A) \[\frac{{{\pi }^{2}}}{8}\]
B) \[\frac{{{\pi }^{2}}}{16}\]
C) \[1\]
D) \[0\]
Correct Answer: B
Solution :
Let\[I=\int_{0}^{\pi /2}{\frac{x\sin x\cos x}{{{\cos }^{4}}x+{{\sin }^{4}}x}}dx\] ? (i) \[\Rightarrow \]\[I=\int_{0}^{\pi /2}{\frac{\left( \frac{\pi }{2}-x \right)\sin x\cos x}{{{\sin }^{4}}x+{{\cos }^{4}}x}d}x\] ... (ii) On adding Eqs. (i) and (ii), we get \[2I=\frac{\pi }{2}\int_{0}^{\pi /2}{\frac{\sin x\cos x}{{{\cos }^{4}}x+{{\sin }^{4}}x}}dx\] \[\Rightarrow \] \[I=\frac{\pi }{4}\int_{0}^{\pi /2}{\frac{\tan x{{\sec }^{2}}x}{1+{{\tan }^{4}}x}}\] (dividing \[Nr\] and \[Dr\] by\[{{\cos }^{4}}x)\] Put,\[{{\tan }^{2}}x=t\Rightarrow 2\tan x{{\sec }^{2}}x\,\,dx=dt\] \[\therefore \] \[I=\frac{\pi }{4}\int_{0}^{\infty }{\frac{1}{2(1+{{t}^{2}})}dt=\frac{\pi }{8}[{{\tan }^{-1}}t]_{0}^{\infty }}\] \[=\frac{\pi }{8}\times \frac{\pi }{2}=\frac{{{\pi }^{2}}}{16}\]
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