Exercise 7.3 Practice
Integration using Trigonometric Identities – NCERT Solutions
Q1
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Integrate the function: $\cos^2(3x+2)$
Use identity $\cos^2 A = \frac{1+\cos 2A}{2}$.
$\cos^2(3x+2) = \frac{1+\cos(6x+4)}{2}$.
$\int \frac{1}{2} dx + \frac{1}{2} \int \cos(6x+4) dx$.
$\boxed{\frac{x}{2} + \frac{1}{12}\sin(6x+4) + C}$
Q2
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Integrate the function: $\sin 2x \cos 5x$
Use $2\sin A \cos B = \sin(A+B) + \sin(A-B)$.
$\sin 2x \cos 5x = \frac{1}{2}[\sin 7x + \sin(-3x)] = \frac{1}{2}[\sin 7x - \sin 3x]$.
Integrate: $\frac{1}{2}[-\frac{\cos 7x}{7} - (-\frac{\cos 3x}{3})]$.
$\boxed{-\frac{1}{14}\cos 7x + \frac{1}{6}\cos 3x + C}$
Q3
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Integrate the function: $\sin x \sin 2x \sin 4x$
$\sin x \sin 2x = \frac{1}{2}(\cos x - \cos 3x)$.
Multiply by $\sin 4x$: $\frac{1}{2}(\cos x \sin 4x - \cos 3x \sin 4x)$.
$\frac{1}{4}[(\sin 5x + \sin 3x) - (\sin 7x + \sin x)]$.
Integrate term by term.
$\boxed{\frac{1}{4}[-\frac{\cos 5x}{5} - \frac{\cos 3x}{3} + \frac{\cos 7x}{7} + \cos x] + C}$
Q4
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Integrate the function: $\cos^3(2x+3)$
Use $\cos^3 \theta = \frac{3\cos \theta + \cos 3\theta}{4}$.
$\cos^3(2x+3) = \frac{1}{4}[3\cos(2x+3) + \cos(6x+9)]$.
Integrate: $\frac{1}{4}[3\frac{\sin(2x+3)}{2} + \frac{\sin(6x+9)}{6}]$.
$\boxed{\frac{3}{8}\sin(2x+3) + \frac{1}{24}\sin(6x+9) + C}$
Q5
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Integrate the function: $\sin^3 2x \cos^3 2x$
$(\sin 2x \cos 2x)^3 = (\frac{\sin 4x}{2})^3 = \frac{1}{8}\sin^3 4x$.
Use $\sin^3 4x = \frac{3\sin 4x - \sin 12x}{4}$.
$\frac{1}{32} \int (3\sin 4x - \sin 12x) dx$.
$\boxed{\frac{1}{32}[-\frac{3}{4}\cos 4x + \frac{1}{12}\cos 12x] + C}$
Q6
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Integrate the function: $\cos x \cos 2x \cos 5x$
$\cos x \cos 2x = \frac{1}{2}(\cos 3x + \cos x)$.
Multiply by $\cos 5x$: $\frac{1}{2}(\cos 3x \cos 5x + \cos x \cos 5x)$.
$\frac{1}{4}[(\cos 8x + \cos 2x) + (\cos 6x + \cos 4x)]$.
Integrate term by term.
$\boxed{\frac{1}{4}[\frac{\sin 8x}{8} + \frac{\sin 2x}{2} + \frac{\sin 6x}{6} + \frac{\sin 4x}{4}] + C}$
Q7
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Integrate the function: $\cos 3x \cos 5x$
$\frac{1}{2}(\cos(3x+5x) + \cos(3x-5x)) = \frac{1}{2}(\cos 8x + \cos 2x)$.
Integrate: $\frac{1}{2}[\frac{\sin 8x}{8} + \frac{\sin 2x}{2}]$.
$\boxed{\frac{1}{16}\sin 8x + \frac{1}{4}\sin 2x + C}$
Q8
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Integrate the function: $\frac{1+\cos 2x}{1-\cos 2x}$
$\frac{2\cos^2 x}{2\sin^2 x} = \cot^2 x$.
$\cot^2 x = \csc^2 x - 1$.
Integrate: $-\cot x - x$.
$\boxed{-\cot x - x + C}$
Q9
00:00
Integrate the function: $\frac{\cos 2x}{1+\cos 2x}$
$\frac{2\cos^2 x - 1}{2\cos^2 x} = 1 - \frac{1}{2}\sec^2 x$.
Integrate term by term.
$x - \frac{1}{2}\tan x$.
$\boxed{x - \frac{1}{2}\tan x + C}$
Q10
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Integrate the function: $\cos^4 x$
$(\cos^2 x)^2 = (\frac{1+\cos 2x}{2})^2 = \frac{1}{4}(1 + 2\cos 2x + \cos^2 2x)$.
$\frac{1}{4}[1 + 2\cos 2x + \frac{1+\cos 4x}{2}]$.
$\frac{1}{8}[3 + 4\cos 2x + \cos 4x]$.
$\boxed{\frac{3x}{8} + \frac{1}{4}\sin 2x + \frac{1}{32}\sin 4x + C}$
Q11
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Integrate the function: $\sin^4 3x$
$(\sin^2 3x)^2 = (\frac{1-\cos 6x}{2})^2 = \frac{1}{4}(1 - 2\cos 6x + \cos^2 6x)$.
$\frac{1}{4}[1 - 2\cos 6x + \frac{1+\cos 12x}{2}]$.
$\frac{1}{8}[3 - 4\cos 6x + \cos 12x]$.
$\boxed{\frac{3x}{8} - \frac{1}{12}\sin 6x + \frac{1}{96}\sin 12x + C}$
Q12
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Integrate the function: $\frac{\cos^2 x}{1-\sin x}$
$\frac{1-\sin^2 x}{1-\sin x} = \frac{(1-\sin x)(1+\sin x)}{1-\sin x} = 1+\sin x$.
Integrate: $x - \cos x$.
$\boxed{x - \cos x + C}$
Q13
00:00
Integrate the function: $\frac{\cos 2x - \cos 2\alpha}{\cos x + \cos \alpha}$
$\cos 2x = 2\cos^2 x - 1$, $\cos 2\alpha = 2\cos^2 \alpha - 1$.
Num: $2(\cos^2 x - \cos^2 \alpha) = 2(\cos x - \cos \alpha)(\cos x + \cos \alpha)$.
Integrand becomes $2(\cos x - \cos \alpha)$.
$\boxed{2\sin x - 2x\cos \alpha + C}$
Q14
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Integrate the function: $\frac{\sin x + \cos x}{\sqrt{1+\sin 2x}}$
$1+\sin 2x = (\sin x + \cos x)^2$.
$\sqrt{1+\sin 2x} = \sin x + \cos x$.
Integrand becomes $\frac{\sin x + \cos x}{\sin x + \cos x} = 1$.
$\boxed{x + C}$
Q15
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Integrate the function: $\tan^3 3x \sec 3x$
$\tan^2 3x \cdot \sec 3x \tan 3x = (\sec^2 3x - 1) \sec 3x \tan 3x$.
Let $\sec 3x = t \Rightarrow 3\sec 3x \tan 3x dx = dt$.
$\frac{1}{3} \int (t^2 - 1) dt = \frac{1}{3}(\frac{t^3}{3} - t) + C$.
$\boxed{\frac{1}{9}\sec^3 3x - \frac{1}{3}\sec 3x + C}$
Q16
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Integrate the function: $\cot^4 x$
$\cot^2 x (\csc^2 x - 1) = \cot^2 x \csc^2 x - \cot^2 x$.
$\cot^2 x \csc^2 x - (\csc^2 x - 1)$.
Integrate: $-\frac{\cot^3 x}{3} - (-\cot x) + x$.
$\boxed{-\frac{1}{3}\cot^3 x + \cot x + x + C}$
Q17
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Integrate the function: $\frac{\sin^3 x - \cos^3 x}{\sin^2 x \cos^2 x}$
$\frac{\sin^3 x}{\sin^2 x \cos^2 x} - \frac{\cos^3 x}{\sin^2 x \cos^2 x} = \frac{\sin x}{\cos^2 x} - \frac{\cos x}{\sin^2 x}$.
$\tan x \sec x - \cot x \csc x$.
Integrate: $\sec x - (-\csc x)$.
$\boxed{\sec x + \csc x + C}$
Q18
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Integrate the function: $\frac{\cos 2x - 2\cos^2 x}{\sin^2 x}$
$\cos 2x = 2\cos^2 x - 1$.
Num: $(2\cos^2 x - 1) - 2\cos^2 x = -1$.
$\int \frac{-1}{\sin^2 x} dx = \int -\csc^2 x dx$.
$\boxed{\cot x + C}$
Q19
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Integrate the function: $\frac{1}{\sin x \cos x}$
Multiply num/den by 2: $\frac{2}{2\sin x \cos x} = \frac{2}{\sin 2x}$.
$2\csc 2x$.
$\boxed{\log|\tan x| + C}$
Q20
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Integrate the function: $\frac{\cos 2x}{(\cos x - \sin x)^2}$
$\cos 2x = \cos^2 x - \sin^2 x = (\cos x - \sin x)(\cos x + \sin x)$.
$\frac{\cos x + \sin x}{\cos x - \sin x}$.
Let $\cos x - \sin x = t \Rightarrow (-\sin x - \cos x) dx = dt \Rightarrow (\sin x + \cos x) dx = -dt$.
$\int -\frac{1}{t} dt = -\log|t| + C$.
$\boxed{-\log|\cos x - \sin x| + C}$
Q21
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Integrate the function: $\cos^{-1}(\sin x)$
$\sin x = \cos(\frac{\pi}{2} - x)$.
$\cos^{-1}(\cos(\frac{\pi}{2} - x)) = \frac{\pi}{2} - x$.
Integrate: $\frac{\pi}{2}x - \frac{x^2}{2} + C$.
$\boxed{\frac{\pi x}{2} - \frac{x^2}{2} + C}$
Q22
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Integrate the function: $\frac{1}{\sin(x-a)\sin(x-b)}$
Multiply/divide by $\sin(a-b)$.
$\frac{1}{\sin(a-b)} \int \frac{\sin((x-b)-(x-a))}{\sin(x-a)\sin(x-b)} dx$.
Expand $\sin(A-B)$: $\sin(x-b)\cos(x-a) - \cos(x-b)\sin(x-a)$.
$\frac{1}{\sin(a-b)} [\int \cot(x-a) dx - \int \cot(x-b) dx]$.
$\boxed{\frac{1}{\sin(a-b)} \log|\frac{\sin(x-a)}{\sin(x-b)}| + C}$
Q23
00:00
$\int \frac{\cos^2 x - \sin^2 x}{\sin^2 x \cos^2 x} dx$ equals:
(A) $-\cot x - \tan x + C$
(B) $\tan x + \cot x + C$
(C) $\tan x - \cot x + C$
(D) $\cot x - \tan x + C$
(A) $-\cot x - \tan x + C$
(B) $\tan x + \cot x + C$
(C) $\tan x - \cot x + C$
(D) $\cot x - \tan x + C$
$\int (\csc^2 x - \sec^2 x) dx$.
$-\cot x - \tan x + C$.
$\boxed{\text{(A) } -\cot x - \tan x + C}$
Q24
00:00
$\int \frac{e^x(1+x)}{\sin^2(xe^x)} dx$ equals:
(A) $-\cot(xe^x) + C$
(B) $\tan(xe^x) + C$
(C) $\tan(e^x) + C$
(D) $\cot(e^x) + C$
(A) $-\cot(xe^x) + C$
(B) $\tan(xe^x) + C$
(C) $\tan(e^x) + C$
(D) $\cot(e^x) + C$
Let $xe^x = t \Rightarrow (e^x + xe^x) dx = dt \Rightarrow e^x(1+x) dx = dt$.
$\int \frac{1}{\sin^2 t} dt = \int \csc^2 t dt = -\cot t + C$.
$\boxed{\text{(A) } -\cot(xe^x) + C}$