Question

$ \int_{0}^{\pi /2}{\frac{x\sin x\cdot \cos x}{{{\cos }^{4}}x+{{\sin }^{4}}x}}dx $ is equal to

• A. $ \frac{{{\pi }^{2}}}{8} $
• B. $ \frac{{{\pi }^{2}}}{16} $
• C. $ 1 $
• D. $ 0 $

Answer 1

Answer: (B)

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} $