Question

The value of definite integral $\frac{4}{\pi} \int\limits_0^\pi \frac{|x| \sin ^2 x}{1+2|\cos x| \sin x} d x$ is equal to

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (B)

Let $I=\frac{4}{\pi} \int\limits_0^\pi \frac{x \sin ^2 x}{1+2|\cos x| \sin x} d x$...(1)
Applying king property, we get
$I=\frac{4}{\pi} \int\limits_0^\pi \frac{(\pi-x) \sin ^2 x}{1+2|\cos x| \sin x} d x $...(2)
$\therefore(1)+(2) $
$\Rightarrow 2 I=\frac{4}{\pi} \times \pi \int\limits_0^\pi \frac{\sin ^2 x}{1+2|\cos x| \sin x} d x $
$\Rightarrow I=\frac{4}{2} \times 2 \int\limits_0^{\frac{\pi}{2}} \frac{\sin ^2 x}{1+\sin 2 x} d x$
$\therefore I =4 \int\limits_0^{\frac{\pi}{2}} \frac{\sin ^2 x }{1+\sin 2 x } dx $ [Using Queen Property] ....(3)
Apply king property, we get
$I=4 \int\limits_0^{\frac{\pi}{2}} \frac{\cos ^2 x}{1+\sin 2 x} d x $....(4)
$\therefore(3)+(4)$
$\Rightarrow 2 I =4 \int\limits_0^{\frac{\pi}{2}} \frac{ dx }{(\cos x+\sin x)^2} \Rightarrow I =2 \int\limits_0^{\frac{\pi}{2}} \frac{\sec ^2 x}{(1+\tan x)^2} d x$
So, $I=2 \int\limits_1^{\infty} \frac{ dt }{ t ^2}=-2\left(\frac{1}{ t }\right)_1^{\infty}=-2(0-1)=2$
Hence, I=2.