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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

pie
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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 07 '12 at 19:04
  • See https://math.stackexchange.com/questions/3710279/how-to-solve-int-csc3-x-dx – Travis Willse Apr 13 '24 at 05:16
  • I believe https://math.stackexchange.com/questions/112918/integral-int-csc3x-dx is a better duplicate target. – User Apr 13 '24 at 10:54

3 Answers3

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Yes. Your simplification is correct.

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.

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In the first integral you can substitute u=cos x, this will reduce the problem to the problem of finding the integral $1/(1-u^2)^2$

Amr
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Yes, the simplified integral that you got is correct.

Now, we are left with the integral $$I=-\int\frac{dv}{(1-v^2)^2}=\int\frac{-2vdv}{2v(1-v^2)^2}$$

Using integration by parts (IBP),

$$I=\frac{-1}{2v(1-v^2)}-\frac{1}{2}\int\frac{dv}{v^2(1-v^2)}$$ Let the latter integral be $J$. Then $$J=\int\frac{dv}{v^2(1-v^2)}=\int\frac{dv}{v^2}+\int\frac{dv}{1-v^2}$$ $$J=\frac{-1}{v}+\frac{1}{2}\ln\bigg(\frac{1+v}{1-v}\bigg)$$

Therefore, the original integral $\int\frac{1}{\sin^3x}dx$ $$=\frac{-1}{2\cos x(1-\sin^2x)}-\frac{1}{2}\bigg(\frac{-1}{\cos x}+\frac{1}{2}\ln\bigg(\frac{1+\cos x}{1-\cos x}\bigg)\bigg) + C$$ On simplifying by using trigonometric identities, $$\int\frac{dx}{\sin^3x}=\frac{-1}{2}\bigg(\cot x\csc x+\ln{(|\csc x+\cot x|)}\bigg)+C$$

Sparsh Gupta
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