# Yet another integration

#### Ockonal

Hello, can't understand how to process this one:

integral { x*cos(x)*sin(x) / cos³(2*x) dx }

#### MarkFL

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\)

Thanks