If `int_0^pi 1/(a+bcosx)dx=pi/sqrt(a^2-b^2)`, then `int_0^pi 1/(a+bcosx)^2dx` is

Explanation

Let `f(a)=int_{0}^{pi} frac{1}{a+b cos x} dx`

On differentiating wrt. a we get,

`f^{prime}(a)=-int_{0}^{pi} frac{1}{(a+b cos x)^{2}} d x`

We have, `f(a)=frac{pi}{sqrt{a^{2}-b^{2}}}` on differentiating above function w.r.t a we get, Hence, `f^{prime}(a)=-frac{pi a}{(a^{2}-b^{2})^{3 / 2}}`