We are given:

\(\displaystyle \int\frac{x\cos(x)\sin(x)}{\cos^3(2x)}\,dx\)

Using the double-angle identity for sine, we may write this as:

\(\displaystyle \frac{1}{2}\int\frac{x\sin(2x)}{\cos^3(2x)}\,dx=\frac{1}{2}\int x\tan(2x)\sec^2(2x)\,dx\)

Using integration by parts, let:

\(\displaystyle u=x\:\therefore\:du=dx\)

\(\displaystyle dv=\tan(2x)\sec^2(2x)\,dx\:\therefore\:v=\frac{1}{2}\tan^2(2x)\) and we have:

\(\displaystyle \frac{1}{2}\(\frac{x}{2}\tan^2(2x)-\frac{1}{2}\int\tan^2(2x)\,dx\)=\)

\(\displaystyle \frac{1}{2}\(\frac{x}{2}\tan^2(2x)-\frac{1}{4}\int\sec^2(2x)-1\,2dx\)=\)

\(\displaystyle \frac{1}{2}\(\frac{x}{2}\tan^2(2x)-\frac{1}{4}\(\tan(2x)-2x\)\)+C=\)

\(\displaystyle \frac{1}{2}\(\frac{x}{2}\(\tan^2(2x)+1\)-\frac{1}{4}\(\tan(2x)\)+C=\)

\(\displaystyle \frac{1}{2}\(\frac{x}{2}\(\sec^2(2x)\)-\frac{1}{4}\(\tan(2x)\)+C=\)

\(\displaystyle \frac{x}{4}\(\sec^2(2x)\)-\frac{1}{8}\(\tan(2x)\)+C\)