Question

The value of definite integral $\frac{4}{\pi }\underset{0}{\overset{\pi }{\int }}\frac{|x|{\sin }^{2}x}{1+2|\cos x|\sin x}dx$ is equal to

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (B)

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