Integration problem

**etha** · May 6th 2008, 08:26 PM

Hi everyone. I'm having trouble with an integration problem.
The problem is:
Solve the integral for:

sin(4x) dx
_________
4 + (cos(4x))^2

I put u=cos4x and du=-4sin(4x). However, whenever I continue to solve, I just don't come up with the right answer. Can someone work through this problem for me?

**mr fantastic** · May 6th 2008, 08:46 PM

Originally Posted by etha

Hi everyone. I'm having trouble with an integration problem.
The problem is:
Solve the integral for:

sin(4x) dx
_________
4 + (cos(4x))^2

I put u=cos4x and du=-4sin(4x) dx. However, whenever I continue to solve, I just don't come up with the right answer. Can someone work through this problem for me?

First, the bit in red I added is important - it's omission by you is probably why you're going wrong. Then:

Substitute $dx = - \frac{du}{4 \sin(4x)}$ .

I couldn't see the above latex equation, there might be a problem so I'll give it here too: dx = - du/(4 sin(4x))

**etha** · May 6th 2008, 08:47 PM

no, I've done this. My answer is still wrong.

**o_O** · May 6th 2008, 08:49 PM

Show us what you've done and we'll see if we can try to help. Better than us having to go through the whole thing and finding out that you only need one small step to finish.

**Chris L T521** · May 6th 2008, 08:56 PM

Originally Posted by etha

Hi everyone. I'm having trouble with an integration problem.
The problem is:
Solve the integral for:

sin(4x) dx
_________
4 + (cos(4x))^2

I put u=cos4x and du=-4sin(4x). However, whenever I continue to solve, I just don't come up with the right answer. Can someone work through this problem for me?

Your choice for u is correct: $u=cos(4x)$ . Thus $du=-4sin(4x)dx$ . Solving for $dx$ , we get $dx=\frac{du}{-4sin(4x)}$ . Now make your substitutions:

$\int\frac{sin(4x) du}{-4sin(4x)(4+u^2)}$ .

The integral now becomes

$-\frac{1}{4}\int\frac{du}{4+u^2}$ .

However, recall that

$\int\frac{du}{a^2+u^2}=\frac{1}{a}arctan(\frac{u}{ a})+c$

Therefore,

$-\frac{1}{4}\int\frac{du}{4+u^2}=-\frac{1}{8}\arctan(\frac{u}{2})+c$

Substituting u back into the evaluated integral, we get

$-\frac{1}{8}\arctan(\frac{cos(4x)}{2})+c$ .

Hope this helps you out!

**etha** · May 6th 2008, 08:57 PM

what I've done so far is u= cos(4x) and du=-4sin(4x) dx. I know that I need to put it in 1/1+u^2 so that i can integrate it as a arctan(x) + c. there's a 1/4 in the original problem I have to account for. At this point, I'm at a loss as to what to do.

**etha** · May 6th 2008, 08:58 PM

I understand now. Thank you.

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