Question

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

• A. $0$
• B. $\frac{1}{2}$
• C. $2$
• D. $\frac{3}{2}$

Answer 1

The correct option is C $2$

${\int }_0^{\pi }\frac{dx}{1+sinx}={\int }_0^{\pi }\frac{1-sinx}{co{s}^{2}x}dx={\int }_0^{\pi }(se{c}^{2}x-secxtanx)dx$
$=[tanx-secx{]}_0^{\pi }=[tan\pi -sec\pi +1]=[0+1+1]=2$