\begin{equation*} \int \frac{\cos ^{m}x}{\sin ^{n}x}dx=-\frac{\cos ^{m+1}x}{(n-1)\sin ^{n-1}x}- \frac{m-n+2}{n-1}\int \frac{\cos ^{m}x}{\sin ^{n-2}x}dx+C,\qquad (n\neq 1). \end{equation*}
I`d like to know if someone could help me with this exercise. Maybe there´s a hint I don´t see right now.
On
I guess the reduction formula should be like this: $$ \begin{aligned} I(m, n) :&=\int \frac{\cos ^{m} x}{\sin ^{n} x} d x \\&=-\frac{1}{n-1} \int \cos ^{m-1} x d\left(\frac{1}{\sin ^{n-1} x}\right) \\ &=-\frac{1}{n-1}\left(\frac{\cos ^{m-1} x}{\sin ^{n-1} x}+\int \frac{(m-1) \cos ^{m-2} x}{\sin ^{n-2} x} d x\right) \\ &=-\frac{\cos^{m-1} x}{(n-1) \sin ^{n-1}x}-\frac{m-1}{n-1} I(m-2, n-2) \end{aligned} $$
Since you're just asking for a hint I'll get you started.
You will need to use integration by parts, as is the case for most "reduction formulas". In this case, note that $$\int{\frac{\cos x}{\sin^n x}}\,\mathrm{d}x=\frac{-1}{(n-1)\sin^{n-1}x}$$ So integrate the term $$\frac{\cos x}{\sin^n x}$$ and differentiate the term $$\cos^{m-1}x$$ with integration by parts. (It looks like this will work, but I didn't actually check it myself so let me know if you have trouble).