Question

$$\begin{gathered} \int \frac{1}{\sin (x-a)\sin (x-b)}dxisequalto \\ \left(\mathrm{a}\right)\sin \left(b-a\right)\log \left|\frac{\sin (x-a)}{\sin (x-b)}\right|+C\left(\mathrm{b}\right)\cos ec\left(b-a\right)\log \left|\frac{\sin (x-a)}{\sin (x-b)}\right|+C \\ \left(\mathrm{c}\right)\cos ec\left(b-a\right)\log \left|\frac{\sin (x-b)}{\sin (x-a)}\right|+C\left(\mathrm{d}\right)\sin \left(b-a\right)\log \left|\frac{\sin (x-a)}{\sin (x-b)}\right|+C \end{gathered}$$

Answer 1

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

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

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

$\Rightarrow I=\frac{1}{\sin \left(b-a\right)}\int \frac{\sin \left[\left(x-a\right)-\left(x-b\right)\right]}{\sin \left(x-a\right)\sin \left(x-b\right)}dx$

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

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

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

$\Rightarrow I=\frac{1}{\sin \left(b-a\right)}\log \left|\sin \left(x-b\right)\right|-\frac{1}{\sin \left(b-a\right)}\log \left|\sin \left(x-a\right)\right|+C$

$\Rightarrow I=\frac{1}{\sin \left(b-a\right)}\left(\log \left|\sin \left(x-b\right)\right|-\log \left|\sin \left(x-a\right)\right|\right)+C$

$\Rightarrow I=\mathrm{cosec}\left(b-a\right)\log \left|\frac{\sin \left(x-b\right)}{\sin \left(x-a\right)}\right|+C$

Hence, the correct answer is option (c).