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Q. $ \int\frac{dx}{ sin\left(x-a\right) sin\left(x-b\right)}$ is equal to

Integrals

Solution:

We have, $I = \int \frac{dx}{sin\left(x-a\right)sin\left(x-b\right) }$

$= \frac{1}{sin\left(b - a\right)} \int \frac{sin\left(b-a\right)}{sin\left(x-a\right)sin\left(x-b\right) }dx$

$ = \frac{1}{sin\left(b - a\right)} \int \frac{sin\left(\left(x-a\right)-\left(b-a\right)\right)}{sin\left(x-a\right)sin\left(x-b\right) }dx $

$= \frac{1}{sin\left(b - a\right)} \int \frac{sin\left(x-a\right) \cdot cos\left(x-b\right) - cos \left(x-a\right)sin\left(x-b\right)}{sin\left(x-a\right)sin\left(x-b\right) }dx$

$ = \frac{1}{sin\left(b - a\right)} \int \left(cot\left(x-b\right) -cot\left(x-a\right)\right) dx$

$ = \frac{1}{sin\left(b - a\right)} \left[log\left|sin\left(x-b\right)\right|-log\left|sin\left(x-a\right)\right|\right] + C $

$ = cosec \left(b-a\right)\left[log\left|\frac{sin\left(x-b\right)}{sin\left(x-a\right)}\right|\right] + C $