evaluate the following definite integral

akki ~~ unlucky forever ~~

differentiate wrt b

$\frac{dI}{db}=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\frac{dx}{1+bsinx}$

again by property of definite integral.

$\frac{dI}{db}=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\frac{dx}{1-bsinx}$

adding both,

$\frac{dI}{db}=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\frac{dx}{(1-b^2sin^2x)}$

$\frac{dI}{db}=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\frac{sec^2x\ dx}{(1+tan^2x(1-b^2))}$

put tanx = t

$\frac{dI}{db}=\frac{tan^{-1}(t\sqrt{1-b^2)}}{\sqrt{1-b^2}}$ limit from -infty to infty

$\frac{dI}{db}=\frac{\pi}{\sqrt{(1-b^2)}}$

Integrating,

$I=\pi sin^{-1}b+C$

now see the original equation, at b = 0, I = 0,

so, C = 0

$I=\pi sin^{-1}b$