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Math Help - Integration!

  1. #1
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    Integration!

    How do I integrate sin^4(2x) ?
    Note: The 4 is the power but the 2x is not.

    I tried and I got 3/8x - (1/8)sin4x + (1/64)sin8x + c but the answer is 1/32(48x-4sin4x+sin8x)+c..
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  2. #2
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    Re: Integration!

    Quote Originally Posted by Tutu View Post
    How do I integrate sin^4(2x) ?
    Note: The 4 is the power but the 2x is not.

    I tried and I got 3/8x - (1/8)sin4x + (1/64)sin8x + c but the answer is 1/32(48x-4sin4x+sin8x)+c..
    Here's a quick run-down.

    \int sin^4(2x)~dx

    First a variable change. Let y = 2x, etc
    = \frac{1}{2} \int sin^4(y)~dy

    Clever trick warning!!

    = \frac{1}{2} \int \left ( sin^2(y) \right ) ^2dy

    And from cos(2 \theta ) = 1 - 2~sin^2( \theta ). Plugging this in:

    = \frac{1}{2} \int \left ( \frac{1}{2}(1 - cos(2y) \right ) ^2dy

    = \frac{1}{8} \int (1 - 2~cos(2y) + cos^2(2y))dy

    See what you can do from here.

    -Dan
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  3. #3
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    Re: Integration!

    When I let y=2x why do.I take the 1/2 out? Is it because cause 2x=2sinxcosx?.thanks!
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  4. #4
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    Re: Integration!

    Quote Originally Posted by Tutu View Post
    When I let y=2x why do.I take the 1/2 out? Is it because cause 2x=2sinxcosx?.thanks!
    It is the subsitution rule for integration. dy=2dx --> dy/2=dx
    Normally you will see the variable 'u' used instead of 'y'... but which you use doesn't matter.
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  5. #5
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    Re: Integration!

    Hello, Tutu!

    \int \sin^4(2x)\,dx

    I tried and I got: . \tfrac{3}{8}x - \tfrac{1}{8}\sin4x + \tfrac{1}{64}\sin8x + C . You are right!

    but the answer is: . \tfrac{1}{32}(48x-4\sin4x+\sin8x)+C . They are wrong!
    Here are my steps:

    \int\sin^4(2x)\,dx \;=\;\int\left[\sin^2(2x)\right]^2dx

    . . . . . . . . . . =\;\int\left[\frac{1-\cos4x}{2}\right]^2dx

    . . . . . . . . . . =\;\tfrac{1}{4}\int\big[1 - 2\cos(4x) + \cos^2(4x)\big]\,dx

    . . . . . . . . . . =\;\tfrac{1}{4}\int\left[1 - 2\cos(4x) + \frac{1 + \cos(8x)}{2}\right]\,dx

    . . . . . . . . . . =\;\tfrac{1}{4}\int \left[\tfrac{3}{2} - 2\cos(4x) + \tfrac{1}{2}\cos(8x)\right]\,dx

    . . . . . . . . . . =\;\tfrac{1}{4}\left[\tfrac{3}{2}x - \tfrac{1}{2}\sin(4x) + \tfrac{1}{16}\son(8x)\right] + C

    . . . . . . . . . . =\;\tfrac{3}{8}x - \tfrac{1}{8}\sin(4x) + \tfrac{1}{64}\sin(8x) + C
    Thanks from topsquark
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