In one of my exercises about integration we had to solve the following integral:

\begin{equation} \int \frac{\sin^n(x)}{\cos^m(x)}dx \end{equation}

We had to do this via a recursive integral. I found:

\begin{equation} \mathcal{K_{m,n}} = \frac{\sin^{n-1}(x)}{(m-1)\cdot\cos^{m-1}(x)}-\frac{n-1}{m-1}\cdot\mathcal{K}_{m-2,n-2}, \qquad n,m\geq2 \end{equation}

I know for a fact that this solution is correct because we solved this in class, but the other cases where m and/or n are not $\geq$ 2, were left as an exercise for us at home. I've found a solution for every case except for the case where $m = 1$, which makes the following integral.

\begin{equation} \int \frac{\sin^n(x)}{\cos(x)}dx \end{equation}

I've tried different things, I tried integration by parts with many different u's and v's but none of them seem to work out. I tried for example:

  • $u = \frac{\sin^{n-1}(x)}{\cos(x)}$, $v = \sin(x)$
  • $u = \tan(x)$, $v = \sin^{n-1}(x)$
  • $u = \sin^{n-1}(x)$, $v = \tan(x)$

Would anyone know how to solve this integral? I'm not looking for a complete solution but rather for a method that should work so i can find the solution by myself from that point on.


Denote $I_n=\displaystyle\int \frac{\sin^n(x)}{\cos(x)}~\mathrm dx$. We have \begin{align*} I_{n}-I_{n+2}&=\int \frac{\sin^n(x)(1-\sin^2(x))}{\cos(x)}~\mathrm dx\\ &=\int \sin^n(x)\cos(x)~\mathrm dx, \end{align*} which is the reduction formula for $I_n$.


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