Question

${\int }_0^{\pi }\frac{dx}{a+b\cos x}=\frac{\pi }{\sqrt{a^2-b^2}}$ where $a,b>0$ then find ${\int }_a^{\pi }\frac{dx}{(a+b\cos n{)}^{3/2}}$

Answer 1

According to question........................

$\underset{0}{\overset{\pi }{\int }}\frac{1}{a+b\cos x}dx=\frac{\pi }{\sqrt{a^2-b^2}}$

$I=\underset{0}{\overset{\pi }{\int }}\frac{1}{{(a+b\cos x)}^2}dx=\frac{1}{a^2-b^2}\underset{0}{\overset{\pi }{\int }}\frac{a^2-b^2\times 1}{(a+b\cos x)}dx$

$=\frac{1}{a^2-b^2}\underset{0}{\overset{\pi }{\int }}\frac{a^2-b^2{\cos }^{2}x-b^2{\sin }^{2}xdx}{{(a+b\cos x)}^2}$

$\begin{array}{c}=\frac{1}{a^2-b^2}\left[{}_0^{\pi }\frac{a-b\cos x\left)\right(a+b\cos x)dx}{{(a+b\cos x)}^2}{-}_0^{\pi }\frac{b^2{\sin }^{2}x}{{(a+b\cos x)}^2}dx\right]\\ =\frac{1}{a^2-b^2}\left[\left({}_0^{\pi }\frac{a-b\cos x)dx}{(a+b\cos x)}\right){-}_0^{\pi }\left(b\sin x\right)\frac{b\sin x}{{(a+b\cos x)}^2}dx\right]\end{array}$

$=\frac{1}{a^2-b^2}\left[\left({}_0^{\pi }\frac{2a-a-b\cos x)dx}{(a+b\cos x)}\right){-}_0^{\pi }\left(b\sin x\right)\frac{b\sin x}{{(a+b\cos x)}^2}dx\right]|\begin{array}{c}wedenoteasb\sin x=u\\ \frac{b\sin x}{{(a+b\cos x)}^2}=v\end{array}$

$\begin{array}{c}=\frac{1}{a^2-b^2}\left[2a_0^{\pi }\frac{dx}{(a+b\cos x)}{-}_0^{\pi }\frac{a+b\cos x}{(a+b\cos x)}dx-b\sin x_0^{\pi }\frac{b\sin x}{{(a+b\cos x)}^2}{-}_0^{\pi }\frac{b\sin x}{{(a+b\cos x)}^2}dx\right]\{arrcordingtou,vrole:\\ =\frac{1}{a^2-b^2}\left[2a_0^{\pi }\frac{dx}{(a+b\cos x)}-\pi -[{\frac{b\sin x}{a+b\cos x}}_0^{\pi }{+}_0^{\pi }\frac{b\cos xdx}{(a+b\cos x)}\right]\end{array}$

$\begin{array}{c}=\frac{1}{a^2-b^2}\left[2a_0^{\pi }\frac{dx}{(a+b\cos x)}-\pi {+}_0^{\pi }\frac{a+b\cos x}{(a+b\cos x)}-a_0^{\pi }\frac{dx}{a+b\cos x}\right]\\ =\frac{1}{a^2-b^2}\left[a_0^{\pi }\frac{dx}{(a+b\cos x)}\right]\\ =\frac{a}{a^2-b^2}\left[\frac{\pi }{\sqrt{a^2-b^2}}\right]=\frac{\pi a}{{(a^2-b^2)}^{\frac{3}{2}}}\end{array}$

$So,wefindtheanswer:\frac{\pi a}{{(a^2-b^2)}^{\frac{3}{2}}}$