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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.

Thank you in advance

  • 0
    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* $B \implies A$.2012-11-07

2 Answers 2