<s>[INST] Finding the $\int \frac{dx}{\sin^5x}$ $\int \frac{dx}{\sin^5x}$
I did $t= \cos(x)$, $-\frac{dt}{\sin(x)}=dx$:
$$\int \frac{dx}{\sin^5x} = \int \frac{dt}{\sin^6x} = -\int \frac{dt}{(1-t^2)^3}=-\int (1-t^2)^{-3} = \frac{(1-t^2)^{-3}}{-3} = -\frac{1}{3(1-\cos^2x)}+C$$
Is this correct?
 [/INST] With $t=\cos x$
\begin{align}
\int \frac{1}{\sin^5x} dx= &-\int \frac{1}{(1-t^2)^3}dt
=-\int \frac1{4t^3}\ d\left( \frac{t^4}{(1-t^2)^2}\right)\\
\overset{ibp}= & -\frac t{4(1-t^2)^2}-\frac38\int \frac1{t}\ d\left( \frac{t^2}{1-t^2}\right)\\
\overset{ibp}= & - \frac t{4(1-t^2)^2}-\frac38 \left(
\frac t{1-t^2}+\tanh^{-1}t \right)+C\\
= & -\frac {\cos x}{4\sin^4 x}-
\frac {3\cos x}{8\sin^2x}-\frac38\tanh^{-1}\cos x+C\\
\end{align}
</s>