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

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Question

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

Vedclass · Accepted Answer

$\frac{1}{\sin a} \log |\sin (x-a) \operatorname{cosec} x|+c$

Vedclass · Answer

(C) Let $I = \int \operatorname{cosec}(x-a) \operatorname{cosec} x \, dx$. Multiply and divide by $\sin a$: $I = \int \frac{\sin a}{\sin a \sin(x-a) \sin x} \, dx$. Since $\sin a = \sin(x - (x-a))$,we have: $I = \frac{1}{\sin a} \int \frac{\sin(x - (x-a))}{\sin(x-a) \sin x} \, dx$. Using the identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$: $I = \frac{1}{\sin a} \int \frac{\sin x \cos(x-a) - \cos x \sin(x-a)}{\sin(x-a) \sin x} \, dx$. $I = \frac{1}{\sin a} \int \left( \frac{\cos(x-a)}{\sin(x-a)} - \frac{\cos x}{\sin x} \right) \, dx$. $I = \frac{1}{\sin a} \int (\cot(x-a) - \cot x) \, dx$. Integrating,we get: $I = \frac{1}{\sin a} [\log |\sin(x-a)| - \log |\sin x|] + c$. $I = \frac{1}{\sin a} \log \left| \frac{\sin(x-a)}{\sin x} \right| + c$. This can be written as $\frac{1}{\sin a} \log |\sin(x-a) \operatorname{cosec} x| + c$.

$\int \operatorname{cosec}(x-a) \operatorname{cosec} x \, dx$ is equal to

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