Trig integration

**SuumEorum** · June 20th 2008, 02:49 PM

I need some help with this integral.

$\int{\frac{1}{4}\sin ^{2}2\theta \cos 2\theta }d\theta$

I've tried using integration by parts:

$u=\sin ^{2}2\theta$

$\frac{dv}{d\theta }=\cos 2\theta$

And got the integral as:

$\frac{1}{4}\left( \frac{1}{2}\sin ^{3}2\theta -\int{\sin 4\theta \sin 2\theta } \right)$ .

I integrated by parts the integral again and I seem to be stuck in a loop. I think I might've made a mistake though, any help is appreciated. Thanks

.

**Mathstud28** · June 20th 2008, 02:53 PM

Originally Posted by SuumEorum

I need some help with this integral.

$\int{\frac{1}{4}\sin ^{2}2\theta \cos 2\theta }d\theta$

I've tried using integration by parts:

$u=\sin ^{2}2\theta$

$\frac{dv}{d\theta }=\cos 2\theta$

And got the integral as:

$\frac{1}{4}\left( \frac{1}{2}\sin ^{3}2\theta -\int{\sin 4\theta \sin 2\theta } \right)$ .

I integrated by parts the integral again and I seem to be stuck in a loop. I think I might've made a mistake though, any help is appreciated. Thanks

.

Let $\varphi=\sin(2\theta)\Rightarrow{d\varphi=2\cos(2\ theta)d\theta}\Rightarrow\frac{d\varphi}{2}=\cos(2 \theta)d\theta$

So we have

$\frac{1}{8}\int{\varphi^2d\varphi}$

You can take it from theree.

**SuumEorum** · June 20th 2008, 02:59 PM

Yay I have it now, thanks a lot

. So what I did was completely wrong? D:

**Mathstud28** · June 20th 2008, 07:33 PM

Originally Posted by SuumEorum

Yay I have it now, thanks a lot

. So what I did was completely wrong? D:

I think you had the right idea...but when you apply integration by parts to non integration by parts friendly integrals, the result can be ugly. So right idea for the method, but the wrong method.

**topsquark** · June 20th 2008, 07:58 PM

Originally Posted by SuumEorum

I need some help with this integral.

$\int{\frac{1}{4}\sin ^{2}2\theta \cos 2\theta }d\theta$

I've tried using integration by parts:

$u=\sin ^{2}2\theta$

$\frac{dv}{d\theta }=\cos 2\theta$

And got the integral as:

$\frac{1}{4}\left( \frac{1}{2}\sin ^{3}2\theta -\int{\sin 4\theta \sin 2\theta } \right)$ .

I integrated by parts the integral again and I seem to be stuck in a loop. I think I might've made a mistake though, any help is appreciated. Thanks

.

This question has already been answered satisfactorally, but I'd like to make the observation that if we first put $y = 2 \theta$ the rest of the logic is easier to see, since we don't have those pesky 2's cropping up.

-Dan

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