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$\begingroup$

There's

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

I tried to write it like

$$\int\frac{(\sin^2x+\cos^2x)}{\sin^3x}\,\mathrm dx,$$

and then made partial fractions from it, but it didn't help much, the answer is still incorrect.

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    Multiply top and bottom by sin, and use that identithe on the denominator. – 2017-02-21
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    The incorrect answer you have obtained is? – 2017-02-21
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    it is csc^3 which is do-able by integration by parts I think - look for integration of sec^3 by parts (for clues) – 2017-02-21
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    Write it as $$\int\frac{\sin x dx}{(1-\cos^2 x)\sin^2 x}$$ – 2017-02-21
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    here is a good article on it's cousin sec https://en.wikipedia.org/wiki/Integral_of_secant_cubed#Integration_by_parts – 2017-02-21

4 Answers 4

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Hint:

Bioche's rules lead to to make the substitution $u=\cos x$, $\;\mathrm d u=\sin x \,\mathrm d x$. You obtain $$\int\frac{\mathrm d x}{\sin^3 x}=\int\frac{-\mathrm d u}{\sin^4 x}=\int\frac{-\mathrm d u}{(1-u^2)^2}.$$ There remains to compute the partial fraction decomposition of $\;\dfrac{-1}{(1-u^2)^2}$.

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write your integrand as $$\frac{1}{\sin(x)}+\frac{\cos(x)^2}{\sin(x)^3}$$ and the first as $$\sin(x)+\frac{\cos(x)^2}{\sin(x)}$$ now you can set $$t=\sin(x)$$

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Let $x=u+\frac\pi2$ to get

$$\int\frac{dx}{\sin^3(x)}=\int\frac{du}{\cos^3(u)}=\int\sec^3(u)\ du$$

This is a well known integral. One technique is to apply integration by parts:

$$\int\sec(u)\ du=\sec(u)\tan(u)+\int\sec(u)\ du-\int\sec^3(u)\ du\\=\sec(u)\tan(u)+\ln|\sec(u)+\tan(u)|-\int\sec^3(u)\ du$$

In general, $\int\sec^n(x)\ dx$ should be handled with integration by parts and reduction formulas for odd $n$ and Pythagorean identities for even $n$.

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$$ \begin{aligned} \int \frac{1}{\sin^3x}dx & = \int \frac{\left(t^2+1\right)^2}{4t^3}dt \\& =\frac{1}{4}\int \:\frac{1}{t^3}+t+\frac{2}{t}dt \\& = \color{red}{\frac{1}{4}\left(\frac{1}{2}\tan ^2\left(\frac{x}{2}\right)-\frac{1}{2}\cot ^2\left(\frac{x}{2}\right)+2\ln \left|\tan \left(\frac{x}{2}\right)\right|\right)+C} \end{aligned} $$ Solved by substitution $\color{blue}{t=\tan \left(\frac{x}{2}\right)}$