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

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

I just need a reference point for where to go.

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- Feb 24th 2009, 02:43 PMsaiyanmx89trig integrals
$\displaystyle \int sin^4(2x)dx$

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

I just need a reference point for where to go. - Feb 24th 2009, 03:47 PMReckoner
Use $\displaystyle \sin^2\theta=\frac{1-\cos2\theta}2$ and $\displaystyle \cos^2\theta=\frac{1+\cos2\theta}2\colon$

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

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

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

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

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

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

The rest should be easy. - Feb 24th 2009, 04:12 PMsaiyanmx89
I don't know how to integrate cos4xdx and (1+cos8x)

- Feb 24th 2009, 04:17 PMReckoner
- Feb 24th 2009, 04:37 PMsaiyanmx89
Is the integral of $\displaystyle 1+cosu = cos^2u$ ?

- Feb 24th 2009, 04:37 PMsaiyanmx89
I mean derivative...! not integral.

- Feb 24th 2009, 04:43 PMReckoner