1. ## simple integral

trying ti do a problem

I need the intregal of [(3/2)-2cos(2x)+((1/2)cos(4x)]

i got (3/2)x -4sin(2x) + 2cos(4x)+c

is that right?

2. I'm getting
$\dfrac{3}{2}x - \sin\, 2x + \dfrac{1}{8}\sin\, 4x + C$

When you do u substitution on the second term, for instance:
\begin{aligned}
u &= 2x \\
du &= 2 dx \\
\int 2 \cos\,2x\; dx &= \int \cos\,u\;du \\
&= \sin\,u + C \\
&= \sin\,2x + C
\end{aligned}

3. thanks,

so for the rest of the problem, I take it from the bonds 0 to pi.

I got pi/4 [(3/2)pi-sin92pi)+(1/8)sin(4pi)]- (pi/4)[0-sin(0)-(1/8)sin(0)]

what is the final answer to this?

4. I'm not sure what you're doing, but if this is a definite integral going from 0 to pi, then you would have
$\left( \dfrac{3}{2}(\pi) - \sin\, 2(\pi) + \dfrac{1}{8}\sin\, 4(\pi) \right) - \left( \dfrac{3}{2}(0) - \sin\, 2(0) + \dfrac{1}{8}\sin\, 4(0) \right)$
$= \dfrac{3\pi}{2}$

5. yeah urs looks like what I have, except there is a pi/4 outside the integral. how do I put that in? I have to put multiply it to both ()'s before subtracting them right?

6. Grr, please state the ENTIRE problem next time. So I'm guessing that the problem was this:
$\dfrac{\pi}{4} {\displaystyle \int_0^{\pi}} \dfrac{3}{2} - 2\cos\, 2x + \dfrac{1}{2}\cos\, 4x\; dx$

After taking the integral you get:
$\dfrac{\pi}{4} \left[ \dfrac{3}{2}x - \sin\, 2x + \dfrac{1}{8}\sin\, 4x\; \Big{\vert}_0^{\pi} \right]$

Then evaluate:
$\dfrac{\pi}{4} \left[ \left( \dfrac{3}{2}(\pi) - \sin\, 2(\pi) + \dfrac{1}{8}\sin\, 4(\pi) \right) - \left( \dfrac{3}{2}(0) - \sin\, 2(0) + \dfrac{1}{8}\sin\, 4(0) \right) \right]$
$= \dfrac{\pi}{4} \left( \dfrac{3\pi}{2} \right)$
$= \dfrac{3\pi^2}{8}$