#### Explain Solution R.D Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Multiple Choice Questions Question 41 Maths Textbook Solution.

$\operatorname{cosec}(b-a) \log \left|\frac{\sin (x-b)}{\sin (x-a)}\right|$

Given:

$\int \frac{1}{\sin (x-a) \sin (x-b)} d x$

Hint:

Using  $\sin (x-y)=\sin x \cos y-\cos x \sin y \text { and } \int \frac{1}{x} d x$

Explanation:

Let $I=\int \frac{1}{\sin (x-a) \sin (x-b)} d x$

$=\frac{1}{\sin (b-a)} \int \frac{\sin (b-a)}{\sin (x-a) \sin (x-b)} d x \ldots \ldots \text { multiplying by } \sin (b-a) \text { in num.\& den. }$

$=\frac{1}{\sin (b-a)} \int \frac{\sin ((x-a)-(x-b))}{\sin (x-a) \sin (x-b)} d x$                                                $[\sin (x-y)=\sin x \cos y-\cos x \sin y]$

\begin{aligned} &=\frac{1}{\sin (b-a)} \int \frac{\sin (x-a) \cos (x-b)-\cos (x-a) \sin (x-b)}{\sin (x-a) \sin (x-b)} d x\\ &=\frac{1}{\sin (b-a)}\left[\int \frac{\cos (x-b)}{\sin (x-b)} d x-\int \frac{\cos (x-a)}{\sin (x-a)}\right] d x\\ &=\frac{1}{\sin (b-a)}[\log |\sin (x-b)|-\log |\sin (x-a)|]+C \end{aligned}

$=\operatorname{cosec}(b-a) \log \left|\frac{\sin (x-b)}{\sin (x-a)}\right|$                                                            $\left[\because \log a-\log b=\log \frac{a}{b}\right]$