int operatorname{cosec}(x-a) operatorname{cos | vedclass

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Question

(A) Let $I = \int \operatorname{cosec}(x-a) \operatorname{cosec} x \, dx = \int \frac{dx}{\sin(x-a) \sin x}$.Multiply and divide by $\sin a$:$I = \fra

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $I = \int \operatorname{cosec}(x-a) \operatorname{cosec} x \, dx = \int \frac{dx}{\sin(x-a) \sin x}$.Multiply and divide by $\sin a$:$I = \frac{1}{\sin a} \int \frac{\sin a}{\sin(x-a) \sin x} \, dx$.Since $a = x - (x-a)$,we have:$I = \frac{1}{\sin a} \int \frac{\sin(x - (x-a))}{\sin(x-a) \sin x} \, dx$.Using $\sin(A-B) = \sin A \cos B - \cos A \sin B$:$I = \frac{1}{\sin a} \int \frac{\sin x \cos(x-a) - \cos x \sin(x-a)}{\sin(x-a) \sin x} \, dx$.$I = \frac{1}{\sin a} \left( \int \frac{\cos(x-a)}{\sin(x-a)} \, dx - \int \frac{\cos x}{\sin x} \, dx \right)$.$I = \frac{1}{\sin a} (\log |\sin(x-a)| - \log |\sin x|) + c$.$I = \operatorname{cosec} a \cdot \log \left| \frac{\sin(x-a)}{\sin x} \right| + c$.

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

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