If $\mathrm{I}=\int \frac{\mathrm{d} x}{\sin (x-\mathrm{a}) \sin (x-\mathrm{b})}$, then $\mathrm{I}$ is…

If $\mathrm{I}=\int \frac{\mathrm{d} x}{\sin (x-\mathrm{a}) \sin (x-\mathrm{b})}$, then $\mathrm{I}$ is given by
  1. $\frac{1}{\sin (\mathrm{b}-\mathrm{a})} \log |\sin (x-\mathrm{a}) \sin (x-\mathrm{b})|+\mathrm{c}$ where $\mathrm{c}$ is a constant of integration.
  2. $\log \left|\frac{\sin (x-a)}{\sin (x-b)}\right|+c$, where $\mathrm{c}$ is a cónstant of integration.
  3. $\frac{1}{\sin (\mathrm{b}-\mathrm{a})} \log \left|\frac{\sin (x-\mathrm{a})}{\sin (x-\mathrm{b})}\right|+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
  4. $\frac{1}{\sin (\mathrm{b}-\mathrm{a})} \log \left|\frac{\sin (x-\mathrm{b})}{\sin (x-\mathrm{a})}\right|+\mathrm{c}$, where $\mathrm{c}$ is $\mathrm{a}$ constant of integration.

Solution

$\begin{aligned} & \mathrm{I}=\int \frac{\mathrm{d} x}{\sin (x-\mathrm{a}) \sin (x-\mathrm{b})} \\ & =\frac{1}{\sin (\mathrm{b}-\mathrm{a})} \int \frac{\sin \{(x-\mathrm{a})-(x-\mathrm{b})\}}{\sin (x-\mathrm{a}) \sin (x-\mathrm{b})} \mathrm{d} x \\ & =\frac{1}{\sin (\mathrm{b}-\mathrm{a})} \int \frac{1}{\sin (x-\mathrm{a}) \sin (x-\mathrm{b})}[\sin (x-\mathrm{a}) \cos (x-\mathrm{b}) \\ & =\frac{1}{\sin (\mathrm{b}-\mathrm{a})} \int[\cot (x-\mathrm{b})-\cot (x-\mathrm{a})] \mathrm{d} x \\ & =\frac{1}{\sin (\mathrm{b}-\mathrm{a})}[\log |\sin (x-\mathrm{b})|-\log |\sin (x-\mathrm{a})|]+\mathrm{c} \\ & =\frac{1}{\sin (\mathrm{b}-\mathrm{a})} \log \left|\frac{\sin (x-\mathrm{b})}{\sin (x-\mathrm{a}) \mid}\right|+\mathrm{c}\end{aligned}$

Asked in: MHT CET 2023 (14 May Shift 2)

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