Question

Solve : $\underset{0}{\stackrel{2\pi }{\int }}\frac{sin\theta d\theta }{a-b\cos \theta }$

Answer 1

: $\underset{0}{\stackrel{2\pi }{\int }}\frac{sin\theta d\theta }{a-b\cos \theta }$

let $a-bcos\theta =t$

$+b\sin \theta d\theta =dt$

$I=\int \frac{dt}{b.t}=\frac{1}{b}\int \frac{dt}{e}$

$\frac{1}{b}\left[Int\right]$

$=\frac{1}{b}\left[in\right(a-b\cos \theta \left)\right]{}_0^{2\pi }$

$=\frac{1}{b}\left[in\right(a-b)-in(a+b\left)\right]$

$=0$