A) 0
B) \[\frac{\pi }{8}\]
C) \[\frac{{{\pi }^{2}}}{8}\]
D) \[\frac{{{\pi }^{2}}}{16}\]
Correct Answer: D
Solution :
\[I=\int_{0}^{\pi /2}{\frac{x\sin x\cos x}{{{\cos }^{4}}x+{{\sin }^{4}}x}dx}\] .....(i) \[=\int_{0}^{\pi /2}{\frac{\left( \frac{\pi }{2}-x \right)\cos x\sin x}{{{\sin }^{4}}x+{{\cos }^{4}}x}}\] .....(ii) By adding (i) and (ii), we get \[2I=\frac{\pi }{2}\int_{0}^{\pi /2}{\frac{\cos x\sin x}{{{\cos }^{4}}x+{{\sin }^{4}}x}}\]dx Þ \[I=\frac{\pi }{4}\int_{0}^{\pi /2}{\frac{\tan x{{\sec }^{2}}x}{1+{{\tan }^{4}}x}dx}\] Now, Put \[{{\tan }^{2}}x=t\], we get \[I=\frac{\pi }{8}\int_{0}^{\infty }{\frac{dt}{1+{{t}^{2}}}=\frac{\pi }{8}[{{\tan }^{-1}}t]_{0}^{\infty }=\frac{{{\pi }^{2}}}{16}}\].
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