Question

$\int \frac{1}{\sin \left(x-a\right)\sin \left(x-b\right)}dx$

Answer 1

$$\begin{gathered} \int \frac{1}{\sin \left(x-a\right)·\sin \left(x-b\right)}dx \\ =\frac{1}{\sin \left(b-a\right)}\int \frac{\sin \left(b-a\right)}{\sin \left(x-a\right)·\sin \left(x-b\right)}dx \\ =\frac{1}{\sin \left(b-a\right)}\int \frac{\sin \left[\left(x-a\right)-\left(x-b\right)\right]}{\sin \left(x-a\right)·\sin \left(x-b\right)}dx \\ =\frac{1}{\sin \left(b-a\right)}\int \frac{\sin \left(x-a\right)·\cos \left(x-b\right)-\cos \left(x-a\right)\sin \left(x-b\right)}{\sin \left(x-a\right)·\sin \left(x-b\right)}dx \\ =\frac{1}{\sin \left(b-a\right)}\int \left[\frac{\sin \left(x-a\right)·\cos \left(x-b\right)}{\sin \left(x-a\right)·\sin \left(x-b\right)}-\frac{\cos \left(x-a\right)\sin \left(x-b\right)}{\sin \left(x-a\right)\sin \left(x-b\right)}\right]dx \\ =\frac{1}{\sin \left(b-a\right)}\int \left[\cot \left(x-b\right)-\cot \left(x-a\right)\right]dx \\ =\frac{1}{\sin \left(b-a\right)}\int \cot \left(x-b\right)dx-\int \cot \left(x-a\right)dx \\ =\frac{1}{\sin \left(b-a\right)}\left[\ln \left|\sin \left(x-b\right)\right|-\ln \left|\sin \left(x-a\right)\right|\right]+C \\ =\frac{1}{\sin \left(b-a\right)}\left[\ln \left|\frac{\sin \left(x-b\right)}{\sin \left(x-a\right)}\right|\right]+C \\ =\frac{-1}{\sin \left(a-b\right)}\left[\ln \left|\frac{\sin \left(x-b\right)}{\sin \left(x-a\right)}\right|\right]+C \\ =\frac{1}{\sin \left(a-b\right)}\ln \left|\frac{\sin \left(x-a\right)}{\sin \left(x-b\right)}\right|+C \end{gathered}$$