How would I integrate the following:
.$$\int\frac{\sin^3x}{\cos x}\,\mathrm dx.$$
I am not sure what to do. I could split $\sin^3x=(1-\cos^2x)(\sin x)$
Then get $\int \tan(x)(1-\cos^2x)$
But would this be the right way to go.
[Math] Integrating $\int\frac{\sin^3x}{\cos x}\,\mathrm dx.$
calculusindefinite-integralsintegration
Best Answer
Let $u = \cos x \implies du = -\sin x\,dx$
$$\begin{align} \int \dfrac{\sin^3x}{\cos x} \,dx & = -\int \dfrac{-\sin x\sin^2 x}{\cos x}\,dx \\ \\ & = -\int \dfrac{-\sin x(1 - \cos^2 x)}{\cos x}\,dx \\ \\ & = -\int \dfrac{(1 - u^2)}{u}\,du \\ \\ & = \int u \,du - \int\dfrac 1u \,du\\ \\ & = \frac{u^2}{2} - \ln |u| + C \\ \\ & = \frac 12 \cos^2 x - \ln |\cos x| + C\end{align}$$