1. ## trig 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.

2. 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 $\displaystyle \sin^2\theta=\frac{1-\cos2\theta}2$ 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.

3. I don't know how to integrate cos4xdx and (1+cos8x)

4. Originally Posted by saiyanmx89
I don't know how to integrate cos4xdx and (1+cos8x)
Really? Let $\displaystyle u=4x$ (and $\displaystyle u=8x$ for the second). Substitute.

And you should know that $\displaystyle \int\cos x\,dx=\sin x+C$ (if not, brush up!)

5. Is the integral of $\displaystyle 1+cosu = cos^2u$ ?

6. I mean derivative...! not integral.

7. Originally Posted by saiyanmx89
I mean derivative...! not integral.
Huh? $\displaystyle \frac d{dx}\left[1+\cos u\right]=-(\sin u)\frac{du}{dx}\text.$

What does that have to do with your problem?