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?
 A: 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}
A: If you're familiar with cosecants, you can derive a reduction formula for
$$I_n=\int \frac{1}{\sin^{n}(x)}\mathrm{d}x=\int \csc^{n}(x)\mathrm{d}x$$
We only use integration by parts and the identity $1+\cot^2(x)=\csc^2(x)$.
$$\begin{align}
I_n&=\csc^{n-2}(x)(-\cot(x))-\int (n-2)\csc^{n-3}(x)(-\csc(x)\cot(x))(-\cot(x))\mathrm{d}x\\
&=-\csc^{n-2}(x)\cot(x)-(n-2)\int \csc^{n-2}(x)\cot^2(x)\mathrm{d}x \\
&=-\csc^{n-2}(x)\cot(x)-(n-2)\int \csc^{n-2}(x)(\csc^2(x)-1)\mathrm{d}x \\
&=-\csc^{n-2}(x)\cot(x)-(n-2)\int\csc^{n}(x)\mathrm{d}x+(n-2)\int\csc^{n-2}(x)\mathrm{d}x
\end{align}$$
which gives you the formula
$$I_n=-\frac{1}{n-1}\csc^{n-2}(x)\cot(x)+\frac{n-2}{n-1}I_{n-2} $$
In your case, apply it for $n=5$ to obtain an expression in terms of $I_3$, and then apply it again for $n=3$. The final step is to use (proved by direct differentiation)
$$I_1=\int \csc(x)\mathrm{d}x=\ln|\csc(x)-\cot(x)|+C $$ 
