Exercise 7.3, Question 1
\(\displaystyle\int \sin^2(2x + 5)\,dx\)
Show solution
- \[\sin^2(2x + 5) = \tfrac12[1 - \cos(4x + 10)]\]
Answer: \[\dfrac{x}{2} - \dfrac{\sin(4x + 10)}{8} + C\]
NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.3
Exercise 7.3: Integration using trigonometric identities. Reduce powers and products to single sines and cosines: \(\sin^2 x = \tfrac{1 - \cos 2x}{2}\), \(\cos^2 x = \tfrac{1 + \cos 2x}{2}\), \(\sin^3 x = \tfrac{3\sin x - \sin 3x}{4}\); \(2\sin A\cos B = \sin(A + B) + \sin(A - B)\), \(2\cos A\cos B = \cos(A + B) + \cos(A - B)\), \(2\sin A\sin B = \cos(A - B) - \cos(A + B)\). Odd powers: keep one factor for dt.
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Exercise 7.3 questions and solutions
Exercise 7.3, Question 2
\(\displaystyle\int \sin 3x\cos 4x\,dx\)
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- \[\begin{aligned}\sin 3x\cos 4x &= \tfrac12[\sin 7x + \sin(-x)] \\ &= \tfrac12[\sin 7x - \sin x]\end{aligned}\]
Answer: \[\dfrac{\cos x}{2} - \dfrac{\cos 7x}{14} + C\]
Exercise 7.3, Question 3
\(\displaystyle\int \cos 2x\cos 4x\cos 6x\,dx\)
Show solution
- \[\cos 2x\cos 4x = \tfrac12(\cos 6x + \cos 2x)\]
- Times \(\cos 6x\): \[\tfrac12(\cos^2 6x + \cos 6x\cos 2x) = \tfrac14(1 + \cos 12x) + \tfrac14(\cos 8x + \cos 4x)\]
Answer: \[\tfrac14\left[x + \dfrac{\sin 4x}{4} + \dfrac{\sin 8x}{8} + \dfrac{\sin 12x}{12}\right] + C\]
Exercise 7.3, Question 4
\(\displaystyle\int \sin^3(2x + 1)\,dx\)
Show solution
- \[\sin^3\theta = (1 - \cos^2\theta)\sin\theta\]; \(t = \cos(2x + 1)\), \(dt = -2\sin(2x + 1)\,dx\).
- \[-\tfrac12\int(1 - t^2)\,dt = -\tfrac12\left(t - \tfrac{t^3}{3}\right)\]
Answer: \[-\tfrac12\cos(2x + 1) + \tfrac16\cos^3(2x + 1) + C\]
Exercise 7.3, Question 5
\(\displaystyle\int \sin^3 x\cos^3 x\,dx\)
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- \[\sin^3 x\cos^2 x \cdot \cos x = \sin^3 x(1 - \sin^2 x)\cos x\]; \(t = \sin x\): \(\int(t^3 - t^5)\,dt\).
Answer: \[\dfrac{\sin^4 x}{4} - \dfrac{\sin^6 x}{6} + C\]
Exercise 7.3, Question 6
\(\displaystyle\int \sin x\sin 2x\sin 3x\,dx\)
Show solution
- \[\sin x\sin 3x = \tfrac12(\cos 2x - \cos 4x)\]
- Times \(\sin 2x\): \[\tfrac12(\sin 2x\cos 2x - \sin 2x\cos 4x) = \tfrac14\sin 4x - \tfrac14(\sin 6x - \sin 2x)\]
Answer: \[\tfrac14\left[\dfrac{\cos 6x}{6} - \dfrac{\cos 4x}{4} - \dfrac{\cos 2x}{2}\right] + C\]
Exercise 7.3, Question 7
\(\displaystyle\int \sin 4x\sin 8x\,dx\)
Show solution
- \[\sin 4x\sin 8x = \tfrac12[\cos 4x - \cos 12x]\]
Answer: \[\tfrac12\left[\dfrac{\sin 4x}{4} - \dfrac{\sin 12x}{12}\right] + C\]
Exercise 7.3, Question 8
\(\displaystyle\int \dfrac{1 - \cos x}{1 + \cos x}\,dx\)
Show solution
- \[\begin{aligned}\dfrac{2\sin^2\frac{x}{2}}{2\cos^2\frac{x}{2}} &= \tan^2\dfrac{x}{2} \\ &= \sec^2\dfrac{x}{2} - 1\end{aligned}\]
Answer: \(2\tan\dfrac{x}{2} - x + C\)
Exercise 7.3, Question 9
\(\displaystyle\int \dfrac{\cos x}{1 + \cos x}\,dx\)
Show solution
- \[1 - \dfrac{1}{1 + \cos x} = 1 - \tfrac12\sec^2\dfrac{x}{2}\]
Answer: \(x - \tan\dfrac{x}{2} + C\)
Exercise 7.3, Question 10
\(\displaystyle\int \sin^4 x\,dx\)
Show solution
- \[\begin{aligned}\sin^4 x &= \left(\dfrac{1 - \cos 2x}{2}\right)^2 \\ &= \tfrac14\left(1 - 2\cos 2x + \dfrac{1 + \cos 4x}{2}\right) \\ &= \tfrac38 - \tfrac12\cos 2x + \tfrac18\cos 4x\end{aligned}\]
Answer: \[\dfrac{3x}{8} - \dfrac{\sin 2x}{4} + \dfrac{\sin 4x}{32} + C\]
Exercise 7.3, Question 11
\(\displaystyle\int \cos^4 2x\,dx\)
Show solution
- \[\begin{aligned}\cos^4 2x &= \tfrac14(1 + \cos 4x)^2 \\ &= \tfrac38 + \tfrac12\cos 4x + \tfrac18\cos 8x\end{aligned}\]
Answer: \[\dfrac{3x}{8} + \dfrac{\sin 4x}{8} + \dfrac{\sin 8x}{64} + C\]
Exercise 7.3, Question 12
\(\displaystyle\int \dfrac{\sin^2 x}{1 + \cos x}\,dx\)
Show solution
- \[\dfrac{1 - \cos^2 x}{1 + \cos x} = 1 - \cos x\]
Answer: \(x - \sin x + C\)
Exercise 7.3, Question 13
\(\displaystyle\int \dfrac{\cos 2x - \cos 2\alpha}{\cos x - \cos\alpha}\,dx\)
Show solution
- \[\cos 2x - \cos 2\alpha = 2(\cos^2 x - \cos^2\alpha)\], so the integrand is \(2(\cos x + \cos\alpha)\).
Answer: \(2\sin x + 2x\cos\alpha + C\)
Exercise 7.3, Question 14
\(\displaystyle\int \dfrac{\cos x - \sin x}{1 + \sin 2x}\,dx\)
Show solution
- \(1 + \sin 2x = (\sin x + \cos x)^2\); \(t = \sin x + \cos x\), \(dt = (\cos x - \sin x)\,dx\): \(\int t^{-2}\,dt\).
Answer: \(-\dfrac{1}{\sin x + \cos x} + C\)
Exercise 7.3, Question 15
\(\displaystyle\int \tan^3 2x\sec 2x\,dx\)
Show solution
- \((\sec^2 2x - 1)\sec 2x\tan 2x\); \(t = \sec 2x\), \(dt = 2\sec 2x\tan 2x\,dx\): \(\tfrac12\int(t^2 - 1)\,dt\).
Answer: \[\dfrac{\sec^3 2x}{6} - \dfrac{\sec 2x}{2} + C\]
Exercise 7.3, Question 16
\(\displaystyle\int \tan^4 x\,dx\)
Show solution
- \[\begin{aligned}\tan^4 x &= \tan^2 x(\sec^2 x - 1) \\ &= \tan^2 x\sec^2 x - \sec^2 x + 1\end{aligned}\]
Answer: \(\dfrac{\tan^3 x}{3} - \tan x + x + C\)
Exercise 7.3, Question 17
\(\displaystyle\int \dfrac{\sin^3 x + \cos^3 x}{\sin^2 x\cos^2 x}\,dx\)
Show solution
- Split: \[\dfrac{\sin x}{\cos^2 x} + \dfrac{\cos x}{\sin^2 x} = \sec x\tan x + \csc x\cot x\]
Answer: \(\sec x - \csc x + C\)
Exercise 7.3, Question 18
\(\displaystyle\int \dfrac{\cos 2x + 2\sin^2 x}{\cos^2 x}\,dx\)
Show solution
- \(\cos 2x + 2\sin^2 x = 1\), so the integrand is \(\sec^2 x\).
Answer: \(\tan x + C\)
Exercise 7.3, Question 19
\(\displaystyle\int \dfrac{1}{\sin x\cos^3 x}\,dx\)
Show solution
- \[\begin{aligned}\dfrac{1}{\sin x\cos^3 x} &= \dfrac{\sec^4 x}{\tan x} \\ &= \dfrac{(1 + \tan^2 x)\sec^2 x}{\tan x}\end{aligned}\]
- \(t = \tan x\): \(\int\left(\dfrac1t + t\right)dt\).
Answer: \(\log|\tan x| + \dfrac{\tan^2 x}{2} + C\)
Exercise 7.3, Question 20
\(\displaystyle\int \dfrac{\cos 2x}{(\cos x + \sin x)^2}\,dx\)
Show solution
- \[\cos 2x = (\cos x + \sin x)(\cos x - \sin x)\], so the integrand is \(\dfrac{\cos x - \sin x}{\cos x + \sin x}\), a derivative over its function.
Answer: \(\log|\cos x + \sin x| + C\)
Exercise 7.3, Question 21
\(\displaystyle\int \sin^{-1}(\cos x)\,dx\)
Show solution
- \[\cos x = \sin\left(\tfrac{\pi}{2} - x\right)\], and for \(0 \le x \le \pi\), \(\tfrac{\pi}{2} - x\) is in the principal branch: \(\sin^{-1}(\cos x) = \tfrac{\pi}{2} - x\).
Answer: \(\dfrac{\pi x}{2} - \dfrac{x^2}{2} + C\)
Exercise 7.3, Question 22
\(\displaystyle\int \dfrac{1}{\cos(x - a)\cos(x - b)}\,dx\)
Show solution
- Write \[\begin{aligned}1 &= \dfrac{\sin(a - b)}{\sin(a - b)} \\ &= \dfrac{\sin[(x - b) - (x - a)]}{\sin(a - b)}\end{aligned}\] and expand.
- The integrand becomes \[\dfrac{1}{\sin(a - b)}[\tan(x - b) - \tan(x - a)]\]
- \(\int\tan(x - b)\,dx = -\log|\cos(x - b)|\) and \(\int\tan(x - a)\,dx = -\log|\cos(x - a)|\).
Answer: \[\dfrac{1}{\sin(a - b)}\log\left|\dfrac{\cos(x - a)}{\cos(x - b)}\right| + C\]
Exercise 7.3, Question 23
\(\displaystyle\int\dfrac{\sin^2 x - \cos^2 x}{\sin^2 x\cos^2 x}\,dx\) is: (A) \(\tan x + \cot x + C\) (B) \(\tan x + \csc x + C\) (C) \(-\tan x + \cot x + C\) (D) \(\tan x + \sec x + C\)
Show solution
- \[\dfrac{1}{\cos^2 x} - \dfrac{1}{\sin^2 x} = \sec^2 x - \csc^2 x\], which integrates to \(\tan x + \cot x\).
Answer: (A)
Exercise 7.3, Question 24
\(\displaystyle\int\dfrac{e^x(1 + x)}{\cos^2(e^x x)}\,dx\) equals: (A) \(-\cot(ex^x) + C\) (B) \(\tan(xe^x) + C\) (C) \(\tan(e^x) + C\) (D) \(\cot(e^x) + C\)
Show solution
- \(t = xe^x\), \(dt = e^x(1 + x)\,dx\): \(\int\sec^2 t\,dt\).
Answer: (B)
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