Question

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

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

• A. $1$
• B. $2$
• C. $4$
• D. $-2$

Answer 1

The correct option is B $2$

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

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

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

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

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

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

$I=[0+2]$

$I=2$