# Complex integration

• Aug 8th 2010, 06:20 AM
bones
Complex integration
Hi guys! Just working on a puzzle and I realized I have no idea how to integrate this. Help would be much appreciated. Thanks!

$\int _{0} ^{\pi} (2\sin^2\alpha + (\pi-2\alpha)\sin\alpha\cos\alpha)d\alpha$
• Aug 8th 2010, 08:19 AM
Amer
$\sin ^2 x = \frac{1- \cos 2x }{2}$

$\sin x \cos x = \frac{\sin 2x}{2}$
• Aug 8th 2010, 08:38 AM
Soroban
Hello, bones!

If you "break it down", the solution will come to you . . .

Quote:

$\displaystyle{ \int _{0}^{\pi} (2\sin^2\!x + (\pi-2x)\sin x\cos x)\,dx}$

We have: . $\displaystyle{ \int^{\pi}_0\!\bigg(2\sin^2\!x + \pi\sin x\cos x - 2x\sin x\cos x\bigg)\,dx }$

. . . . $\displaystyle{=\;\int^{\pi}_0\!\! 2\sin^2\!x\,dx + \pi\!\int^{\pi}_0\!\! \sin x\cos x\,dx - 2\!\int^{\pi}_0 \!\!x\sin x\cos x\,dx }$

. . . . $\displaystyle{=\; \int^{\pi}_0\!(1-\cos2x)\,dx + \tfrac{\pi}{2}\!\int^{\pi}_0\!\!\sin2x\,dx - \int^{\pi}_0\!\!x\sin2x\,dx }$

You should be able to integrate the first two.

The third requires Integration by Parts:

. . $\begin{array}{ccccccc}u &=& x && dv &=& \sin2x\,dx \\ du &=& dx && v &=&\text{-}\frac{1}{2}\cos2x \end{array} \quad \hdots\;\;\text{ etc.}$

Edit: Amer beat me to it . . . *sigh*
• Aug 8th 2010, 08:50 AM
chisigma
With the substitution $\alpha= \frac{\theta}{2}$ is...

$\displaystyle 2\ \sin^{2} \alpha= 1 - \cos \theta$

$\displaystyle \sin \alpha\ \cos \alpha = \frac{1}{2}\ \sin \theta$ (1)

... so that is...

$\displaystyle \int_{0}^{\pi} \{2 \sin^{2} \alpha + (\pi -2 \alpha)\ \sin a \cos \alpha \}\ d \alpha =$

$\displaystyle = \frac{1}{2} \int_{0}^{2 \pi} (1 - \cos \theta + \frac{\pi}{2}\ \sin \theta - \frac{\theta}{2}\ \sin \theta)\ d \theta =$

$\displaystyle = \frac{1}{2} \int_{0}^{2 \pi} (1 - \frac{\theta}{2}\ \sin \theta)\ d \theta = \frac{1}{2}\ |\theta - \frac{1}{2}\ \sin \theta +\frac{\theta}{2}\ \cos \theta |_{0}^{2 \pi} = \frac{3}{2}\ \pi$ (2)

Kind regards

$\chi$ $\sigma$