Trying to evaluate $\int \frac{1}{\sin(x)\cos^3(x)} \,dx$ and got stuck

Write the integral as $$\int \sec^3(x)\csc(x)\,dx$$ Make the substitution $u=\tan x$, then $du=\sec^2(x) \,dx$, and it is $$\int \sec(x)\csc(x)\,du$$ $$\int u+\frac{1}{u}\,du ~~~~ \text{(why?)}$$ Can you finish?

You can use the fact that$$\int\frac{\sin(x)}{\cos^3(x)}\,\mathrm dx=-\frac{\cos^{-2}(x)}{-2}=\frac1{2\cos^2(x)}$$and that\begin{align}\int\frac1{\sin(x)\cos(x)}\,\mathrm dx&=\int\frac{\sin(x)}{(1-\cos^2(x))\cos(x)}\,\mathrm dx\\&=-\log|\cos x|+\frac12\log(1-\cos^2(x))\\&=-\log|\cos x|+\log|\sin x|\\&=\log|\tan x|.\end{align}

With your great answers I was able to evaluate this integral as follows:

$$ \int \frac{1}{\sin(x)\cos^3(x)} \,dx = \int \sec^3(x)\csc(x) \,dx =\\ \int \sec(x)\csc(x) \sec^2(x) \,dx = \int \frac{1}{\cos(x)\sin(x)} \sec^2(x) \,dx = \\ \int \frac{\sin^2(x)+\cos^2(x)}{\cos(x)\sin(x)} \sec^2(x) \,dx = \int \frac{\sin(x)}{\cos(x)} + \frac{\cos(x)}{\sin(x)} \sec^2(x) \,dx = \\ \int \tan(x) +\cot(x) \sec^2(x) \,dx = \int \tan(x) + \frac{1}{\tan(x)} \sec^2(x) \,dx\\ u=\tan(x)\\du=\sec^2(x)dx\\ \int u +\frac{1}{u} \,du = \frac{u^2}{2}+\ln|u|+c =\\ \frac{1}{2}\tan^2(x)+\ln|\tan(x)| +c $$

Thank you everybody for your helpful answers and comments.