Question

Consider $I={\int }_0^{\pi }\frac{xdx}{1+\sin x}$.

What is ${\int }_0^{\pi }\frac{(\pi -x)dx}{1+\sin x}$ equal to?

• A. $\pi$
• B. $\frac{\pi }{2}$
• C. $0$
• D. $2\pi$

Answer 1

The correct option is A $\pi$

$I={\int }_0^{\pi }\frac{(\pi -x)dx}{1+\sin x}$

by property,

$I={\int }_0^{\pi }\frac{xdx}{1+\sin x}$

$2I={\int }_0^{\pi }\frac{\pi dx}{1+\sin x}$

$2I={\int }_0^{\pi }\frac{\pi }{1+\sin x}\times \frac{1-\sin x}{1-\sin x}dx$

$2I={\int }_0^{\pi }\frac{\pi (1-\sin x)dx}{{\cos }^{2}x}$

$I=\frac{\pi }{2}{\int }_0^{\pi }\left[{\sec }^{2}x\tan x\sec x\right]dx$

$I=\frac{\pi }{2}[\tan x-\sec x{]}_0^{\pi }$

$I=\frac{\pi }{2}[0+2]$

$I=\pi$