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

$\int \operatorname{cosec}(x-a) \operatorname{cosec} x d x=$
  1. $\operatorname{cosec} a \cdot \log [\sin (x-a) \operatorname{cosec} x]+c$
  2. $\operatorname{cosec} a \log [\sin (x-a) \sin x]+c$
  3. $\sin a \log [\sin (\mathrm{x}-\mathrm{a}) \sin \mathrm{x}]+\mathrm{c}$
  4. $\operatorname{cosec} \mathrm{a} \cdot \log [\operatorname{cosec}(\mathrm{x}-\mathrm{a}) \sin \mathrm{x}]+\mathrm{c}$

Solution

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

Asked in: MHT CET 2021 (20 Sep Shift 2)

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