A) \[\frac{2}{3}\]
B) \[\frac{1}{4}\]
C) \[\frac{1}{3}\]
D) \[\frac{1}{6}\]
Correct Answer: D
Solution :
Let \[I=\int_{0}^{\pi /8}{{{\cos }^{3}}4\theta \,d\theta =\int_{0}^{\pi /8}{\,{{\cos }^{2}}4\theta .\cos 4\theta \,d\theta }}\] \[I=\int_{0}^{\pi /8}{\,(1-{{\sin }^{2}}4\theta )\cos 4\theta \,d\theta }\] Put \[\sin 4\theta =t\Rightarrow \cos 4\theta \,d\theta =\frac{dt}{4}\] When \[\theta =0\to \frac{\pi }{8},\] then \[t=0\to 1\] \ \[I=\frac{1}{4}\int_{0}^{1}{(1-{{t}^{2}})dt=\frac{1}{4}}\left[ t-\frac{{{t}^{3}}}{3} \right]_{0}^{1}=\frac{1}{6}\].
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