Question

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

Answer 1

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

$puttingx+b=t$

$dx=dt$

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

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

$=\int \left[\frac{\sin t\cos \left(a-b\right)}{\sin t}+\frac{\cos t\sin \left(a-b\right)}{\sin t}\right]dt$

$=\int \cos \left(a-b\right)dt+\sin \left(a-b\right)\int \cot tdt$

$=\cos \left(a-b\right)dt+\sin \left(a-b\right)\log \sin \left|t\right|+c$

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

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