$\quad \int \operatorname{cosec}(x-a) \cdot \operatorname{cosec} x d x=$

$\quad \int \operatorname{cosec}(x-a) \cdot \operatorname{cosec} x d x=$
  1. $\frac{-1}{\sin \mathrm{a}} \log (\sin (x-\mathrm{a}) \sin x)+\mathrm{c}$, where c is a constant of integration.
  2. $\frac{1}{\sin \mathrm{a}} \log (\sin (x-\mathrm{a}) \sin x)+\mathrm{c}$, where c is a constant of integration.
  3. $\frac{1}{\sin \mathrm{a}} \log (\sin (x-\mathrm{a}) \cdot \operatorname{cosec} x)+\mathrm{c}$, where c is a constant of integration.
  4. $\frac{-1}{\sin \mathrm{a}} \log (\operatorname{cosec}(x-\mathrm{a}) \cdot \sin x)+\mathrm{c}$, where c is a constant of integration.

Solution

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

Asked in: MHT CET 2024 (15 May Shift 2)

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