Solution

$= \frac{1}{x ^{3}} 1 + \frac{1}{x ^{2}} lo g (1 + \frac{1}{x ^{2}})$ . Put $t = 1 + \frac{1}{x ^{2}}$ , $d t = - \frac{2}{x ^{3}} d x$ :

$- \frac{1}{2} \int t^{1/2} lo g t d t = - \frac{1}{2} [\frac{2}{3} t^{3/2} lo g t - \frac{4}{9} t^{3/2}] = - \frac{1}{3} t^{3/2} [lo g t - \frac{2}{3}]$

$- \frac{1}{3} (1 + \frac{1}{x ^{2}})^{3/2} [lo g (1 + \frac{1}{x ^{2}}) - \frac{2}{3}] + C$

Evaluate

\int_{π /2}^{π} e^{x} (\frac{1 - sin x}{1 - cos x}) d x

Solution

$\frac{1 - sin x}{1 - cos x} = \frac{1}{2} cosec^{2} \frac{x}{2} - cot \frac{x}{2}$ : form $e^{x} [f + f^{'}]$ with $f = - cot \frac{x}{2}$ .

$[- e^{x} cot \frac{x}{2}]_{π /2}^{π} = - e^{π} \cdot 0 + e^{π /2} \cdot 1$

$e^{π /2}$

Evaluate

\int_{0}^{π /4} \frac{sin x cos x}{cos ^{4} x + sin ^{4} x} d x

Solution

Divide by $cos^{4} x$ : $\frac{tan x sec ^{2} x}{1 + tan ^{4} x}$ . Put $t = tan^{2} x$ , $d t = 2 tan x sec^{2} x d x$ ; t: 0→1:

$\frac{1}{2} \int_{0}^{1} \frac{d t}{1 + t ^{2}} = \frac{1}{2} \cdot \frac{π}{4}$

$\frac{π}{8}$

Evaluate

\int_{0}^{π /2} \frac{cos ^{2} x d x}{cos ^{2} x + 4 sin ^{2} x}

Solution

Divide by $cos^{2} x$ and put $t = tan x$ ( $d x = \frac{d t}{1 + t ^{2}}$ ; t: 0→∞): $\int_{0}^{\infty} \frac{d t}{( 1 + t ^{2} ) ( 1 + 4 t ^{2} )} = \frac{1}{3} \int_{0}^{\infty} [\frac{4}{1 + 4 t ^{2}} - \frac{1}{1 + t ^{2}}] d t$ .

$= \frac{1}{3} [2 tan^{- 1} 2 t - tan^{- 1} t]_{0}^{\infty} = \frac{1}{3} (π - \frac{π}{2})$

$\frac{π}{6}$

Evaluate

\int_{π /6}^{π /3} \frac{sin x + cos x}{sin 2 x} d x

Solution

Put $t = sin x - cos x$ , $d t = (cos x + sin x) d x$ , $sin 2 x = 1 - t^{2}$ . Limits: $\frac{π}{6} \to \frac{1 - 3}{2}$ , $\frac{π}{3} \to \frac{3 - 1}{2}$ .

$\int \frac{d t}{1 - t ^{2}} = sin^{- 1} \frac{3 - 1}{2} - sin^{- 1} \frac{1 - 3}{2}$

$2 sin^{- 1} (\frac{3 - 1}{2})$

Evaluate

\int_{0}^{1} \frac{d x}{1 + x - x}

Solution

Rationalise (denominator becomes $(1 + x) - x = 1$ ): $\int_{0}^{1} (1 + x + x) d x = \frac{2}{3} [(1 + x)^{3/2} + x^{3/2}]_{0}^{1} = \frac{2}{3} (22 + 1 - 1)$ .

$\frac{4 2}{3}$

Evaluate

\int_{0}^{π /4} \frac{sin x + cos x}{9 + 16 sin 2 x} d x

Solution

Put $t = sin x - cos x$ , $sin 2 x = 1 - t^{2}$ ; t: −1→0. Denominator $= 25 - 16 t^{2}$ .

$\int_{- 1}^{0} \frac{d t}{25 - 16 t ^{2}} = \frac{1}{16} \cdot \frac{1}{2 \cdot \frac{5}{4}} [lo g \frac{\frac{5}{4} + t}{\frac{5}{4} - t}]_{- 1}^{0} = \frac{1}{40} [0 - lo g \frac{1}{9}]$

$\frac{1}{40} lo g 9$

Evaluate

\int_{0}^{π /2} sin 2 x tan^{- 1} (sin x) d x

Solution

Put $t = sin x$ : $sin 2 x d x = 2 t d t$ ; t: 0→1.

$\int_{0}^{1} 2 t tan^{- 1} t d t = [t^{2} tan^{- 1} t]_{0}^{1} - \int_{0}^{1} \frac{t ^{2}}{1 + t ^{2}} d t = \frac{π}{4} - [t - tan^{- 1} t]_{0}^{1} = \frac{π}{4} - 1 + \frac{π}{4}$

$\frac{π}{2} - 1$

Evaluate

\int_{1}^{4} (∣ x - 1∣ + ∣ x - 2∣ + ∣ x - 3∣) d x

Solution

$\int_{1}^{4} ∣ x - 1∣ = \frac{9}{2}$ ; $\int_{1}^{4} ∣ x - 2∣ = \int_{1}^{2} (2 - x) + \int_{2}^{4} (x - 2) = \frac{1}{2} + 2 = \frac{5}{2}$ ; $\int_{1}^{4} ∣ x - 3∣ = \int_{1}^{3} (3 - x) + \int_{3}^{4} (x - 3) = 2 + \frac{1}{2} = \frac{5}{2}$ .

$\frac{19}{2}$

Prove that

\int_{1}^{3} \frac{d x}{x ^{2} ( x + 1 )} = \frac{2}{3} + lo g \frac{2}{3}

Solution

$\frac{1}{x ^{2} ( x + 1 )} = - \frac{1}{x} + \frac{1}{x ^{2}} + \frac{1}{x + 1}$ (from $1 = A x (x + 1) + B (x + 1) + C x^{2}$ : $B = 1$ , $C = 1$ , $A = - 1$ ).

$[lo g \frac{x + 1}{x} - \frac{1}{x}]_{1}^{3} = (lo g \frac{4}{3} - \frac{1}{3}) - (lo g 2 - 1) = lo g \frac{2}{3} + \frac{2}{3}$

Proved.

Prove that

\int_{0}^{1} x e^{x} d x = 1

Solution

$[e^{x} (x - 1)]_{0}^{1} = 0 - (- 1) = 1$

Proved.

Prove that

\int_{- 1}^{1} x^{17} cos^{4} x d x = 0

Solution

$f (- x) = (- x)^{17} cos^{4} (- x) = - f (x)$ : odd function on [−1, 1].

Proved.

Prove that

\int_{0}^{π /2} sin^{3} x d x = \frac{2}{3}

Solution

$sin^{3} x = (1 - cos^{2} x) sin x$ ; put $t = cos x$ : $\int_{0}^{1} (1 - t^{2}) d t = 1 - \frac{1}{3}$ .

Proved.

Prove that

\int_{0}^{π /4} 2 tan^{3} x d x = 1 - lo g 2

Solution

$2 tan^{3} x = 2 tan x (sec^{2} x - 1)$ : $\int = [tan^{2} x + 2 lo g cos x]_{0}^{π /4} = 1 + 2 lo g \frac{1}{2} = 1 - lo g 2$ .

Proved.

Prove that

\int_{0}^{1} sin^{- 1} x d x = \frac{π}{2} - 1

Solution

By parts: $[x sin^{- 1} x + 1 - x^{2}]_{0}^{1} = \frac{π}{2} + 0 - (0 + 1)$

Proved.

\int \frac{d x}{e ^{x} + e ^{- x}}

is equal to (A) tan⁻¹ (eˣ) + C (B) tan⁻¹ (e⁻ˣ) + C (C) log (eˣ – e⁻ˣ) + C (D) log (eˣ + e⁻ˣ) + C

Solution

$= \int \frac{e ^{x}}{e ^{2 x} + 1} d x$ ; put $t = e^{x}$ : $\int \frac{d t}{1 + t ^{2}} = tan^{- 1} e^{x}$ .

(A)

\int \frac{cos 2 x}{( sin x + cos x ) ^{2}} d x

is equal to (A)

\frac{- 1}{s i n x + c o s x} + C

(B) log |sin x + cos x| + C (C) log |sin x – cos x| + C (D)

\frac{1}{( s i n x + c o s x ) ^{2}}

Solution

$\frac{cos 2 x}{( sin x + cos x ) ^{2}} = \frac{cos x - sin x}{cos x + sin x}$ , whose integral is $lo g ∣ sin x + cos x ∣$ .

(B)

If f (a + b – x) = f (x), then

\int_{a}^{b} x f (x) d x

is equal to (A)

\frac{a + b}{2} \int_{a}^{b} f (b - x) d x

(B)

\frac{a + b}{2} \int_{a}^{b} f (b + x) d x

(C)

\frac{b - a}{2} \int_{a}^{b} f (x) d x

(D)

\frac{a + b}{2} \int_{a}^{b} f (x) d x

Solution

$I = \int_{a}^{b} (a + b - x) f (a + b - x) d x = \int_{a}^{b} (a + b - x) f (x) d x = (a + b) \int_{a}^{b} f (x) d x - I$ .

So $I = \frac{a + b}{2} \int_{a}^{b} f (x) d x$ .

(D)

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