<s>[INST] Additional methods for integral reduction formula So I have successfully found a reduction formula for 
$$I_{m,n}=\int\frac{dx}{\sin^m(ax)\cos^n(ax)}$$
Went as follows:
$$\int\frac{dx}{\sin^m(ax)\cos^n(ax)}=\int\csc^m(ax)\sec^n(ax)dx=\int\csc^m(ax)\sec^{n-2}(ax)\sec^2(ax)dx\\
\begin{vmatrix}u=\csc^m(ax)\sec^{n-2}(ax)\\du=-am\csc^m(ax)\cot(ax)\sec^{n-2}(ax)+a(n-2)\csc^m(ax)\sec^{n-2}(ax)\tan(ax)dx\end{vmatrix}\\
\begin{vmatrix}dv=\sec^2(ax)\quad v=\frac{1}{a}\tan(ax)\end{vmatrix}\\
\int udv=uv-\int vdu\\
=\frac{1}{a}\csc^m(ax)\sec^{n-2}(ax)\tan(ax)-\int(n-2)\csc^m(ax)\sec^{n-2}(ax)\tan^2(ax)\\
-m\csc^m(ax)\sec^{n-2}(ax)dx\\
=\frac{1}{a}\csc^m(ax)\sec^{n-2}(ax)\tan(ax)-\int(n-2)\csc^m(ax)\sec^n(ax)\\
-(n-2)\csc^m(ax)\sec^{n-2}(ax)-m\csc^m(ax)\sec^{n-2}(ax)dx\\
=\frac{1}{a}\csc^m(ax)\sec^{n-2}(ax)\tan(ax)-(n-2)\int\csc^m(ax)\sec^n(ax)dx\\
+(m+n-2)\int\csc^m(ax)\sec^{n-2}(ax)dx\\
\int\frac{dx}{\sin^m(ax)\cos^n(ax)}=\frac{1}{a}\csc^m(ax)\sec^{n-2}(ax)\tan(ax)-(n-2)I_{m,n}+(m+n-2)I_{m,n-2}\\
\int\frac{dx}{\sin^m(ax)\cos^n(ax)}=\frac{\csc^m(ax)\sec^{n-2}(ax)\tan(ax)}{a(n-1)}+\left(\frac{m+n-2}{n-1}\right)I_{m,n-2}\\
\int\frac{dx}{\sin^m(ax)\cos^n(ax)}=\frac{\csc^{m-1}(ax)\sec^{n-1}(ax)}{a(n-1)}+\left(\frac{m+n-2}{n-1}\right)I_{m,n-2}\\
\int\frac{dx}{\sin^m(ax)\cos^n(ax)}=\frac{1}{a(n-1)\sin^{m-1}(ax)\cos^{n-1}(ax)}+\left(\frac{m+n-2}{n-1}\right)I_{m,n-2}$$
This is definitely not a check my proof question, but I was wondering if there are more elegant or creative ways of finding this reduction formula, or if IBP, u-sub, and identities are the only ways of going with formulas like these. 
 [/INST] The idea of this trick is similar to IBP but this ruduces a lot in calculations (because differentiation is "easier" than integration). 
Take $f_{m,n}(x)= \csc^m(ax)\sec^n(ax)$. Take the derivative with respect to $x$:
\begin{align*}
f'_{m-1,n-1}&=a(n-1)\underbrace{(1-\cos^2(ax))f_{m,n}}_{f_{m-2,n}}-a(m-1)f_{m,n-2}\\
f'_{m-1,n-1}&=a(n-1)f_{m,n}-a(m+n-2)f_{m,n-2}.\\
\end{align*}
Integrate on both sides, we arrive the reduction formula

\begin{align*}
f_{m-1, n-1}=a(n-1) \color{blue}{I_{m,n}}-a(m+n-2) I_{m, n-2},
\end{align*}

equivalent to the last line of your question.
Similarly, we can find

$$f_{m+1, n-1}=a(n-1) \color{blue}{I_{m,n}}-a(m+1) I_{m+2, n-2}.$$

</s>