# Calculus 2 Integral Question (difficult)

1. Sep 24, 2008

### hamburgler

1. The problem statement, all variables and given/known data
$$\int$$cos3xdx

2. Relevant equations
cos3x = cos2x * cosx

3. The attempt at a solution

Let u= cos2x
du=(x/2) + (cos2x/4x)
v=-sinx
dv=cos

uv-$$\int$$vdu

-cos2xsinx - $$\int$$-sinx(x/2 + cos2x/4x)dx

-cos2xsinx -$$\int$$xsinx/2 + (sinxcos2x/4x)

-cos2xsinx - (1/2)xsinx +(sinxcos2x/4x)

-cos2xsinx -(sinx-xcosx/2) + (sinxcos2x/4x)

....

aaaand that's about where I get lost/stuck. i can't figure out how to finish off the problem. i get to the last line on the far right and get confused. any help would be much appreciated.

2. Sep 24, 2008

### ||spoon||

Hey,

Why don't you try and use the trig identity$$\cos^2x +\sin^2x = 1$$ and then try substituting $$\sin x = u$$

Hope that helps,

-Spoon

3. Sep 24, 2008

### hamburgler

hm ill give it a shot

4. Sep 24, 2008

### ||spoon||

No worries, glad to have helped.

- Spoon

5. Sep 24, 2008

### HallsofIvy

Staff Emeritus
Here's your first error. The derivative of cos2 x, using the chain rule, is (2 cos x)(cos x)'= - 2sin x cos x. I have absolutely no idea how you got your du!

You started wrong by trying to use integration by parts when it is not necessary. cos3 x dx= (cos2 x) (cos x dx)= (1- sin2 x) (cos x dx). Now use the obvious substitution.