Question

The value of definite integral $\underset{0}{\overset{\pi }{\int }}\frac{\pi \tan x}{\sec x+\tan x}dx$ is equal to

• A. $\pi (1-\pi )$
• B. $\pi (\pi -2)$
• C. $\pi (2-\pi )$
• D. $\pi (\pi -1)$

Answer 1

Answer: (B)

Let $I=\underset{0}{\overset{\pi }{\int }}\frac{\pi \sin x}{1+\sin x}dx=2\pi \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{\sin x}{1+\sin x}dx=2\pi \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{\cos x}{1+\cos x}dx$ (Using king property)
$\therefore I=2\pi \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{(1+\cos x)-1}{1+\cos x}dx=2\pi \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(1-\frac{1}{2{\cos }^2\frac{x}{2}}\right)dx$
$=2\pi \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(1-\frac{1}{2}{\sec }^2\frac{x}{2}\right)dx=2\pi {\left(x-\tan \frac{x}{2}\right)}_0^{\frac{\pi }{2}}=2\pi \left(\frac{\pi }{2}-1\right)=\pi (\pi -2)$.