Question

Solve:

$\int \frac{1}{sin(x-a)sin(x-b)}dx.$

Answer 1

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

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

using $\left[sin\right(A-B)=sinAcosB-cosAsinB]$

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

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

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

$\frac{1}{sin(b-a)}[ln\left|sin\right(x-b\left)\right|-ln\left|sin\right(x-a\left)\right|]+C$ $[\because \int cotx=ln\left|sinx\right|]$

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