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

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

Thanks !

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marked as duplicate by Fabian, froggie, Davide Giraudo, Dan Brumleve, Did Jan 7 '13 at 12:52

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

If this is homework, you should add the homework tag... – Fabian Jan 7 '13 at 9:20
Wolfram Alpha – Calvin Lin Jan 7 '13 at 9:24

2 Answers 2

up vote 4 down vote accepted


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?

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Indeed. Thanks ! – theMissingIngredient Jan 7 '13 at 9:30
@theMissingIngredient: if the answer did help you, don't forget to accept the answer (and potentially upvote it). Do so also for the other questions which you have asked... – Fabian Jan 7 '13 at 9:31
Its the same integral as $$\int{csc^3(x)dx} $$ easy integration by parts – KGTW Nov 27 '13 at 21:51

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.

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