How to take the integral $\int\frac{dx}{\sin^3x}$ 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.
 A: 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 $\;\dfrac{-1}{(1-u^2)^2}$.
A: 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 $$t=\sin(x)$$
A: 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 $n$.
A: $$
\begin{aligned}
\int \frac{1}{\sin^3x}dx 
& = \int \frac{\left(t^2+1\right)^2}{4t^3}dt 
\\& =\frac{1}{4}\int \:\frac{1}{t^3}+t+\frac{2}{t}dt
\\& = \color{red}{\frac{1}{4}\left(\frac{1}{2}\tan ^2\left(\frac{x}{2}\right)-\frac{1}{2}\cot ^2\left(\frac{x}{2}\right)+2\ln \left|\tan \left(\frac{x}{2}\right)\right|\right)+C}
\end{aligned}
$$
Solved by substitution $\color{blue}{t=\tan \left(\frac{x}{2}\right)}$
