3
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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 $I_n$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.