Question

Integrate : $\frac{\sin x}{\sin (a+x)}$ with respect to $x$.

Answer 1

Consider, $I=\int \frac{\sin xdx}{\sin (a+x)}$

$=\int \frac{\sin (a+x-a)dx}{\sin (a+x)}$

$=\int \frac{\sin (a+x)\cos a-\sin a\cos (x+a)}{\sin (a+x)}dx$

$=\cos a\int dx-\sin a\int \frac{\cos (x+a)}{\sin (x+a)}dx$

Let $t=\sin (x+a)\Rightarrow dt=\cos (x+a)dx$

$=x\cos a-\sin a\int \frac{dt}{t}$

$=x\cos a-\sin a\ln t+c$ where $c$ is the constant of integration

$=x\cos a-\sin a\ln \sin (x+a)+c$ where

$t=\sin (x+a)$