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How to find following integral $$\int \csc^3 x\ dx=?$$

my work:

i substitute $\csc x =t$, $dx=-\frac{dt}{\csc x\cot x}=-\frac{dt}{t\sqrt{t^2-1}}$ $$\int t^3\frac{-dt}{t\sqrt{t^2-1}}$$

$$=-\int \frac{t^2dt}{\sqrt{t^2-1}}$$ substitute $t=\sec\theta$, $dt=\sec\theta\tan\theta\ d\theta$ $$=-\int \frac{\sec^2\theta\sec\theta\tan\theta\ d\theta}{\tan\theta}$$ $$=-\int \sec^3\theta d\theta$$

i got the similar cubic trigonometric function. I am not able to solve this integration. please help me solve it. thanks.

Arturo Magidin
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    Yet another https://math.stackexchange.com/questions/112918/integrate-csc3x-dx OR https://math.stackexchange.com/questions/1637752/how-do-you-integrate-csc3x?noredirect=1&lq=1 – Anurag A Jun 07 '20 at 23:12
  • No, that is not my question. My question is different from these questions. –  Jun 09 '20 at 02:53
  • You want to solve $\int \sec^3 x , dx$ or $\int \csc^3 x , dx$ and these questions that I have referred to give you exactly that. How's your question different, beats me? – Anurag A Jun 09 '20 at 03:42

2 Answers2

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Use integration by parts $$I=\int \csc^3 x\ dx$$ $$I=\int \csc x\ \csc^2 x\ dx$$ $$I=\csc x\int \csc^2 x\ dx-\int (-\csc x\cot x) (-\cot x)\ dx$$ $$I=-\csc x\cot x-\int \csc x\cot^2 x\ dx$$ $$I=-\csc x\cot x-\int \csc x (\csc^2 x-1)\ dx$$ $$I=-\csc x\cot x-\int (\csc^3x -\csc x)\ dx$$ $$I=-\csc x\cot x-\int \csc^3 x\ dx+\int \csc x\ dx$$ $$I=-\csc x\cot x-I+\int \csc x\ dx$$ $$2I=-\csc x\cot x+\ln\left|\tan\frac x2\right|$$ $$I=-\frac12\csc x\cot x+\frac12\ln\left|\tan\frac x2\right|+C$$

Harish Chandra Rajpoot
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  • The constant of integration $C$ must go at least since the penultimate equality. The other equalities are well-computed. – John M-D94 Jun 08 '20 at 00:10
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$$\int \dfrac{dx}{\sin^{3}x} = \int \dfrac{\sin x dx}{\sin^{4}x} = - \int \dfrac{ d \cos x}{\sin^{4}x} = - \int \dfrac{dy}{(1-y^2)^2} $$

zkutch
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