A) \[\frac{\pi }{2}\]
B) \[\frac{\pi }{4}\]
C) \[\frac{\pi }{6}\]
D) \[\frac{\pi }{8}\]
Correct Answer: D
Solution :
Put \[{{\sin }^{2}}x=t\Rightarrow dt=2\sin x\cos x\,dx\] Now\[\int_{0}^{\pi /2}{\frac{\sin x\cos x}{1+{{\sin }^{4}}x}dx=\frac{1}{2}\int_{0}^{1}{\frac{1}{1+{{t}^{2}}}dt=\frac{1}{2}[{{\tan }^{-1}}t]_{0}^{1}=\frac{\pi }{8}}}\].
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