Question

If ${\int }_0^{\pi }\frac{1}{a+b\cos x}dx=\frac{\pi }{\sqrt{a^2-b^2}}$, then ${\int }_0^{\pi }\frac{1}{(a+b\cos x{)}^2}dx=?$

(Note : $a>0$)

• A. $-\frac{\pi a}{(a^2-b^2{)}^{3/2}}$
• B. $\frac{\pi b}{(a^2-b^2{)}^{3/2}}$
• C. $\frac{\pi }{(a^2-b^2{)}^{3/2}}$
• D. $\pi$

Answer 1

The correct option is A $-\frac{\pi a}{(a^2-b^2{)}^{3/2}}$

Let $f\left(a\right)={\int }_0^{\pi }\frac{1}{a+b\cos x}dx$

On differentiating wrt. $a$ we get,
$f^{\prime }\left(a\right)=-{\int }_0^{\pi }\frac{1}{(a+b\cos x{)}^2}dx$

We have, $f\left(a\right)=\frac{\pi }{\sqrt{a^2-b^2}}$

on differentiating above function w.r.t a we get,

Hence, $f^{\prime }\left(a\right)=-\frac{\pi a}{(a^2-b^2{)}^{3/2}}$