# Calculating the integral

I need to calculate the following integral:

$\int \frac{dx}{\sin^3x}$

I noticed that $\int \frac{dx}{\sin^3x}=\int \frac{dx}{\sin x \sin^2x}=-\int \frac{dx}{-\sin x (1-\cos^2x)} (A)$

Let $v=\cos u \Leftrightarrow dv=-\sin u du$

Therefore: $(A)= -\int \frac{dv}{(1-v^2)^2}$ Is that correct ?

How do I calculate then the final integral.

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Your use of $\iff$ is incorrect. If $dv = -\sin u \ du$ it does not necessarily follow that $v = \cos u$. For example, if $v = \cos u + 1$ then $dv = -\sin u \, du.$ The correct statement is that $v = \cos u \implies dv = -\sin u \, du$. Remember that $A \iff B$ means that both $A \implies B$ and $B \implies A$. – Fly by Night Nov 7 '12 at 19:04

Once you have your integrand in the form $\dfrac1{(1-v^2)^2}$. make use of partial fractions to rewrite the integrand as $$\dfrac1{(1-v^2)^2}= \dfrac14 \left( \dfrac1{1+v} + \dfrac1{(1+v)^2} + \dfrac1{1-v} + \dfrac1{(1-v)^2}\right)$$ Now you should be able to integrate it out easily.
In the first integral you can substitute u=cos x, this will reduce the problem to the problem of finding the integral $1/(1-u^2)^2$