Use \(\displaystyle \cos^{2}\theta=\dfrac{1+\cos 2\theta}{2}\) twice. With \(\displaystyle \theta=2x\),
\[\cos^{4}2x=\left(\frac{1+\cos 4x}{2}\right)^{2}=\frac{1+2\cos 4x+\cos^{2}4x}{4}.\]
Reduce the leftover square, \(\displaystyle \cos^{2}4x=\dfrac{1+\cos 8x}{2}\):
\[\cos^{4}2x=\frac{1}{4}\left[1+2\cos 4x+\frac{1+\cos 8x}{2}\right]=\frac{3}{8}+\frac{\cos 4x}{2}+\frac{\cos 8x}{8}.\]
Integrating,
\[\int\cos^{4}2x\,dx=\frac{3x}{8}+\frac{1}{2}\cdot\frac{\sin 4x}{4}+\frac{1}{8}\cdot\frac{\sin 8x}{8}+\mathrm{C}.\]
Final answer: \(\displaystyle \dfrac{3x}{8}+\dfrac{\sin 4x}{8}+\dfrac{\sin 8x}{64}+\mathrm{C}\).
Exercise 12
1+cosxsin2x
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NCERT’s answer
\(\displaystyle x-\sin x+\mathrm{C}\)
Use the Pythagorean identity \(\displaystyle \sin^{2}x=1-\cos^{2}x\) and factor the difference of squares, so the denominator cancels:
\[\frac{\sin^{2}x}{1+\cos x}=\frac{1-\cos^{2}x}{1+\cos x}=\frac{(1-\cos x)(1+\cos x)}{1+\cos x}=1-\cos x,\]
the cancellation being legitimate wherever \(\displaystyle 1+\cos x\neq0\).
Hence
\[\int\frac{\sin^{2}x}{1+\cos x}\,dx=\int(1-\cos x)\,dx=x-\sin x+\mathrm{C}.\]
Final answer: \(\displaystyle x-\sin x+\mathrm{C}\).
Exercise 13
cosx−cosαcos2x−cos2α
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Write both cosines of the double angle in the same form, \(\displaystyle \cos 2\theta=2\cos^{2}\theta-1\):
\[\cos 2x-\cos 2\alpha=(2\cos^{2}x-1)-(2\cos^{2}\alpha-1)=2(\cos^{2}x-\cos^{2}\alpha).\]
Factor the difference of squares so the denominator cancels:
\[\frac{2(\cos x-\cos\alpha)(\cos x+\cos\alpha)}{\cos x-\cos\alpha}=2(\cos x+\cos\alpha),\qquad \cos x\neq\cos\alpha.\]
Here \(\displaystyle \alpha\) is a constant, so \(\displaystyle \cos\alpha\) integrates like a number:
\[\int\frac{\cos 2x-\cos 2\alpha}{\cos x-\cos\alpha}\,dx=2\int(\cos x+\cos\alpha)\,dx=2\sin x+2x\cos\alpha+\mathrm{C}.\]
Final answer: \(\displaystyle 2(\sin x+x\cos\alpha)+\mathrm{C}\).
Exercise 14
1+sin2xcosx−sinx
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NCERT’s answer
\(\displaystyle -\frac{1}{\cos x+\sin x}+C\)
Recognise the denominator as a perfect square: since \(\displaystyle \sin^{2}x+\cos^{2}x=1\) and \(\displaystyle \sin 2x=2\sin x\cos x\),
\[1+\sin 2x=\sin^{2}x+\cos^{2}x+2\sin x\cos x=(\sin x+\cos x)^{2}.\]
The numerator is exactly the derivative of \(\displaystyle \sin x+\cos x\), so substitute \(\displaystyle t=\sin x+\cos x\), \(\displaystyle dt=(\cos x-\sin x)\,dx\):
\[\int\frac{\cos x-\sin x}{1+\sin 2x}\,dx=\int\frac{dt}{t^{2}}=-\frac{1}{t}+\mathrm{C}.\]
Final answer: \(\displaystyle -\dfrac{1}{\sin x+\cos x}+\mathrm{C}\).
Exercise 15
tan32xsec2x
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This solution has not been cross-checked against the answer printed in NCERT.
Keep one factor of \(\displaystyle \sec 2x\tan 2x\) aside, because that is the derivative pattern of \(\displaystyle \sec 2x\), and convert the rest with \(\displaystyle \tan^{2}\theta=\sec^{2}\theta-1\):
\[\tan^{3}2x\sec 2x=\tan^{2}2x\,(\sec 2x\tan 2x)=(\sec^{2}2x-1)\,(\sec 2x\tan 2x).\]
Put \(\displaystyle u=\sec 2x\); then \(\displaystyle du=2\sec 2x\tan 2x\,dx\), so \(\displaystyle \sec 2x\tan 2x\,dx=\tfrac12\,du\):
\[\int\tan^{3}2x\sec 2x\,dx=\frac{1}{2}\int(u^{2}-1)\,du=\frac{1}{2}\left[\frac{u^{3}}{3}-u\right]+\mathrm{C}.\]
Substituting back \(\displaystyle u=\sec 2x\):
Final answer: \(\displaystyle \dfrac{\sec^{3}2x}{6}-\dfrac{\sec 2x}{2}+\mathrm{C}\).
Exercise 16
tan4x
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Peel off \(\displaystyle \tan^{2}x\) using \(\displaystyle \tan^{2}x=\sec^{2}x-1\), which manufactures the \(\displaystyle \sec^{2}x\,dx\) needed for a substitution:
\[\tan^{4}x=\tan^{2}x(\sec^{2}x-1)=\tan^{2}x\sec^{2}x-\tan^{2}x.\]
For the first piece put \(\displaystyle t=\tan x,\;dt=\sec^{2}x\,dx\), giving \(\displaystyle \int t^{2}dt=\dfrac{\tan^{3}x}{3}\). For the second piece apply the identity once more:
\[\int\tan^{2}x\,dx=\int(\sec^{2}x-1)\,dx=\tan x-x.\]
Therefore
\[\int\tan^{4}x\,dx=\frac{\tan^{3}x}{3}-(\tan x-x)+\mathrm{C}.\]
Final answer: \(\displaystyle \dfrac{\tan^{3}x}{3}-\tan x+x+\mathrm{C}\).
Exercise 17
sin2xcos2xsin3x+cos3x
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Split the fraction over the two terms of the numerator and cancel:
\[\frac{\sin^{3}x+\cos^{3}x}{\sin^{2}x\cos^{2}x}=\frac{\sin^{3}x}{\sin^{2}x\cos^{2}x}+\frac{\cos^{3}x}{\sin^{2}x\cos^{2}x}=\frac{\sin x}{\cos^{2}x}+\frac{\cos x}{\sin^{2}x}.\]
Each piece is a standard derivative in disguise:
\[\frac{\sin x}{\cos^{2}x}=\frac{1}{\cos x}\cdot\frac{\sin x}{\cos x}=\sec x\tan x,\qquad \frac{\cos x}{\sin^{2}x}=\mathrm{cosec}\,x\cot x.\]
Using \(\displaystyle \int\sec x\tan x\,dx=\sec x\) and \(\displaystyle \int\mathrm{cosec}\,x\cot x\,dx=-\,\mathrm{cosec}\,x\) (mind that minus sign):
\[\int\frac{\sin^{3}x+\cos^{3}x}{\sin^{2}x\cos^{2}x}\,dx=\sec x-\mathrm{cosec}\,x+\mathrm{C}.\]
Final answer: \(\displaystyle \sec x-\mathrm{cosec}\,x+\mathrm{C}\).
Exercise 18
cos2xcos2x+2sin2x
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NCERT’s answer
\(\displaystyle \tan x+\mathrm{C}\)
Choose the form of \(\displaystyle \cos 2x\) that matches the other term in the numerator, namely \(\displaystyle \cos 2x=1-2\sin^{2}x\):
\[\cos 2x+2\sin^{2}x=(1-2\sin^{2}x)+2\sin^{2}x=1.\]
The whole numerator collapses to \(\displaystyle 1\), so
\[\int\frac{\cos 2x+2\sin^{2}x}{\cos^{2}x}\,dx=\int\frac{dx}{\cos^{2}x}=\int\sec^{2}x\,dx.\]
Final answer: \(\displaystyle \tan x+\mathrm{C}\).
Exercise 19
sinxcos3x1
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Force the integrand into a function of \(\displaystyle \tan x\) by dividing numerator and denominator by \(\displaystyle \cos^{4}x\). Since \(\displaystyle \sin x\cos^{3}x=\cos^{4}x\tan x\),
\[\frac{1}{\sin x\cos^{3}x}=\frac{\sec^{4}x}{\tan x}=\frac{(1+\tan^{2}x)\sec^{2}x}{\tan x},\]
using \(\displaystyle \sec^{2}x=1+\tan^{2}x\) on one of the two factors of \(\displaystyle \sec^{2}x\) and keeping the other for the substitution.
Put \(\displaystyle t=\tan x,\;dt=\sec^{2}x\,dx\):
\[\int\frac{dx}{\sin x\cos^{3}x}=\int\frac{1+t^{2}}{t}\,dt=\int\left(\frac{1}{t}+t\right)dt=\log|t|+\frac{t^{2}}{2}+\mathrm{C}.\]
Final answer: \(\displaystyle \log|\tan x|+\dfrac{\tan^{2}x}{2}+\mathrm{C}\).
Exercise 20
(cosx+sinx)2cos2x
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NCERT’s answer
\(\displaystyle \log |\cos x+\sin x|+\mathrm{C}\)
Expand \(\displaystyle \cos 2x\) as a difference of squares, \(\displaystyle \cos 2x=\cos^{2}x-\sin^{2}x=(\cos x-\sin x)(\cos x+\sin x)\). Then one factor of the denominator cancels:
\[\frac{\cos 2x}{(\cos x+\sin x)^{2}}=\frac{(\cos x-\sin x)(\cos x+\sin x)}{(\cos x+\sin x)^{2}}=\frac{\cos x-\sin x}{\cos x+\sin x}.\]
The numerator is now the derivative of the denominator, so put \(\displaystyle t=\cos x+\sin x\), \(\displaystyle dt=(\cos x-\sin x)\,dx\):
\[\int\frac{\cos 2x}{(\cos x+\sin x)^{2}}\,dx=\int\frac{dt}{t}=\log|t|+\mathrm{C}.\]
Final answer: \(\displaystyle \log|\cos x+\sin x|+\mathrm{C}\).