Definite Integration Question 453
Question: $ \int_0^{\pi /2}{\frac{x\sin x\cos x}{{{\cos }^{4}}x+{{\sin }^{4}}x}}dx= $
[IIT 1985]
Options:
A) 0
B) $ \frac{\pi }{8} $
C) $ \frac{{{\pi }^{2}}}{8} $
D) $ \frac{{{\pi }^{2}}}{16} $
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Answer:
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{( \frac{\pi }{2}-x )\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
therefore $ 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}} $ .
BETA
