Question

${\int }_0^{\pi }\frac{1}{1+sinx}dx$ equals

• A. 0
• B. 1/2
• C. 2
• D. 3/2

Answer 1

The correct option is C 2

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

Sub $u=\tan \left(\frac{x}{2}\right)$

$dv=\frac{1}{2}{\sec }^2\left(\frac{x}{2}\right)dx$

$dx=\frac{2du}{u^2+1}$

$={\int }_0^{\infty }\frac{2}{(u^2+1)}(\frac{2u}{u^2+1}+1)du$

$={\int }_0^{\infty }\frac{2}{u^2+2u+1}du$

$=2{\int }_0^{\infty }\frac{1}{u^2+2u+1}du$

$=2{\int }_0^{\infty }\frac{1}{(u+1{)}^2}du$

Sub $s=u+1$

$ds=du$

$\Rightarrow 2{\int }_1^{\infty }\frac{1}{5^2}dx$

$\Rightarrow 2\left(\frac{1}{5}\right){\int }_1^{\infty }$

$\Rightarrow 2$.