Question

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

• A. $sin\left(b-a\right)log\left|\frac{sin\left(x-b\right)}{sin\left(x-a\right)}\right|+C$
• B. $cosec\left(b-a\right)log\left|\frac{sin\left(x-a\right)}{sin\left(x-b\right)}\right|+C$
• C. $cosec\left(b-a\right)log\left|\frac{sin\left(x-b\right)}{sin\left(x-a\right)}\right|+C$
• D. $sin\left(b-a\right)log\left|\frac{sin\left(x-a\right)}{sin\left(x-b\right)}\right|+C$

Answer 1

Answer: (C)

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

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

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

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

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

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

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