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

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Question

(C) Let $I = \int \operatorname{cosec}(x-a) \cdot \operatorname{cosec} x \, dx$. We can write this as $I = \int \frac{1}{\sin (x-a) \sin x} \, dx$. Mu

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $I = \int \operatorname{cosec}(x-a) \cdot \operatorname{cosec} x \, dx$. We can write this as $I = \int \frac{1}{\sin (x-a) \sin x} \, dx$. Multiply and divide by $\sin a$: $I = \frac{1}{\sin a} \int \frac{\sin a}{\sin (x-a) \sin x} \, dx$. Express $a$ as $x - (x-a)$: $I = \frac{1}{\sin a} \int \frac{\sin (x - (x-a))}{\sin (x-a) \sin x} \, dx$. Using the formula $\sin(A-B) = \sin A \cos B - \cos A \sin B$: $I = \frac{1}{\sin a} \int \frac{\sin x \cos (x-a) - \cos x \sin (x-a)}{\sin (x-a) \sin x} \, dx$. $I = \frac{1}{\sin a} \int \left( \frac{\sin x \cos (x-a)}{\sin (x-a) \sin x} - \frac{\cos x \sin (x-a)}{\sin (x-a) \sin x} \right) \, dx$. $I = \frac{1}{\sin a} \int (\cot (x-a) - \cot x) \, dx$. Integrating,we get: $I = \frac{1}{\sin a} [\log |\sin (x-a)| - \log |\sin x|] + c$. Using the property $\log m - \log n = \log(\frac{m}{n})$: $I = \frac{1}{\sin a} \log \left| \frac{\sin (x-a)}{\sin x} \right| + c$. Since $\frac{1}{\sin x} = \operatorname{cosec} x$,this is equivalent to: $I = \frac{1}{\sin a} \log |\sin (x-a) \cdot \operatorname{cosec} x| + c$.

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

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