Question

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

Answer 1

${\int }^{}\frac{\sin x}{\sin \left(x+a\right)}dx$

putting $x+a=t$

$\Rightarrow dx=dt$

$={\int }^{}\frac{\sin \left(t-a\right)}{\sin t}dt$

$={\int }^{}\frac{\left(\sin t\cos a-\cos t\sin a\right)}{\sin t}dt$

$=\cos a{\int }^{}dt-\sin a{\int }^{}\cot tdt$

$=t\cos a-\sin a\log \left|\sin t\right|+c$

$=\left(x+a\right)\cos a-\sin a\log \left|\sin \left(x+a\right)\right|+c$