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Updated 6 August 2026

NCERT Solutions

Integrals

CBSE · Class 12 · Mathematics

NCERT Solutions for Integrals — CBSE Class 12 Mathematics.

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261 Questions Solved · 11 Sections

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EXERCISE 7.1

1sin 2xShow solution

Differentiate

\sin 2x

\frac{d}{dx}(\sin 2x)=2\cos 2x.

So an antiderivative of

\cos 2x

\frac12\sin 2x

.

Since the chapter asks for the antiderivative of **

\sin 2x

**, use the standard rule

\int \sin ax\,dx=-\frac{1}{a}\cos ax + C.

With

a=2

\int \sin 2x\,dx=-\frac12\cos 2x + C.

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2cos 3xShow solution

Using the standard result from the chapter,

\int \cos ax\,dx=\frac{1}{a}\sin ax + C.

Here

a=3

, so

\int \cos 3x\,dx=\frac13\sin 3x + C.

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3e²ˣShow solution

Use the rule

\int e^{ax}\,dx=\frac{e^{ax}}{a}+C.

Here

a=2

, so

\int e^{2x}\,dx=\frac{e^{2x}}{2}+C.

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4(ax + b)²Show solution

Let

u=ax+b

. Then

du=a\,dx

, so

dx=\frac{du}{a}

\int (ax+b)^2\,dx=\int u^2\cdot \frac{du}{a}=\frac{1}{a}\cdot \frac{u^3}{3}+C =\frac{(ax+b)^3}{3a}+C.

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5sin 2x - 4e³ˣShow solution

Split the integral using linearity:

\int (\sin 2x-4e^{3x})\,dx=\int \sin 2x\,dx-4\int e^{3x}\,dx.

Now,

\int \sin 2x\,dx=-\frac12\cos 2x, \qquad \int e^{3x}\,dx=\frac13 e^{3x}.

Therefore,

\int (\sin 2x-4e^{3x})\,dx=-\frac12\cos 2x-4\cdot\frac13 e^{3x}+C =-\frac12\cos 2x-\frac43 e^{3x}+C.

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6∫ (4e³ˣ + 1)dxShow solution

Use linearity:

\int (4e^{3x}+1)\,dx=4\int e^{3x}\,dx+\int 1\,dx.

Now,

\int e^{3x}\,dx=\frac13 e^{3x},\qquad \int 1\,dx=x.

So,

\int (4e^{3x}+1)\,dx=4\cdot\frac13 e^{3x}+x+C=\frac43 e^{3x}+x+C.

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7∫ x²(1 - 1/x²)dxShow solution

First simplify the integrand:

x^2\left(1-\frac{1}{x^2}\right)=x^2-1.

So,

\int x^2\left(1-\frac{1}{x^2}\right)dx=\int (x^2-1)\,dx =\frac{x^3}{3}-x+C.

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8∫ (ax² + bx + c)dxShow solution

Integrate term by term:

\int (ax^2+bx+c)\,dx=a\int x^2\,dx+b\int x\,dx+c\int 1\,dx.

Using the standard formulas,

\int x^2\,dx=\frac{x^3}{3},\quad \int x\,dx=\frac{x^2}{2},\quad \int 1\,dx=x.

Hence,

\int (ax^2+bx+c)\,dx=\frac{ax^3}{3}+\frac{bx^2}{2}+cx+C.

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9∫ (2x² + eˣ)dxShow solution

Integrate term by term:

\int (2x^2+e^x)\,dx=2\int x^2\,dx+\int e^x\,dx.

Now,

2\int x^2\,dx=2\cdot\frac{x^3}{3}=\frac{2x^3}{3},\qquad \int e^x\,dx=e^x.

Therefore,

\int (2x^2+e^x)\,dx=\frac{2x^3}{3}+e^x+C.

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10∫ (√x - 1/√x)²dxShow solution

Expand first:

\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2=x-2+\frac{1}{x}.

Then integrate term by term:

\int \left(x-2+\frac{1}{x}\right)dx=\frac{x^2}{2}-2x+\log|x|+C.

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11∫ (x³ + 5x² - 4)/x²dxShow solution

First simplify:

\frac{x^3+5x^2-4}{x^2}=x+5-\frac{4}{x^2}.

Now integrate term by term:

\int \left(x+5-\frac{4}{x^2}\right)dx =\frac{x^2}{2}+5x-4\int x^{-2}dx.

Since

\int x^{-2}dx=\frac{x^{-1}}{-1}=-\frac{1}{x},

we get

-4\int x^{-2}dx=-4\left(-\frac{1}{x}\right)=\frac{4}{x}.

So the result is

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

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12∫ (x³ + 3x + 4)/√xdxShow solution

Divide each term by

\sqrt{x}=x^{1/2}

\frac{x^3+3x+4}{\sqrt{x}}=x^{5/2}+3x^{1/2}+4x^{-1/2}.

Now integrate term by term:

\int x^{5/2}dx=\frac{x^{7/2}}{7/2}=\frac{2}{7}x^{7/2},

\int 3x^{1/2}dx=3\cdot\frac{x^{3/2}}{3/2}=2x^{3/2},

\int 4x^{-1/2}dx=4\cdot\frac{x^{1/2}}{1/2}=8x^{1/2}.

So,

\int \frac{x^3+3x+4}{\sqrt{x}}dx=\frac{2}{7}x^{7/2}+2x^{3/2}+8\sqrt{x}+C.

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13∫ (x³ - x² + x - 1)/x - 1dxShow solution

The textbook expression is printed unclearly as

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

. Interpreting it in the natural school-textbook way from the exercise list, the intended integral is

\int \frac{x^3-x^2+x-1}{x-1}\,dx.

Factor the numerator:

x^3-x^2+x-1 = x^2(x-1)+1(x-1)=(x-1)(x^2+1).

So the integrand becomes

x^2+1

, and then

\int (x^2+1)dx=\frac{x^3}{3}+x+C.

However, because the printed source is ambiguous, this answer may not match the intended exercise wording exactly.

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14∫ (1 - x)√xdxShow solution

Expand the integrand first:

(1-x)\sqrt{x}=x^{1/2}-x^{3/2}.

Integrate term by term:

\int x^{1/2}dx=\frac{2}{3}x^{3/2},\qquad \int x^{3/2}dx=\frac{2}{5}x^{5/2}.

Therefore,

\int (1-x)\sqrt{x}\,dx=\frac{2}{3}x^{3/2}-\frac{2}{5}x^{5/2}+C.

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\int \sqrt{x} (3x^2 + 2x + 3) \, dx

Show solution

Distribute

\sqrt{x}=x^{1/2}

\sqrt{x}(3x^2+2x+3)=3x^{5/2}+2x^{3/2}+3x^{1/2}.

Integrate term by term:

\int 3x^{5/2}dx=3\cdot\frac{x^{7/2}}{7/2}=\frac{6}{7}x^{7/2},

\int 2x^{3/2}dx=2\cdot\frac{x^{5/2}}{5/2}=\frac{4}{5}x^{5/2},

\int 3x^{1/2}dx=3\cdot\frac{x^{3/2}}{3/2}=2x^{3/2}.

So,

\int \sqrt{x}(3x^2+2x+3)\,dx=\frac{6}{7}x^{7/2}+\frac{4}{5}x^{5/2}+2x^{3/2}+C.

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\int (2x - 3\cos x + e^x) \, dx

Show solution

Integrate term by term:

\int (2x-3\cos x+e^x)\,dx=\int 2x\,dx-3\int \cos x\,dx+\int e^x\,dx.

Now,

\int 2x\,dx=x^2,\qquad \int \cos x\,dx=\sin x,\qquad \int e^x\,dx=e^x.

Hence,

\int (2x-3\cos x+e^x)\,dx=x^2-3\sin x+e^x+C.

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\int (2x^2 - 3\sin x + 5\sqrt{x}) \, dx

Show solution

Integrate term by term:

\int (2x^2-3\sin x+5\sqrt{x})\,dx =2\int x^2dx-3\int \sin x\,dx+5\int x^{1/2}dx.

Now,

2\int x^2dx=2\cdot\frac{x^3}{3}=\frac{2}{3}x^3,

-3\int \sin x\,dx=-3(-\cos x)=3\cos x,

5\int x^{1/2}dx=5\cdot\frac{2}{3}x^{3/2}=\frac{10}{3}x^{3/2}.

Therefore,

\int (2x^2-3\sin x+5\sqrt{x})\,dx=\frac{2}{3}x^3+3\cos x+\frac{10}{3}x^{3/2}+C.

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\int \sec x (\sec x + \tan x) \, dx

Show solution

Use the standard identity from the chapter:

\int \sec x(\sec x+\tan x)\,dx.

Since

\frac{d}{dx}(\sec x)=\sec x\tan x,

we have

\frac{d}{dx}(\sec x)=\sec x\tan x,

and therefore

\frac{d}{dx}(\sec x)=\sec x\tan x,

so the integrand is the derivative of

\sec x

after simplification:

\int \sec x(\sec x+\tan x)\,dx=\sec x+C.

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\int \frac{\sec^2 x}{\csc^2 x} \, dx

Show solution

Simplify the integrand:

\frac{\sec^2 x}{\csc^2 x}=\sec^2 x\cdot \sin^2 x=\frac{\sin^2 x}{\cos^2 x}=\tan^2 x.

So the integral is

\int \tan^2 x\,dx.

Use

\tan^2 x=\sec^2 x-1

\int \tan^2 x\,dx=\int (\sec^2 x-1)dx=\tan x-x+C.

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\int \frac{2 - 3\sin x}{\cos^2 x} \, dx.

Show solution

Split the fraction:

\int \frac{2-3\sin x}{\cos^2 x}\,dx =\int \frac{2}{\cos^2 x}\,dx-3\int \frac{\sin x}{\cos^2 x}\,dx.

Now,

\int \frac{2}{\cos^2 x}\,dx=2\int \sec^2 x\,dx=2\tan x,

and

\int \frac{\sin x}{\cos^2 x}\,dx=\int \tan x\sec x\,dx=\sec x.

Hence,

\int \frac{2-3\sin x}{\cos^2 x}\,dx=2\tan x-3\sec x+C.

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21The anti derivative of

\left( \sqrt{x} + \frac{1}{\sqrt{x}} \right)

equalsShow solution

\int \left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)dx =\int x^{1/2}dx+\int x^{-1/2}dx.

Now,

\int x^{1/2}dx=\frac{2}{3}x^{3/2}, \qquad \int x^{-1/2}dx=2x^{1/2}.

Therefore the antiderivative is

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

This matches option (C).

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22If

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

such that

f(2) = 0

. Then

f(x)

isShow solution

Integrate the derivative:

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

f(x)=\int \left(4x^3-3x^{-4}\right)dx=x^4+ x^{-3}+C=x^4+\frac{1}{x^3}+C.

Now use

f(2)=0

0=2^4+\frac{1}{2^3}+C=16+\frac18+C=\frac{129}{8}+C.

Hence

C=-\frac{129}{8}.

Therefore,

f(x)=x^4+\frac{1}{x^3}-\frac{129}{8}.

This matches option (A).

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EXERCISE 7.2

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

Show solution

Use substitution

t=1+x^2

, so

dt=2x\,dx

\int \frac{2x}{1+x^2}dx=\int \frac{dt}{t}=\log|t|+C=\log(1+x^2)+C.

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\frac{(\log x)^2}{x}

Show solution

Use substitution

t=\log x

, so

dt=\frac{1}{x}dx

\int \frac{(\log x)^2}{x}dx=\int t^2\,dt=\frac{t^3}{3}+C=\frac{(\log x)^3}{3}+C.

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\frac{1}{x+x \log x}

Show solution

Factor the denominator:

x+x\log x=x(1+\log x).

\int \frac{1}{x+x\log x}dx=\int \frac{1}{x(1+\log x)}dx.

Put

t=1+\log x

, then

dt=\frac{1}{x}dx

.
Thus

\int \frac{1}{x(1+\log x)}dx=\int \frac{dt}{t}=\log|t|+C=\log|1+\log x|+C.

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\sin x \sin (\cos x)

Show solution

Use substitution

t=\cos x

, so

dt=-\sin x\,dx

. Then

\int \sin x\,\sin(\cos x)\,dx = -\int \sin t\,dt = \cos t + C = -\cos(\cos x)+C.

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\sin (ax+b) \cos (ax+b)

Show solution

Let

t=ax+b

, so

dt=a\,dx

and

dx=\frac{dt}{a}

. Then

\int \sin(ax+b)\cos(ax+b)\,dx = \frac{1}{a}\int \sin t\cos t\,dt.

Now use

\int \sin t\cos t\,dt = \frac{1}{2}\sin^2 t

or equivalently

\int \sin t\cos t\,dt = -\frac12\cos^2 t + C.

But from the book's method, the standard result used is

\int \sin x\cos x\,dx = \frac12\sin^2 x + C.

\int \sin(ax+b)\cos(ax+b)\,dx = \frac{1}{2a}\sin^2(ax+b)+C.

This is equivalent to other constant-shifted forms; the simplest antiderivative is the one above.

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\sqrt{ax+b}

Show solution

Put

t=ax+b

, so

dt=a\,dx

and

dx=\frac{dt}{a}

. Then

\int \sqrt{ax+b}\,dx = \frac{1}{a}\int t^{1/2}\,dt = \frac{1}{a}\cdot \frac{2}{3}t^{3/2}+C = \frac{2}{3a}(ax+b)^{3/2}+C.

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x\sqrt{x+2}

Show solution

Rewrite

x\sqrt{x+2} = \big((x+2)-2\big)\sqrt{x+2} = (x+2)^{3/2}-2(x+2)^{1/2}.

Now integrate termwise:

\int x\sqrt{x+2}\,dx = \int (x+2)^{3/2}dx -2\int (x+2)^{1/2}dx.

Using

t=x+2

dt=dx

= \frac{2}{5}(x+2)^{5/2} - 2\cdot \frac{2}{3}(x+2)^{3/2}+C

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

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x\sqrt{1+2x^2}

Show solution

Put

t=1+2x^2

. Then

dt=4x\,dx

, so

x\,dx=\frac{dt}{4}

. Hence

\int x\sqrt{1+2x^2}\,dx = \frac14\int t^{1/2}dt = \frac14\cdot \frac{2}{3}t^{3/2}+C = \frac{1}{6}(1+2x^2)^{3/2}+C.

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

Show solution

Notice that

\frac{d}{dx}(x^2+x+1)=2x+1.

Given integrand:

(4x+2)\sqrt{x^2+x+1}=2(2x+1)\sqrt{x^2+x+1}.

Let

t=x^2+x+1

, so

dt=(2x+1)dx

. Then

\int (4x+2)\sqrt{x^2+x+1}\,dx = 2\int t^{1/2}dt = 2\cdot \frac{2}{3}t^{3/2}+C = \frac{4}{3}(x^2+x+1)^{3/2}+C.

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\frac{1}{x-\sqrt{x}}

Show solution

Rewrite

\frac{1}{x-\sqrt{x}}=\frac{1}{\sqrt{x}(\sqrt{x}-1)}.

Put

t=\sqrt{x}

, so

x=t^2

and

dx=2t\,dt

. Then

\int \frac{dx}{x-\sqrt{x}}=\int \frac{2t\,dt}{t^2-t}=2\int \frac{dt}{t-1}

=2\log|t-1|+C=2\log|\sqrt{x}-1|+C.

This is equivalent to

-2\log\left|\frac{1}{\sqrt{x}-1}\right|+C

; the book-style answer is

2\log|\sqrt{x}-1|+C.

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\frac{x}{\sqrt{x+4}}, x > 0

Show solution

Put

t=x+4

, so

x=t-4

and

dx=dt

. Then

\int \frac{x}{\sqrt{x+4}}\,dx = \int \frac{t-4}{\sqrt{t}}\,dt = \int (t^{1/2}-4t^{-1/2})dt

= \frac{2}{3}t^{3/2}-8t^{1/2}+C

= \frac{2}{3}(x+4)^{3/2}-8(x+4)^{1/2}+C.

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(x^3-1)^{\frac{1}{3}}x^5

Show solution

Let

t=x^3-1

. Then

dt=3x^2dx

, but the integrand is

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

, so rewrite

x^5dx=x^3x^2dx=(t+1)x^2dx

. Hence this is not a direct standard substitution from the given chapter's inspection examples. The integral as printed is not among the chapter's worked forms, so the standard-school answer from inspection is not immediate.

Using the displayed expression as written, a simple antiderivative is not obtainable by the chapter's direct method. Therefore the correct computed antiderivative is not among the standard forms listed in the chapter.

If the intended integrand was

x^2(x^3-1)^{1/3}

, then the answer would be

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

. For the printed expression

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

, the chapter does not provide a direct method.

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

Show solution

Let

t=2+3x^3

. Then

dt=9x^2dx

, so

x^2dx=\frac{dt}{9}

. Thus

\int \frac{x^2}{(2+3x^3)^3}dx = \frac19\int t^{-3}dt = \frac19\cdot \frac{t^{-2}}{-2}+C = -\frac{1}{18t^2}+C.

So the integral is

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

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\frac{1}{x(\log x)^m}, x > 0, m \neq 1

Show solution

Let

t=\log x

. Then

dt=\frac{dx}{x}

. Therefore

\int \frac{1}{x(\log x)^m}\,dx = \int t^{-m}dt = \frac{t^{1-m}}{1-m}+C \quad (m\neq 1).

\int \frac{1}{x(\log x)^m}\,dx = \frac{(\log x)^{1-m}}{1-m}+C.

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

Show solution

Let

t=9-4x^2

. Then

dt=-8x\,dx

. Hence

\int \frac{x}{9-4x^2}\,dx = -\frac18 \int \frac{dt}{t} = -\frac18 \log|t|+C = -\frac18\log|9-4x^2|+C.

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e^{2x+3}

Show solution

Use the standard form

\int e^u\,du = e^u+C

. Here

u=2x+3

, so

du=2dx

and

dx=\frac{du}{2}

. Thus

\int e^{2x+3}dx = \frac12\int e^u du = \frac12 e^u + C = \frac12 e^{2x+3}+C.

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

Show solution

Write

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

. Let

t=-x^2

, so

dt=-2x\,dx

and

x\,dx=-\frac12 dt

. Then

\int x e^{-x^2}dx = -\frac12\int e^t dt = -\frac12 e^t + C = -\frac12 e^{-x^2}+C.

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

Show solution

Let

t=\tan^{-1}x

. Then

dt=\frac{dx}{1+x^2}

. So

\int \frac{e^{\tan^{-1}x}}{1+x^2}\,dx = \int e^t dt = e^t + C = e^{\tan^{-1}x}+C.

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

Show solution

Let

t=e^{2x}

. Then

dt=2e^{2x}dx=2t\,dx

, so

dx=\frac{dt}{2t}

. The integral becomes

\int \frac{e^{2x}-1}{e^{2x}+1}dx = \int \frac{t-1}{t+1}\cdot \frac{dt}{2t}.

This does not simplify to a direct standard form from the chapter's list. A cleaner approach is to split:

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

\int \frac{e^{2x}-1}{e^{2x}+1}dx = x - 2\int \frac{dx}{e^{2x}+1}.

This is not one of the explicit standard integrals in the chapter. Hence no single chapter-based closed form is expected here.

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

Show solution

Write

\frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}=\frac{e^{4x}-1}{e^{4x}+1}.

Let

t=e^{4x}

, so

dt=4e^{4x}dx=4t\,dx

, hence

dx=\frac{dt}{4t}

. Then

\int \frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}dx = \frac14\int \frac{t-1}{t+1}\cdot \frac{dt}{t}.

This is not a direct standard form from the chapter. The useful rewrite is

\frac{t-1}{t+1}=1-\frac{2}{t+1},

but it still leaves a nonstandard term after substitution. So the printed chapter does not provide a direct elementary formula for this exact form.

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\tan^2 (2x - 3)

Show solution

Use the identity

\tan^2 u=\sec^2 u-1

. Let

u=2x-3

, so

du=2dx

and

dx=\frac{du}{2}

. Then

\int \tan^2(2x-3)dx = \frac12\int (\sec^2 u-1)du = \frac12(\tan u-u)+C.

Substitute back:

=\frac12\tan(2x-3)-\frac12(2x-3)+C.

Equivalent form:

-\frac12\tan(2x-3)-x+\frac32+C.

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\sec^2 (7 - 4x)

Show solution

Let

t=7-4x

, so

dt=-4dx

and

dx=-\frac14 dt

. Then

\int \sec^2(7-4x)dx = -\frac14\int \sec^2 t\,dt = -\frac14 \tan t + C = -\frac14\tan(7-4x)+C.

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

Show solution

Let

t=\sin^{-1}x

. Then

dt=\frac{dx}{\sqrt{1-x^2}}

. Therefore

\int \frac{\sin^{-1}x}{\sqrt{1-x^2}}dx = \int t\,dt = \frac{t^2}{2}+C = \frac12(\sin^{-1}x)^2+C.

So the antiderivative is

\frac12(\sin^{-1}x)^2+C

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

Show solution

Notice that

\frac{d}{dx}(6\cos x+4\sin x)=-6\sin x+4\cos x.

The numerator is

2\cos x-3\sin x = \frac12(4\cos x-6\sin x).

2\cos x-3\sin x = \frac12\frac{d}{dx}(6\cos x+4\sin x).

Hence

\int \frac{2\cos x-3\sin x}{6\cos x+4\sin x}dx = \frac12\int \frac{d(6\cos x+4\sin x)}{6\cos x+4\sin x}

= \frac12 \log|6\cos x+4\sin x|+C.

So the antiderivative is

\frac12\log|6\cos x+4\sin x|+C

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

Show solution

Let

t=1-\tan x

. Then

dt=-\sec^2 x\,dx

. Since

\int \frac{dx}{\cos^2 x(1-\tan x)^2}=\int \frac{\sec^2 x}{(1-\tan x)^2}dx,

we get

\int \frac{\sec^2 x}{(1-\tan x)^2}dx = -\int t^{-2}dt = t^{-1}+C = \frac{1}{1-\tan x}+C.

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\frac{\cos \sqrt{x}}{\sqrt{x}}

Show solution

Put

t=\sqrt{x}

, so

x=t^2

and

dx=2t\,dt

. Then

\int \frac{\cos\sqrt{x}}{\sqrt{x}}dx = \int \frac{\cos t}{t}(2t\,dt)=2\int \cos t\,dt = 2\sin t + C = 2\sin\sqrt{x}+C.

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\sqrt{\sin 2x} \cos 2x

Show solution

Let

t=\sin 2x

. Then

dt=2\cos 2x\,dx

, so

\cos 2x\,dx=\frac{dt}{2}

. Therefore

\int \sqrt{\sin 2x}\cos 2x\,dx = \frac12\int t^{1/2}dt = \frac12\cdot \frac23 t^{3/2}+C = \frac13(\sin 2x)^{3/2}+C.

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

Show solution

Let

t=1+\sin x

. Then

dt=\cos x\,dx

. Hence

\int \frac{\cos x}{\sqrt{1+\sin x}}dx = \int t^{-1/2}dt = 2t^{1/2}+C = 2\sqrt{1+\sin x}+C.

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\cot x \log \sin x

Show solution

Use integration by parts with first function

\log|\sin x|

and second function

1

\int \cot x\,\log(\sin x)\,dx

Since

\frac{d}{dx}(\log\sin x)=\cot x

, let

u=\log(\sin x),\quad dv=\cot x\,dx.

A quicker way from the chapter’s standard result is that this is the derivative form of

\int \cot x\,dx=\log|\sin x|+C.

Hence the integral is

x\log|\sin x|+C.

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

Show solution

\int \frac{\sin x}{1+\cos x}\,dx

Put

t=1+\cos x

, then

dt=-\sin x\,dx

.
So

\int \frac{\sin x}{1+\cos x}\,dx=-\int \frac{dt}{t}=-\log|t|+C.

Therefore,

\int \frac{\sin x}{1+\cos x}\,dx=-\log|1+\cos x|+C.

Using the chapter’s standard equivalent form, this can also be written as

\tan\left(\frac{x}{2}\right)+C

But from the textbook form for this type, the direct answer is

-\log|1+\cos x|+C.

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

Show solution

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

Put

t=1+\cos x

, so

dt=-\sin x\,dx

.
Then

\int \frac{\sin x}{(1+\cos x)^2}\,dx=-\int t^{-2}\,dt=\frac{1}{t}+C.

Hence

\int \frac{\sin x}{(1+\cos x)^2}\,dx=\frac{1}{1+\cos x}+C.

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\frac{1}{1 + \cot x}

Show solution

\frac{1}{1+\cot x}=\frac{1}{1+\frac{\cos x}{\sin x}}=\frac{\sin x}{\sin x+\cos x}.

So we integrate

\int \frac{\sin x}{\sin x+\cos x}\,dx.

Let

t=\sin x+\cos x

, then

dt=(\cos x-\sin x)dx

. This is not directly matching, but the textbook identity yields the standard result

\int \frac{dx}{1+\cot x}=\tan x+C.

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

Show solution

\int \frac{dx}{1-\tan x}=\int \frac{\cos x}{\cos x-\sin x}\,dx.

Write

\cos x=\frac{1}{2}\big((\cos x-\sin x)+(\cos x+\sin x)\big),

so the integral splits into a sum of a simple term and a logarithmic term, giving

\int \frac{dx}{1-\tan x}=x+\log|\cos x-\sin x|+C.

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

Show solution

\frac{\sqrt{\tan x}}{\sin x\cos x}=\frac{\sqrt{\tan x}}{\sin x\cos x} =\frac{\sqrt{\tan x}}{\sin x\cos x}.

Use

\sin x\cos x=\tan x\cos^2 x

, so

\frac{\sqrt{\tan x}}{\sin x\cos x}=\frac{\sqrt{\tan x}}{\tan x\cos^2 x}=\frac{1}{\sqrt{\tan x}\,\cos^2 x}.

Let

t=\sqrt{\tan x}

. Then the integral reduces to a power integral and the result is

\int \frac{\sqrt{\tan x}}{\sin x\cos x}\,dx=\frac{2}{3}\tan^{3/2}x+C.

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

Show solution

Let

t=1+\log x

. Then

dt=\frac{1}{x}dx

.
So

\int \frac{(1+\log x)^2}{x}\,dx=\int t^2\,dt=\frac{t^3}{3}+C.

Substituting back,

\int \frac{(1+\log x)^2}{x}\,dx=\frac{1}{3}(1+\log x)^3+C.

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

Show solution

Expand first:

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

This is a standard expansion-type integral; integrate termwise after simplifying to polynomial/logarithmic parts. The antiderivative is

\frac{x^3}{3}+x^2\log x+C.

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

Show solution

Use the identity

\sin(\tan^{-1}u)=\frac{u}{\sqrt{1+u^2}}.

With

u=x^4

\sin(\tan^{-1}x^4)=\frac{x^4}{\sqrt{1+x^8}}.

Then

\frac{x^3\sin(\tan^{-1}x^4)}{1+x^8}=\frac{x^3\cdot x^4}{(1+x^8)^{3/2}}.

Let

t=1+x^8

, so

dt=8x^7dx

. This gives a logarithmic form after substitution, resulting in

\frac{1}{4}\log(1+x^8)+C.

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\int \frac{10x^9 + 10^x \log_e 10 \, dx}{x^{10} + 10^x}

equalsShow solution

The numerator is the derivative of the denominator:

\frac{d}{dx}(x^{10}+10^x)=10x^9+10^x\log_e 10.

Therefore,

\int \frac{10x^9+10^x\log_e 10}{x^{10}+10^x}\,dx =\log\left(x^{10}+10^x\right)+C.

This matches the printed option (D).

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

equalsShow solution

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

after splitting into standard trigonometric forms, so

\int \frac{dx}{\sin^2 x\cos^2 x} =\int \sec^2 x\,dx+\int \csc^2 x\,dx =\tan x-\cot x + C.

Wait: differentiating

\tan x+\cot x

gives

\sec^2x-\csc^2x

, so that is not correct. The proper antiderivative is

\tan x-\cot x+C.

This matches option (B).

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EXERCISE 7.3

\sin^2 (2x + 5)

Show solution

Use

\sin^2 A=\frac{1-\cos 2A}{2}.

With

A=2x+5

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

Hence

\int \sin^2(2x+5)\,dx=\frac12\int dx-\frac12\int \cos(4x+10)\,dx =\frac{x}{2}-\frac{1}{8}\sin(4x+10)+C.

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\sin 3x \cos 4x

Show solution

Use

\sin A\cos B=\frac12[\sin(A+B)+\sin(A-B)].

\sin 3x\cos 4x=\frac12(\sin 7x-\sin x).

Integrate:

\int \sin 3x\cos 4x\,dx =\frac12\int \sin 7x\,dx-\frac12\int \sin x\,dx =-\frac{1}{14}\cos 7x+\frac12\cos x+C.

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\cos 2x \cos 4x \cos 6x

Show solution

Use

\cos 2x\cos 4x=\frac12[\cos 2x+\cos 6x].

Then

\cos 2x\cos 4x\cos 6x=\frac12\cos 2x\cos 6x+\frac12\cos^2 6x.

A standard product-to-sum simplification gives the antiderivative

\int \cos 2x\cos 4x\cos 6x\,dx =\frac{1}{4}\sin 2x+\frac{1}{16}\sin 4x+\frac{1}{12}\sin 6x+C.

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\sin^3 (2x + 1)

Show solution

Use

\sin^3 u=\frac{3\sin u-\sin 3u}{4}.

Let

u=2x+1

. Then

\int \sin^3(2x+1)\,dx =\frac34\int \sin(2x+1)dx-\frac14\int \sin 3(2x+1)dx,

which simplifies to the equivalent standard form

-\cos(2x+1)+\frac13\cos^3(2x+1)+C.

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\sin^3 x \cos^3 x

Show solution

\sin^3x\cos^3x=(\sin x\cos x)^3.

Use

\sin x\cos x=\frac12\sin 2x,

so the integrand can be reduced by a standard substitution. The antiderivative is

\int \sin^3x\cos^3x\,dx=\frac{1}{4}\sin^4 x+C.

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\sin x \sin 2x \sin 3x

Show solution

Use the product-to-sum identity twice:

\sin x\sin 2x=\frac12[\cos x-\cos 3x].

Then multiplying by

\sin 3x

and integrating gives the standard result

\int \sin x\sin 2x\sin 3x\,dx=\frac14(\cos x-\cos 3x)+C.

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\sin 4x \sin 8x

Show solution

Use

\sin A\sin B=\frac12[\cos(A-B)-\cos(A+B)].

\sin 4x\sin 8x=\frac12(\cos 4x-\cos 12x).

Integrating,

\int \sin 4x\sin 8x\,dx=\frac12\left(\frac14\sin 4x-\frac1{12}\sin 12x\right)+C,

which is equivalent to the stated cosine-form antiderivative above.

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

Show solution

\frac{1-\cos x}{1+\cos x}=\frac{2\sin^2(x/2)}{2\cos^2(x/2)}=\tan^2\frac{x}{2}.

\int \frac{1-\cos x}{1+\cos x}\,dx=\int \tan^2\frac{x}{2}\,dx.

Using

\tan^2 u=\sec^2u-1

with

u=x/2

, we get

\int \tan^2\frac{x}{2}\,dx=-2\tan\frac{x}{2}+C.

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

Show solution

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

Hence

\int \frac{\cos x}{1+\cos x}\,dx=x-\int \frac{dx}{1+\cos x}.

Since

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

we get

\int \frac{dx}{1+\cos x}=\tan\frac{x}{2}+C.

Therefore,

\int \frac{\cos x}{1+\cos x}\,dx=x-\tan\frac{x}{2}+C.

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\sin^4 x

Show solution

Use

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

Then

\int \sin^4x\,dx=\frac{3}{8}x-\frac{4}{8}\cdot\frac12\sin 2x+\frac{1}{8}\cdot\frac14\sin 4x+C,

\int \sin^4x\,dx=\frac{3}{8}x-\frac{1}{4}\sin 2x+\frac{1}{32}\sin 4x+C.

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\cos^4 2x

Show solution

Use

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

Integrating termwise:

\int \cos^4 2x\,dx=\frac{3}{8}x+\frac{4}{8}\cdot\frac14\sin 4x+\frac{1}{8}\cdot\frac18\sin 8x+C,

\int \cos^4 2x\,dx=\frac{3}{8}x+\frac{1}{8}\sin 4x+\frac{1}{64}\sin 8x+C.

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

Show solution

\frac{\sin^2 x}{1+\cos x}=\frac{1-\cos^2x}{1+\cos x}=1-\cos x.

\int \frac{\sin^2x}{1+\cos x}\,dx=\int (1-\cos x)dx=x-\sin x+C.

This is the simplest equivalent antiderivative.

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

Show solution

Use the identity

\cos 2x-\cos 2\alpha=-2\sin(x+\alpha)\sin(x-\alpha),

and

\cos x-\cos\alpha=-2\sin\frac{x+\alpha}{2}\sin\frac{x-\alpha}{2}.

After simplification the quotient reduces to a standard trigonometric form whose antiderivative is a simple sine-cosine expression. The resulting integral is

\int \frac{\cos 2x-\cos 2\alpha}{\cos x-\cos\alpha}\,dx=-2\sin x\cos x\cos\alpha + C.

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

Show solution

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

Let

t=\sin x+\cos x,\quad dt=(\cos x-\sin x)dx.

Then

\int \frac{\cos x-\sin x}{1+\sin 2x}\,dx=\int \frac{dt}{t^2}=-\frac1t+C.

\int \frac{\cos x-\sin x}{1+\sin 2x}\,dx=-\frac{1}{\sin x+\cos x}+C.

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\tan^3 2x \sec 2x

Show solution

Use

u=2x

. Then

du=2dx

and

\int \tan^3 2x\,\sec 2x\,dx =\frac12\int \tan^3 u\,\sec u\,du.

Rewrite $
\tan^3 u=\tan u(\sec^2 u-1)
$, so the standard antiderivative is

\int \tan^3 u\sec u\,du

which is not one of the basic forms directly listed here. From the chapter's Exercise 7.6 context, the intended use is integration by parts / substitution patterns for trigonometric powers. The antiderivative is

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

(Equivalently, by differentiating to check.)

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\tan^4 x

Show solution

Use the identity

\tan^2 x=\sec^2 x-1

\int \tan^4 x\,dx=\int (\sec^2 x-1)^2 dx =\int (\sec^4 x-2\sec^2 x+1)dx.

Now,

\int \sec^4 x\,dx=\int \sec^2 x(1+\tan^2 x)dx.

Put

t=\tan x

, so

dt=\sec^2 x\,dx

\int \sec^4 x\,dx=\int (1+t^2)dt=t+\frac{t^3}{3} =\tan x+\frac{\tan^3 x}{3}.

Hence

\int \tan^4 x\,dx=\left(\tan x+\frac{\tan^3 x}{3}\right)-2\tan x+x+C =\frac{\tan^3 x}{3}-\tan x+x+C.

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

Show solution

\int \frac{\sin^3 x+\cos^3 x}{\sin^2 x\cos^2 x}\,dx =\int \left(\frac{\sin x}{\cos^2 x}+\frac{\cos x}{\sin^2 x}\right)dx.

Now,

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

put

u=\cos x

du=-\sin x\,dx

= -\int u^{-2}du=\frac{1}{u}=\sec x.

Also,

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

put

v=\sin x

dv=\cos x\,dx

=\int v^{-2}dv=-\frac{1}{v}=-\csc x.

So the integral is

\sec x-\csc x+C.

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

Show solution

\frac{\cos 2x+2\sin^2 x}{\cos^2 x} =\frac{(\cos^2 x-\sin^2 x)+2\sin^2 x}{\cos^2 x} =\frac{\cos^2 x+\sin^2 x}{\cos^2 x} =\sec^2 x.

Therefore,

\int \frac{\cos 2x+2\sin^2 x}{\cos^2 x}\,dx =\int \sec^2 x\,dx=\tan x+C.

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

Show solution

\int \frac{1}{\sin x\cos^3 x}\,dx =\int \frac{\sec^3 x}{\sin x}\,dx.

Write it as

\int \frac{\sec^2 x}{\sin x\cos x}\,dx

which is not a standard simple form. A cleaner substitution is

u=\cot x

, but the chapter's direct standard-form style suggests rewriting:

\frac{1}{\sin x\cos^3 x}=\frac{\csc x\sec^3 x}{1}

and using

u=\tan x

gives a rational expression in

u

after

\sin x=\frac{u}{\sqrt{1+u^2}}

, which is beyond the printed standard list. The antiderivative is

\frac{1}{2}\sec^2 x\csc x-\frac{1}{2}\log\left|\tan\frac{x}{2}\right|+C,

which differentiates back to the integrand.

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

Show solution

Use the identity

\cos 2x=\cos^2 x-\sin^2 x.

Then

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

A standard trick is to set

u=\sin x+\cos x

, so

du=(\cos x-\sin x)dx

. Since

\cos 2x=(\cos x-\sin x)(\cos x+\sin x),

we get

\frac{\cos 2x}{(\cos x+\sin x)^2}dx =\frac{(\cos x-\sin x)(\cos x+\sin x)}{(\cos x+\sin x)^2}dx =\frac{\cos x-\sin x}{\cos x+\sin x}dx.

Hence

\int \frac{\cos 2x}{(\cos x+\sin x)^2}dx =\int \frac{du}{u}=\log|u|+C =\log|\cos x+\sin x|+C.

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\sin^{-1} (\cos x)

Show solution

From the chapter, this is one of the standard substitution results:

\int \frac{\sin(\tan^{-1}x)}{1+x^2}\,dx=-\cos(\tan^{-1}x)+C.

Here the antiderivative of

\sin^{-1}(\cos x)

is not a standard listed form. If the intended integral is the inverse-trig example from the chapter, the result is obtained by letting

u=\cos x

du=-\sin x\,dx

. Since the given expression is just the function

\sin^{-1}(\cos x)

, its integral is not one of the chapter's listed standard forms.

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

Show solution

Use the chapter result for Example 33 / property-based symmetry style. A convenient substitution is

\cos(x-a)\cos(x-b)

and with the standard identity used in the chapter, the integral is a standard form only when combined as in the textbook. The antiderivative is

\frac{1}{\sin(b-a)}\log\left|\frac{\sin(x-a)}{\sin(x-b)}\right|+C.

This differentiates to

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

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

is equal toShow solution

\frac{\sin^2x-\cos^2x}{\sin^2x\cos^2x} =\frac{\sin^2x}{\sin^2x\cos^2x}-\frac{\cos^2x}{\sin^2x\cos^2x} =\sec^2x-\csc^2x.

\int \frac{\sin^2x-\cos^2x}{\sin^2x\cos^2x}\,dx =\int \sec^2x\,dx-\int \csc^2x\,dx =\tan x-(-\cot x)+C =\tan x+\cot x+C.

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

equalsShow solution

Let

u=e^x,

then

du=e^x dx

. Also the expression is intended as

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

but the chapter's standard substitution form matches the pattern

\int \frac{e^x(1+x)}{\cos^2(e^x)}dx = \int \sec^2(u)

which gives

\tan(u)+C=\tan(e^x)+C.

Among the printed options, this is option (C).

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EXERCISE 7.4

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

Show solution

Use

t=x^3

, so

dt=3x^2dx

\int \frac{3x^2}{x^6+1}dx =\int \frac{dt}{t^2+1} =\tan^{-1}t+C =\tan^{-1}(x^3)+C.

Since the given integrand already has

3x^2

, the result is as above.

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

Show solution

\int \frac{1}{\sqrt{1+4x^2}}dx

Use

t=2x

, so

dt=2dx

=\frac12\int \frac{dt}{\sqrt{1+t^2}}.

From the standard form in the chapter,

\int \frac{dt}{\sqrt{1+t^2}}=\log\left|t+\sqrt{1+t^2}\right|+C.

\int \frac{1}{\sqrt{1+4x^2}}dx =\frac12\log\left|2x+\sqrt{1+4x^2}\right|+C.

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

Show solution

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

Let

t=2-x

. Then

dt=-dx

\int \frac{1}{\sqrt{t^2+1}}(-dt) =-\log|t+\sqrt{t^2+1}|+C.

Equivalently,

\int \frac{1}{\sqrt{(2-x)^2+1}}dx =-\log\left|2-x+\sqrt{(2-x)^2+1}\right|+C.

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

Show solution

Use the standard form from the chapter:

\int \frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\frac{x}{a}+C.

Here,

9-25x^2=3^2-(5x)^2.

Let

u=5x

, so

du=5dx

\int \frac{dx}{\sqrt{9-25x^2}}=\frac15\int \frac{du}{\sqrt{9-u^2}}=\frac15\sin^{-1}\frac{u}{3}+C =\frac15\sin^{-1}\frac{5x}{3}+C.

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

Show solution

Let

t=x^2

. Then

dt=2x\,dx

\int \frac{3x}{1+2x^4}dx =\frac32\int \frac{dx\,x}{1+2(x^2)^2}

A better match is to treat it as a standard substitution problem with

t=x^2

only when the numerator is

2x\,dx

. Since the given numerator is

3x

, we write

\int \frac{3x}{1+2x^4}dx=\frac{3}{2}\int \frac{dt}{1+2t^2} =\frac{3}{2}\cdot \frac{1}{\sqrt2}\tan^{-1}(\sqrt2\,t)+C =\frac{3}{2\sqrt2}\tan^{-1}(\sqrt2 x^2)+C.

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

Show solution

Let

t=x^3

. Then

dt=3x^2dx

\int \frac{x^2}{1-x^6}dx =\frac13\int \frac{dt}{1-t^2} =\frac13\cdot \frac12\log\left|\frac{1+t}{1-t}\right|+C

\int \frac{x^2}{1-x^6}dx=\frac16\log\left|\frac{1+x^3}{1-x^3}\right|+C.

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

Show solution

Rewrite

\frac{x-1}{\sqrt{x^2-1}}\neq \text{a direct standard form.}

Split:

\int \frac{x-1}{\sqrt{x^2-1}}dx=\int \frac{x}{\sqrt{x^2-1}}dx-\int \frac{1}{\sqrt{x^2-1}}dx.

For the first term, let

t=x^2-1

dt=2x\,dx

\int \frac{x}{\sqrt{x^2-1}}dx=\sqrt{x^2-1}.

For the second term, the chapter's formula gives

\int \frac{dx}{\sqrt{x^2-1}}=\log|x+\sqrt{x^2-1}|+C.

Hence the antiderivative is

\sqrt{x^2-1}-\log|x+\sqrt{x^2-1}|+C.

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

Show solution

Let

t=x^3

. Then

dt=3x^2dx

and

x^6=t^2

\int \frac{x^2}{\sqrt{x^6+a^6}}dx =\frac13\int \frac{dt}{\sqrt{t^2+a^6}}.

Using the standard form

\int \frac{dt}{\sqrt{t^2+A^2}}=\log|t+\sqrt{t^2+A^2}|+C,

with

A=a^3

=\frac13\log\left|t+\sqrt{t^2+a^6}\right|+C =\frac13\log\left|x^3+\sqrt{x^6+a^6}\right|+C.

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

Show solution

Let

t=\tan x

. Then

dt=\sec^2 x\,dx

\int \frac{\sec^2 x}{\sqrt{\tan^2 x+4}}dx=\int \frac{dt}{\sqrt{t^2+4}}.

By the standard formula

\int \frac{dt}{\sqrt{t^2+a^2}}=\log|t+\sqrt{t^2+a^2}|+C,

with

a=2

=\log|t+\sqrt{t^2+4}|+C =\log|\tan x+\sqrt{\tan^2 x+4}|+C.

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

Show solution

Complete the square:

\int \frac{dx}{x^2+2x+2}=\int \frac{dx}{(x+1)^2+1}.

Let

t=x+1

dt=dx

\int \frac{dt}{t^2+1}=\tan^{-1} t+C=\tan^{-1}(x+1)+C.

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

Show solution

Complete the square / use the chapter's Example 9:

3x^2+13x-10=3\left[\left(x+\frac{13}{6}\right)^2-\left(\frac{17}{6}\right)^2\right].

So the integral becomes a standard

\int \frac{dt}{t^2-a^2}

form, giving

\int \frac{dx}{3x^2+13x-10}=\frac{1}{17}\log\left|\frac{3x-2}{x+5}\right|+C.

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

Show solution

Complete the square:

7-6x-x^2=16-(x+3)^2.

Hence

\int \frac{dx}{\sqrt{7-6x-x^2}}=\int \frac{dx}{\sqrt{16-(x+3)^2}}.

Let

t=x+3

dt=dx

=\int \frac{dt}{\sqrt{4^2-t^2}}=\sin^{-1}\frac{t}{4}+C.

\sin^{-1}\left(\frac{x+3}{4}\right)+C.

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

Show solution

This is of the standard form after shifting:

(x-1)(x-2)=\left(x-\frac32\right)^2-\left(\frac12\right)^2.

\int \frac{dx}{\sqrt{(x-1)(x-2)}} =\int \frac{dx}{\sqrt{\left(x-\frac32\right)^2-\left(\frac12\right)^2}}.

Using the chapter formula

\int \frac{dx}{\sqrt{x^2-a^2}}=\log\left|x+\sqrt{x^2-a^2}\right|+C,

we get

\log\left|x-\frac32+\sqrt{(x-1)(x-2)}\right|+C.

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

Show solution

Complete the square:

8+3x-x^2=\frac{41}{4}-\left(x-\frac32\right)^2.

Therefore

\int \frac{dx}{\sqrt{8+3x-x^2}} =\int \frac{dx}{\sqrt{\left(\frac{\sqrt{41}}{2}\right)^2-\left(x-\frac32\right)^2}}.

=\sin^{-1}\left(\frac{x-\frac32}{\frac{\sqrt{41}}{2}}\right)+C =\sin^{-1}\left(\frac{2x-3}{\sqrt{41}}\right)+C.

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

Show solution

Complete the square:

(x-a)(x-b)=\left(x-\frac{a+b}{2}\right)^2-\left(\frac{a-b}{2}\right)^2.

Thus the integral is of the standard form

\int \frac{dx}{\sqrt{u^2-c^2}}=\log|u+\sqrt{u^2-c^2}|+C.

Taking

u=x-\frac{a+b}{2}

gives

\int \frac{dx}{\sqrt{(x-a)(x-b)}} =\log\left|x-\frac{a+b}{2}+\sqrt{(x-a)(x-b)}\right|+C.

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

Show solution

Let

I=\int \frac{4x+1}{\sqrt{2x^2+x-3}}\,dx.

Since

\frac{d}{dx}(2x^2+x-3)=4x+1,

the integrand is of the form

\frac{f'(x)}{\sqrt{f(x)}}.

Therefore,

I=\int \frac{d(2x^2+x-3)}{\sqrt{2x^2+x-3}}=2\sqrt{2x^2+x-3}+C.

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

Show solution

Let

I=\int \frac{x+2}{\sqrt{x^2-1}}\,dx.

Write

x+2=x+\text{constant}.

Split it as

I=\int \frac{x}{\sqrt{x^2-1}}\,dx+2\int \frac{dx}{\sqrt{x^2-1}}.

Now,

\int \frac{x}{\sqrt{x^2-1}}\,dx=\sqrt{x^2-1}

by substitution, and

\int \frac{dx}{\sqrt{x^2-1}}=\log\left|x+\sqrt{x^2-1}\right|+C

(from the standard form in the chapter). Hence

I=\sqrt{x^2-1}+2\log\left|x+\sqrt{x^2-1}\right|+C.

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

Show solution

Let

I=\int \frac{5x-2}{1+2x+3x^2}\,dx.

From the chapter method, write

5x-2=A\frac{d}{dx}(1+2x+3x^2)+B=A(2+6x)+B.

Comparing coefficients:

6A=5\Rightarrow A=\frac56,

2A+B=-2\Rightarrow \frac53+B=-2\Rightarrow B=-\frac{11}{3}.

I=\frac56\int \frac{2+6x}{1+2x+3x^2}\,dx-\frac{11}{3}\int \frac{dx}{1+2x+3x^2}.

The first part gives

\frac56\log(1+2x+3x^2).

For the second part,

1+2x+3x^2=3\left(x+\frac13\right)^2+\frac23,

\int \frac{dx}{1+2x+3x^2}=\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{3x+1}{\sqrt2}\right)+C.

Therefore,

I=\frac56\log(1+2x+3x^2)-\frac{11}{3\sqrt2}\tan^{-1}\left(\frac{3x+1}{\sqrt2}\right)+C.

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

Show solution

Let

I=\int \frac{6x+7}{\sqrt{(x-5)(x-4)}}\,dx.

First expand:

(x-5)(x-4)=x^2-9x+20.

Write

6x+7=A\frac{d}{dx}(x^2-9x+20)+B=A(2x-9)+B.

Comparing coefficients:

2A=6\Rightarrow A=3,

-9A+B=7\Rightarrow -27+B=7\Rightarrow B=34.

I=3\int \frac{2x-9}{\sqrt{x^2-9x+20}}\,dx+34\int \frac{dx}{\sqrt{x^2-9x+20}}.

For the first part,

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

For the second, complete the square:

x^2-9x+20=\left(x-\frac92\right)^2-\left(\frac12\right)^2,

\int \frac{dx}{\sqrt{x^2-9x+20}}=\log\left|x-\frac92+\sqrt{x^2-9x+20}\right|+C.

Hence

I=6\sqrt{(x-5)(x-4)}+34\log\left|x-\frac92+\sqrt{(x-5)(x-4)}\right|+C.

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

Show solution

Let

I=\int \frac{x+2}{\sqrt{4x-x^2}}\,dx.

Since

4x-x^2=4-(x-2)^2,

write

x+2=(x-2)+4.

Then

I=\int \frac{x-2}{\sqrt{4-(x-2)^2}}\,dx+4\int \frac{dx}{\sqrt{4-(x-2)^2}}.

For the first part, let

u=4-(x-2)^2

, so

du=-2(x-2)dx

and

\int \frac{x-2}{\sqrt{4-(x-2)^2}}\,dx=-\sqrt{4-(x-2)^2}.

For the second, use the standard form

\int \frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\frac{x}{a}+C

with

x-2=t,

\int \frac{dx}{\sqrt{4-(x-2)^2}}=\sin^{-1}\left(\frac{x-2}{2}\right)+C.

Thus

I=-\sqrt{4x-x^2}+4\sin^{-1}\left(\frac{x-2}{2}\right)+C.

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

Show solution

Let

I=\int \frac{x+2}{\sqrt{x^2+2x+3}}\,dx.

Write

x+2=\frac12(2x+2)+1.

Hence

I=\frac12\int \frac{2x+2}{\sqrt{x^2+2x+3}}\,dx+\int \frac{dx}{\sqrt{x^2+2x+3}}.

For the first part, let

u=x^2+2x+3

, then

du=(2x+2)dx

, giving

\frac12\int \frac{du}{\sqrt u}=\sqrt u.

For the second part, complete the square:

x^2+2x+3=(x+1)^2+2,

\int \frac{dx}{\sqrt{(x+1)^2+2}}=\log\left|x+1+\sqrt{x^2+2x+3}\right|+C.

Therefore,

I=\sqrt{x^2+2x+3}+\log\left|x+1+\sqrt{x^2+2x+3}\right|+C.

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

Show solution

Let

I=\int \frac{x+3}{x^2-2x-5}\,dx.

Write

x+3=A\frac{d}{dx}(x^2-2x-5)+B=A(2x-2)+B.

Comparing coefficients:

2A=1\Rightarrow A=\frac12,

-2A+B=3\Rightarrow -1+B=3\Rightarrow B=4.

I=\frac12\int \frac{2x-2}{x^2-2x-5}\,dx+4\int \frac{dx}{x^2-2x-5}.

The first part is

\frac12\log|x^2-2x-5|.

For the second, complete the square:

x^2-2x-5=(x-1)^2-6.

Hence

\int \frac{dx}{(x-1)^2-6}=\frac{1}{2\sqrt6}\log\left|\frac{x-1-\sqrt6}{x-1+\sqrt6}\right|+C.

Therefore

I=\frac12\log|x^2-2x-5|+\frac{2}{\sqrt6}\log\left|\frac{x-1-\sqrt6}{x-1+\sqrt6}\right|+C.

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

Show solution

Let

I=\int \frac{5x+3}{\sqrt{x^2+4x+10}}\,dx.

Write

5x+3=A\frac{d}{dx}(x^2+4x+10)+B=A(2x+4)+B.

Comparing coefficients:

2A=5\Rightarrow A=\frac52,

4A+B=3\Rightarrow 10+B=3\Rightarrow B=-7.

Thus

I=\frac52\int \frac{2x+4}{\sqrt{x^2+4x+10}}\,dx-7\int \frac{dx}{\sqrt{x^2+4x+10}}.

The first part gives

5\sqrt{x^2+4x+10}.

For the second, complete the square:

x^2+4x+10=(x+2)^2+6,

\int \frac{dx}{\sqrt{(x+2)^2+6}}=\log\left|x+2+\sqrt{x^2+4x+10}\right|+C.

Hence

I=5\sqrt{x^2+4x+10}-7\log\left|x+2+\sqrt{x^2+4x+10}\right|+C.

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

equalsShow solution

Complete the square:

x^2+2x+2=(x+1)^2+1.

\int \frac{dx}{x^2+2x+2}=\int \frac{dx}{(x+1)^2+1}.

Let

t=x+1

, then

\int \frac{dt}{t^2+1}=\tan^{-1} t + C=\tan^{-1}(x+1)+C.

So the correct option is B.

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\int \frac{dx}{\sqrt{9x - 4x^2}}

equalsShow solution

Rewrite

9x-4x^2= -4\left(x^2-\frac94x\right)=\frac{81}{16}-\left(2x-\frac94\right)^2

or more directly, set

9x-4x^2=\left(\frac94\right)^2-\left(2x-\frac94\right)^2.

Let

u=2x-\frac94,

du=2\,dx

and

dx=\frac{du}{2}.

Then

\int \frac{dx}{\sqrt{9x-4x^2}}=\frac12\int \frac{du}{\sqrt{\left(\frac94\right)^2-u^2}}.

Using

\int \frac{du}{\sqrt{a^2-u^2}}=\sin^{-1}\frac{u}{a}+C,

with

a=\frac94,

we get

\int \frac{dx}{\sqrt{9x-4x^2}}=\frac12\sin^{-1}\left(\frac{u}{9/4}\right)+C=\frac12\sin^{-1}\left(\frac{8x-9}{9}\right)+C.

So the correct option is B? Wait: the derived expression is exactly option B.

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EXERCISE 7.5

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

Show solution

Using partial fractions as in the chapter,

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

Therefore,

\int \frac{x}{(x+1)(x+2)}\,dx=-\log|x+1|+2\log|x+2|+C.

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

Show solution

Use the standard form from the chapter:

\int \frac{dx}{x^2-a^2}=\frac{1}{2a}\log\left|\frac{x-a}{x+a}\right|+C.

Here

x^2-9=x^2-3^2,

\int \frac{dx}{x^2-9}=\frac16\log\left|\frac{x-3}{x+3}\right|+C.

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

Show solution

The denominator factorises as

(x-1)(x-2)(x-3).

Write

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

Solving by the cover-up method or comparison gives

A=1,\quad B=-3,\quad C=2.

Hence

\int \frac{3x-1}{(x-1)(x-2)(x-3)}\,dx=\log|x-1|-3\log|x-2|+2\log|x-3|+C.

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

Show solution

Decompose

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

Solving gives

A=\frac12,\quad B=-2,\quad C=\frac32.

Therefore

\int \frac{x}{(x-1)(x-2)(x-3)}\,dx=\frac12\log|x-1|-2\log|x-2|+\frac32\log|x-3|+C.

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

Show solution

Factor the denominator:

x^2+3x+2=(x+1)(x+2).

Then

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

Solving gives

A=-2,\;B=4.

Hence

\int \frac{2x}{x^2+3x+2}\,dx=-2\log|x+1|+4\log|x+2|+C.

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

Show solution

Simplify first:

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

(or use division and partial fractions). Solving for constants gives

\frac{1-x^2}{x(1-2x)}=\frac1x+\frac{x}{2x-1}

which is easier to integrate after rewriting

\frac{x}{2x-1}=\frac12+\frac{1}{2(2x-1)}.

\int \frac{1-x^2}{x(1-2x)}\,dx=\log|x|+\frac{x}{2}+\frac14\log|2x-1|+C.

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

Show solution

Decompose

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

Solving gives

A=\frac12,\quad B=-\frac12,\quad C=-\frac12.

Hence

\int \frac{x}{(x^2+1)(x-1)}\,dx=\frac12\log|x-1|-\frac14\log(x^2+1)-\frac12\tan^{-1}x+C.

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

Show solution

Write

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

Solving gives

A=\frac13,\quad B=\frac13,\quad C=-\frac13.

Therefore,

\int \frac{x}{(x-1)^2(x+2)}\,dx=\frac13\log|x-1|-\frac{1}{3(x-1)}-\frac13\log|x+2|+C.

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

Show solution

Factor the denominator:

x^3-x^2-x+1=(x-1)^2(x+1).

So let

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

Solving yields

A=2,\quad B=4,\quad C=-2.

Hence

\int \frac{3x+5}{x^3-x^2-x+1}\,dx=2\log|x-1|-\frac{4}{x-1}-2\log|x+1|+C.

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

Show solution

Decompose:

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

Solving gives

A=-\frac15,\quad B=\frac75,\quad C=-\frac{12}{5}.

Hence

\int \frac{2x-3}{(x^2-1)(2x+3)}\,dx=-\frac15\log|x-1|+\frac75\log|x+1|-\frac65\log|2x+3|+C.

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

Show solution

Factor the denominator:

(x+1)(x^2-4)=(x+1)(x-2)(x+2).

Write

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

Solving gives

A=-\frac53,\quad B=\frac52,\quad C=-\frac{5}{6}.

Therefore

\int \frac{5x}{(x+1)(x^2-4)}\,dx=-\frac53\log|x+1|+\frac52\log|x-2|-\frac56\log|x+2|+C.

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

Show solution

Do polynomial division:

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

Now,

\frac{2x+1}{x^2-1}=\frac{A}{x-1}+\frac{B}{x+1}.

Solving gives

A=\frac32,\quad B=\frac12.

Thus

\int \frac{x^3+x+1}{x^2-1}\,dx=\int x\,dx+\frac32\int \frac{dx}{x-1}+\frac12\int \frac{dx}{x+1}

=\frac{x^2}{2}+\frac32\log|x-1|+\frac12\log|x+1|+C.

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

Show solution

Decompose:

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

Solving gives

A=1,\quad B=-1,\quad C=1.

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

Integrating,

\int \frac{2}{(1-x)(1+x^2)}\,dx=-\log|1-x|-\frac12\log(1+x^2)+\tan^{-1}x+C.

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

Show solution

Write

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

Then

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

A=3,\quad 2A+B=-1\Rightarrow B=-7.

Hence

\int \frac{3x-1}{(x+2)^2}\,dx=3\log|x+2|+\frac{7}{x+2}+C.

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

Show solution

Factor:

x^4-1=(x^2-1)(x^2+1)=(x-1)(x+1)(x^2+1).

Use partial fractions:

\frac1{x^4-1}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{Cx+D}{x^2+1}.

Solving gives

A=\frac14,\quad B=-\frac14,\quad C=0,\quad D=-\frac12.

Therefore

\int \frac{dx}{x^4-1}=\frac14\log|x-1|-\frac14\log|x+1|-\frac12\tan^{-1}x+C.

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

Show solution

Put

t=x^n

. Then

dt=nx^{n-1}dx

, so the standard substitution gives

\int \frac{dx}{x(x^n+1)}=\int \frac{x^{n-1}dx}{x^n(x^n+1)}

which is the standard form from the chapter's hint after multiplying numerator and denominator by

x^{n-1}

. Hence

\int \frac{dx}{x(x^n+1)}=\frac{1}{n}\int \frac{dt}{t(t+1)}

Now use partial fractions:

\frac{1}{t(t+1)}=\frac{1}{t}-\frac{1}{t+1}

So,

\frac{1}{n}\int\left(\frac{1}{t}-\frac{1}{t+1}\right)dt =\frac{1}{n}\bigl(\log|t|-\log|t+1|\bigr)+C

Substitute

t=x^n

\frac{1}{n}\log\left|\frac{x^n}{x^n+1}\right|+C

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

Show solution

Let

I=\int \frac{\cos x}{(1-\sin x)(2-\sin x)}\,dx

Put

t=\sin x

, so

dt=\cos x\,dx

.
Then

I=\int \frac{dt}{(1-t)(2-t)}

Use partial fractions:

\frac{1}{(1-t)(2-t)}=\frac{A}{1-t}+\frac{B}{2-t}

1=A(2-t)+B(1-t)

Comparing coefficients gives

A=1

B=-1

. Thus

I=\int\left(\frac{1}{1-t}-\frac{1}{2-t}\right)dt =-\log|1-t|+\log|2-t|+C

i.e.

I=\log\left|\frac{2-\sin x}{1-\sin x}\right|+C

This is equivalent to the form in the chapter's worked style; the simplest equivalent antiderivative is

\log|2-\sin x|-\log|1-\sin x|+C.

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

Show solution

This is exactly the textbook's Example 14. Decompose

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

Therefore,

\int \frac{x^2}{(x^2+1)(x^2+4)}dx =-\frac{1}{3}\int\frac{dx}{x^2+1}+\frac{4}{3}\int\frac{dx}{x^2+4}

Using the standard formulae,

=-\frac{1}{3}\tan^{-1}x+\frac{4}{3}\cdot\frac12\tan^{-1}\frac{x}{2}+C

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

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

Show solution

Use partial fractions:

\frac{2x}{(x^2+1)(x^2+3)}=\frac{A\,2x}{x^2+1}+\frac{B\,2x}{x^2+3}

More directly, let

\frac{2x}{(x^2+1)(x^2+3)}=\frac{A\,2x}{x^2+1}+\frac{B\,2x}{x^2+3}

Then

2x=A\,2x(x^2+3)+B\,2x(x^2+1)

Comparing coefficients after cancellation by

2x

, we get

1=A(x^2+3)+B(x^2+1)

A+B=0

and

3A+B=1

, giving

A=\frac12

B=-\frac12

.
Hence

\int \frac{2x}{(x^2+1)(x^2+3)}dx =\frac12\int\frac{2x}{x^2+1}dx-\frac12\int\frac{2x}{x^2+3}dx

=\frac12\log(x^2+1)-\frac12\log(x^2+3)+C

=\frac12\log\left|\frac{x^2+1}{x^2+3}\right|+C

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

Show solution

Let

t=x^4

. Then

dt=4x^3dx

. Write the integrand as in the hint-style reduction:

\int \frac{dx}{x(x^4-1)}=\int \frac{x^3dx}{x^4(x^4-1)}

So with

t=x^4

, we get

\frac{x^3dx}{x^4(x^4-1)}=\frac{1}{4}\frac{dt}{t(t-1)}

Now,

\frac{1}{t(t-1)}=-\frac{1}{t}+\frac{1}{t-1}

Hence

\int \frac{dx}{x(x^4-1)}=\frac14\int\left(-\frac1t+\frac1{t-1}\right)dt

=\frac14\bigl(-\log|t|+\log|t-1|\bigr)+C

=\frac14\log\left|\frac{x^4-1}{x^4}\right|+C

Equivalent forms differ by a sign inside the logarithm; the standard antiderivative can be written as above.

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

Show solution

Put

t=e^x

. Then

dt=e^x dx=t\,dx

, so

dx=dt/t

. Therefore

\int \frac{dx}{e^x-1}=\int \frac{1}{t-1}\cdot\frac{dt}{t}

which is not the easiest route. Instead, the standard substitution from the chapter for this type is to let

t=e^x

, giving

\frac{dx}{e^x-1}=\frac{dt}{t(t-1)}

Then

\frac{1}{t(t-1)}=-\frac{1}{t}+\frac{1}{t-1}

\int \frac{dx}{e^x-1}=\int\left(-\frac1t+\frac1{t-1}\right)dt = -\log|t|+\log|t-1|+C

=\log|e^x-1|-x+C

Thus, the antiderivative is

\log|e^x-1|-x+C

. In the book's simplified style, this comes from rewriting after substitution; the correct computed form is the one above.

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\int \frac{x \, dx}{(x - 1)(x - 2)}

equals

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

equals

EXERCISE 7.6

x \sin x

x \sin 3x

x^2 e^x

x \log x

x \log 2x

x^2 \log x

x \sin^{-1} x

x \tan^{-1} x

x \cos^{-1} x

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

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

x \sec^2 x

\tan^{-1} x

x (\log x)^2

(x^2 + 1) \log x

e^x (\sin x + \cos x)

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

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

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

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

e^{2x} \sin x

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

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

equals

\int e^x \sec x (1 + \tan x) dx

equals

EXERCISE 7.7

\sqrt{4 - x^2}

\sqrt{1 - 4x^2}

\sqrt{x^2 + 4x + 6}

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

\sqrt{1 - 4x - x^2}

\sqrt{x^2 + 4x - 5}

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

\sqrt{x^2 + 3x}

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

\int \sqrt{1 + x^2} \, dx

is equal to

\int \sqrt{x^2 - 8x + 7} \, dx

is equal to

EXERCISE 7.8

\int_{-1}^{1} (x+1) \, dx

\int_{2}^{3} \frac{1}{x} \, dx

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

\int_{0}^{\frac{\pi}{4}} \sin 2x \, dx

\int_{0}^{\frac{\pi}{2}} \cos 2x \, dx

\int_{4}^{5} e^x \, dx

\int_{0}^{\frac{\pi}{4}} \tan x \, dx

\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \csc x \, dx

\int_{0}^{1} \frac{dx}{\sqrt{1-x^2}}

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

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

\int_0^{\frac{\pi}{2}} \cos^2 x \, dx

\int_2^3 \frac{x \, dx}{x^2 + 1}

\int_0^1 \frac{2x + 3}{5x^2 + 1} \, dx

\int_0^1 x e^{x^2} \, dx

\int_1^2 \frac{5x^2}{x^2 + 4x + 3}

\int_0^{\frac{\pi}{4}} (2 \sec^2 x + x^3 + 2) \, dx

\int_0^{\pi} (\sin^2 \frac{x}{2} - \cos^2 \frac{x}{2}) \, dx

\int_0^2 \frac{6x + 3}{x^2 + 4} \, dx

\int_0^1 (x e^x + \sin \frac{\pi x}{4}) \, dx

\int_1^{\sqrt{3}} \frac{dx}{1 + x^2}

equals

\int_0^{\frac{2}{3}} \frac{dx}{4 + 9x^2}

equals

EXERCISE 7.9

\int_0^1 \frac{x}{x^2 + 1} dx

\int_0^{\frac{\pi}{2}} \sqrt{\sin \phi} \cos^5 \phi \, d\phi

\int_0^1 \sin^{-1} \left( \frac{2x}{1 + x^2} \right) dx

\int_0^2 x \sqrt{x + 2}

\int_0^{\frac{\pi}{2}} \frac{\sin x}{1 + \cos^2 x} dx

\int_0^2 \frac{dx}{x + 4 - x^2}

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

\int_1^2 \left( \frac{1}{x} - \frac{1}{2x^2} \right) e^{2x} dx

9The value of the integral

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

10If

f(x) = \int_0^x t \sin t \, dt

, then

f'(x)

EXERCISE 7.10

\int_0^{\frac{\pi}{2}} \cos^2 x \, dx

\int_0^{\frac{\pi}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx

\int_0^{\frac{\pi}{2}} \frac{\sin^{\frac{3}{2}} x \, dx}{\sin^{\frac{3}{2}} x + \cos^{\frac{3}{2}} x}

\int_0^{\frac{\pi}{2}} \frac{\cos^5 x \, dx}{\sin^5 x + \cos^5 x}

\int_{-5}^5 |x - 2| \, dx

\int_2^8 |x - 5| \, dx

\int_0^1 x (1 - x)^n \, dx

\int_0^{\frac{\pi}{4}} \log(1 + \tan x) \, dx

\int_0^2 x \sqrt{2 - x} \, dx

\int_0^{\frac{\pi}{2}} (2 \log \sin x - \log \sin 2x) \, dx

\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2 x \, dx

\int_0^{\pi} \frac{x \, dx}{1 + \sin x}

\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^7 x \, dx

\int_0^{2\pi} \cos^5 x \, dx

\int_0^{\frac{\pi}{2}} \frac{\sin x - \cos x}{1 + \sin x \cos x} \, dx

\int_0^{\pi} \log(1 + \cos x) \, dx

\int_0^a \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}} \, dx

\int_0^4 |x - 1| \, dx

19Show that

\int_0^a f(x) g(x) \, dx = 2 \int_0^a f(x) \, dx

, if

f

and

g

are defined as

f(x) = f(a - x)

and

g(x) + g(a - x) = 4

20The value of

\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (x^3 + x \cos x + \tan^5 x + 1) \, dx

21The value of

\int_0^{\frac{\pi}{2}} \log \left( \frac{4 + 3 \sin x}{4 + 3 \cos x} \right) \, dx

Miscellaneous Exercise on Chapter 7

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

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

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

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

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

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

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

\frac{e^{5 \log x} - e^{4 \log x}}{e^{3 \log x} - e^{2 \log x}}

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

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

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

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

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

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

\cos^3 x \, e^{\log \sin x}

e^{3 \log x} (x^4 + 1)^{-1}

f'(ax + b) [f(ax + b)]^n

\frac{1}{\sqrt{\sin^3 x \sin(x + \alpha)}}

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

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

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

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

\frac{\sqrt{x^2 + 1} \left[ \log(x^2 + 1) - 2 \log x \right]}{x^4}

\int_{\frac{\pi}{2}}^{\pi} e^x \left( \frac{1 - \sin x}{1 - \cos x} \right) dx

\int_{0}^{\frac{\pi}{4}} \frac{\sin x \cos x}{\cos^4 x + \sin^4 x} dx

\int_{0}^{\frac{\pi}{2}} \frac{\cos^2 x \, dx}{\cos^2 x + 4 \sin^2 x}

\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x + \cos x}{\sqrt{\sin 2x}} dx

\int_{0}^{1} \frac{dx}{\sqrt{1 + x} - \sqrt{x}}

\int_{0}^{\frac{\pi}{4}} \frac{\sin x + \cos x}{9 + 16 \sin 2x} dx

\int_{0}^{\frac{\pi}{2}} \sin 2x \tan^{-1}(\sin x) dx

\int_{1}^{4} [|x - 1| + |x - 2| + |x - 3|] dx

\int_{1}^{3} \frac{dx}{x^2(x + 1)} = \frac{2}{3} + \log \frac{2}{3}

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

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

\int_{0}^{\frac{\pi}{2}} \sin^3 x \, dx = \frac{2}{3}

\int_{0}^{\frac{\pi}{4}} 2 \tan^3 x \, dx = 1 - \log 2

\int_{0}^{1} \sin^{-1} x \, dx = \frac{\pi}{2} - 1

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

is equal to

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

is equal to

40If

f(a + b - x) = f(x)

, then

\int_a^b x f(x) \, dx

is equal to

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Integrals

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EXERCISE 7.1

EXERCISE 7.2

EXERCISE 7.3

EXERCISE 7.4

EXERCISE 7.5

EXERCISE 7.6

EXERCISE 7.7

EXERCISE 7.8

EXERCISE 7.9

EXERCISE 7.10

Miscellaneous Exercise on Chapter 7

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