A) \[\frac{1}{12}\]
B) \[\frac{3}{12}\]
C) \[\frac{5}{12}\]
D) None of these
Correct Answer: C
Solution :
\[\int_{0}^{\pi /4}{{{\sec }^{7}}\theta }.{{\sin }^{3}}\theta \,d\theta \]=\[\int_{0}^{\pi /4}{\frac{{{\sin }^{3}}\theta }{{{\cos }^{3}}\theta }.{{\sec }^{4}}\theta \,d\theta }\] Putting \[\tan \theta =t,\] it reduces to \[\int_{0}^{1}{{{t}^{3}}(1+{{t}^{2}})\,dt}\]= \[\left| \frac{{{t}^{4}}}{4}+\frac{{{t}^{6}}}{6} \right|_{0}^{1}=\frac{5}{12}\].
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