Integral $ \int \frac{\operatorname d\!x}{\sin^3 x} $

Question

Possible Duplicate:
Integral $\int \frac {1}{\sin^3(x)} dx$

Can someone help me compute $$\int \frac {1}{\sin^3 x } dx $$

Thanks !

Answer 1

Hint:

employ the Weierstraß substitution $t=\tan (x/2)$ to bring the integral into a form of a rational function. Note that $\sin x= 2t/(1+t^2)$, $dx = 2 dt/(1+t^2)$, so $$ \int \frac{dx}{\sin^3 x} = \int \frac{(1+t^2)^2}{4 t^3} dt .$$

You can reduce the order by the substitution $u=t^2 =\tan^2 (x/2) =(1-\cos x)/(1+\cos x)$, which yields $$ \int \frac{dx}{\sin^3 x} = \int \frac{(1+u)^2}{8u^2} du = \int \frac{du}{8} + \int\frac{du}{4u} + \int\frac{du}{8 u^2} .$$

Can you take it from there?

Answer 2

Here is another way $$\int\frac{dx}{\sin^3 x}=\int\frac{\sin x dx}{\sin^4 x}=-\int\frac{d(\cos x)}{(1-\cos^2 x)^2}=-\int\frac{dz}{(1+z)^2 (1-z)^2}$$ Now this can be calculated using method of partial fractions.