I have found the reduction formula for the integral of $\sin^nx$
$$\int \sin^{n}x\, dx = -\frac{1}{n}\sin^{n-1}x \cos x + \frac{n-1}{n}\int \sin^{n-2}x \, dx$$
Using this, I am tasked to find the integral $\int \csc^{5}x\,dx$.
What I have now is simply to convert $\csc^{5}x$ to $\sin^{-5}x$ and substitute $n = -5$ in the reduction formula. However, I found that the last integral will simply go into loops of power $-7, -9,\dotsc$ with no end.
Does anyone have tips on how I can solve something like this?
Use the reduction formula
$\int \csc ^{m}(x) d x=-\frac{\cos (x) \csc ^{m-1}(x)}{m-1}+\frac{m-2}{m-1} \int \csc ^{-2+m}(x) d x$
When m=5, you get
$-\frac{1}{4} \cot (x) \csc ^{3}(x)+\frac{3}{4} \int \csc ^{3}(x) d x$
using the formula for m=3 to carry out the integral, you get
$-\frac{3}{8} \cot (x) \csc (x)-\frac{1}{4} \cot (x) \csc ^{3}(x)+\frac{3}{8} \int \csc (x) d x$
\begin{aligned} &\text { Using the fact that integral of } \csc (x) \text { is }-\log (\cot (x)+\csc (x))\\ &-\frac{1}{4} \cot (x) \csc ^{3}(x)-\frac{3}{8} \cot (x) \csc (x)-\frac{3}{8} \log (\cot (x)+\csc (x))+\text { constant } \end{aligned}
Which can be simplified to
$$-\frac{1}{64} \csc ^4\left(\frac{x}{2}\right)-\frac{3}{32} \csc ^2\left(\frac{x}{2}\right)+\frac{1}{64} \sec ^4\left(\frac{x}{2}\right)+\frac{3}{32} \sec ^2\left(\frac{x}{2}\right)+\frac{3}{8} \log \left(\sin \left(\frac{x}{2}\right)\right)-\frac{3}{8} \log \left(\cos \left(\frac{x}{2}\right)\right)+constant$$