trig integrals

**saiyanmx89** · February 24th 2009, 02:43 PM

$\int sin^4(2x)dx$

I don't know what theorem to use or if to separate it...

I just need a reference point for where to go.

**Reckoner** · February 24th 2009, 03:47 PM

Originally Posted by saiyanmx89

I don't know what theorem to use or if to separate it...

I just need a reference point for where to go.

Use $\sin^2\theta=\frac{1-\cos2\theta}2$ and $\cos^2\theta=\frac{1+\cos2\theta}2\colon$

$\int\sin^42x\,dx$

$=\int\left(\sin^22x\right)^2\,dx$

$=\int\left(\frac{1-\cos4x}2\right)^2\,dx$

$=\frac14\int\left(1-2\cos4x+\cos^24x\right)\,dx$

$=\frac14\left[\int dx-2\int\cos4x\,dx+\int\cos^24x\,dx\right]$

$=\frac14\left[\int dx-2\int\cos4x\,dx+\frac12\int(1+\cos8x)\,dx\right]$

The rest should be easy.