Help me out in this integration

Question

Prove that $\sin(wx)\cos(wx)\,dx$ from $0$ to $\pi/w = -2/3w$

I have solved and solved and solved but keep getting zero.

Answer 1

The integral is 0.

$\sin{wx}\cos{wx}=\frac{1}{2}\sin{2wx}$

So the integral is

$$\frac{1}{2}\int_0^{\pi/w} \sin{2 w x} \ dx$$

Let $u=2wx$. Then $du=2wdx$ or $dx=\frac{du}{2w}$

We must also multiply our endpoints by $2w$to get them with respect to $u$.

So we have:

$$\frac{1}{4w}\int_0^{2\pi} \ \sin{u} \ du$$

Integrating the sine from 0 to 2$\pi$ is a full cycle. The region of area is congruent to the area under the curve from $-\pi$ to $\pi$. So we canchange the limits to get :

$$\frac{1}{4w}\int_{-\pi}^{\pi} \sin{u} \ du$$

The sine is an odd function, i.e. $\sin{-x}=-\sin{x}$

So for every bit of area we have above the x axis and the curve there is equal and oppositely signed area on the other side of the y axis. So the sum of the area under the curve must be 0.

We get this solely from geometric considerations, no calculation necessary. So the area is 0.