Question

Evaluate the integral

${\int }_0^{\pi /4}\frac{x^{2}dx}{(xsinx+cosx{)}^2}$

• A. $\frac{4+\pi }{4-\pi }$
• B. $\frac{4+\pi }{2(4-\pi )}$
• C. $\frac{4-\pi }{4+\pi }$
• D. $2\left[\frac{4-\pi }{4+\pi }\right]$

Answer 1

Answer: (C)

$\frac{4-\pi }{4+\pi }$

Let

$I=\int \frac{x^2}{{(x\sin x+\cos x)}^2}dx=\int \left(x\sec x\right)\left(\frac{x\cos x}{{(x\sin x+\cos x)}^2}\right)dx=\left(x\sec x\right)\left(\frac{-1}{x\sin x+\cos x}\right)-\int (\sec x+x\sec x\tan x)\left(\frac{-1}{x\sin x+\cos x}\right)dx=\frac{-x\sec x}{x\sin x+\cos x}+\int {\sec }^{2}xdx=\frac{-x\sec x}{x\sin x+\cos x}+\tan x=\frac{-x}{\cos x(x\sin x+\cos x)}+\tan x=\frac{-x+{\sin }^{2}x(x\sin x+\cos x)}{\cos x(x\sin x+\cos x)}=\frac{-x{\cos }^{2}x+\sin x\cos x}{\cos x(x\sin x+\cos x)}=\frac{\sin x-x\cos x}{x\sin x+\cos x}$

Therefore,

${\int }_0^{\frac{\pi }{4}}\frac{x^2}{{(x\sin x+\cos x)}^2}dx={\left[\frac{\sin x-x\cos x}{x\sin x+\cos x}\right]}_0^{\frac{\pi }{4}}=\frac{4-\pi }{4+\pi }$