TITLE: Evaluate $\int\frac{dx}{\sin(x+a)\sin(x+b)}$
QUESTION [5 upvotes]: Please help me evaluate:
$$
\int\frac{dx}{\sin(x+a)\sin(x+b)}
$$

REPLY [6 votes]: Hint: Multiply and divide by $\sin(b-a)$. 
Further Hint:

 Write $\sin(b-a) = \sin((x-a)-(x-b))$

REPLY [4 votes]: The given integral is: $$\int\frac{dx}{\sin(x+a)\sin(x+b)}$$
The given integral can write:
$$\int\frac{dx}{\sin(x+a)\sin(x+b)}=\int\frac{\sin(x+a)}{\sin(x+b)}\cdot\frac{dx}{\sin^2(x+a)}$$
We substition 
$$\frac{\sin(x+a)}{\sin(x+b)}=t$$
By the substition of the above have:
$$\frac{dx}{\sin^2(x+a)}=\frac{dt}{\sin(a-b)}$$
Now have:
$$\int\frac{dx}{\sin(x+a)\sin(x+b)}=\int\frac{\sin(x+a)}{\sin(x+b)}\cdot\frac{dx}{\sin^2(x+a)}=\frac{1}{\sin(a-b)}\int\frac{dt}{t}=\frac{1}{\sin(a-b)}\ln |t|=\frac{1}{\sin(a-b)}\ln|\frac{\sin(x+a)}{\sin(x+b)}|+C$$