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Math Help - Integration problem

  1. #1
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    Integration problem

    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?
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  2. #2
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    Quote Originally Posted by etha View Post
    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))
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  3. #3
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    no, I've done this. My answer is still wrong.
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  4. #4
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    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.
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by etha View Post
    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!
    Last edited by Chris L T521; May 6th 2008 at 08:58 PM. Reason: couple of typos
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  6. #6
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    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.
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  7. #7
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    I understand now. Thank you.
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