Question

Evaluate :

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

Answer 1

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

$={\int }_0^{\pi }\frac{1-sinx}{1-si{n}^{2}x}dx$ Raticnaising the devotion

$={\int }_0^{\pi }\frac{1-sinx}{co{s}^{2}x}dx$

$={\int }_0^{\pi }(se{c}^{2}x-tanxsecx)dx$

$|tanx-secx{|}_0^{\pi }$

$=tan\pi -tan0-sec\pi +sec0$

$=0-0-(-1)+1$

$\Rightarrow 1+1=2$