trig integrals

• Feb 24th 2009, 02:43 PM
saiyanmx89
trig integrals
$\int sin^4(2x)dx$

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 PM
Reckoner
Quote:

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.
• Feb 24th 2009, 04:12 PM
saiyanmx89
I don't know how to integrate cos4xdx and (1+cos8x)
• Feb 24th 2009, 04:17 PM
Reckoner
Quote:

Originally Posted by saiyanmx89
I don't know how to integrate cos4xdx and (1+cos8x)

Really? Let $u=4x$ (and $u=8x$ for the second). Substitute.

And you should know that $\int\cos x\,dx=\sin x+C$ (if not, brush up!)
• Feb 24th 2009, 04:37 PM
saiyanmx89
Is the integral of $1+cosu = cos^2u$ ?
• Feb 24th 2009, 04:37 PM
saiyanmx89
I mean derivative...! not integral.
• Feb 24th 2009, 04:43 PM
Reckoner
Quote:

Originally Posted by saiyanmx89
I mean derivative...! not integral.

Huh? $\frac d{dx}\left[1+\cos u\right]=-(\sin u)\frac{du}{dx}\text.$

What does that have to do with your problem?