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

Question

Please help me evaluate: $$ \int\frac{dx}{\sin(x+a)\sin(x+b)} $$

Answer 1

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:

$\int\frac{dx}{\sin(x+a)\sin(x+b)}$=$\int\frac{\sin(x+a)}{\sin(x+b)}\cdot\frac{dx}{\sin^2(x+a)}$=$\frac{1}{\sin(a-b)}\int\frac{dt}{t}$=$\frac{1}{\sin(a-b)}\ln |t|$=$\frac{1}{\sin(a-b)}\ln|\frac{\sin(x+a)}{\sin(x+b)}|+C$

Answer 2

Hint: Multiply and divide by $\sin(b-a)$.

Further Hint:

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