int frac{dx}{sin(x - a)sin(x - b)} is | vedclass

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(A) Let $I = \int \frac{dx}{\sin(x - a)\sin(x - b)}$.Multiply and divide by $\sin(a - b)$:$I = \frac{1}{\sin(a - b)} \int \frac{\sin((x - b) - (x - a)

Vedclass · Accepted Answer

$\frac{1}{\sin(a - b)}\log \left| \frac{\sin(x - a)}{\sin(x - b)} \right| + c$

Vedclass · Answer

(A) Let $I = \int \frac{dx}{\sin(x - a)\sin(x - b)}$.Multiply and divide by $\sin(a - b)$:$I = \frac{1}{\sin(a - b)} \int \frac{\sin((x - b) - (x - a))}{\sin(x - a)\sin(x - b)} dx$.Using the identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$:$I = \frac{1}{\sin(a - b)} \int \frac{\sin(x - b)\cos(x - a) - \cos(x - b)\sin(x - a)}{\sin(x - a)\sin(x - b)} dx$.$I = \frac{1}{\sin(a - b)} \left[ \int \frac{\cos(x - a)}{\sin(x - a)} dx - \int \frac{\cos(x - b)}{\sin(x - b)} dx \right]$.$I = \frac{1}{\sin(a - b)} [ \log |\sin(x - a)| - \log |\sin(x - b)| ] + c$.$I = \frac{1}{\sin(a - b)} \log \left| \frac{\sin(x - a)}{\sin(x - b)} \right| + c$.

$\int \frac{dx}{\sin(x - a)\sin(x - b)}$ is

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