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Thread: Indefinite integral of cos^4 2x dx

  1. #1
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    Indefinite integral of cos^4 2x dx

    I tried integrating this on my own but couldn't get the correct answer. I'm not supposed to use any reduction formulas. Any help is much appreciated

    Integral (cos42x dx)
    Integral (1/2(1-cos 4x))2
    Integral ([1/4(1-2cos 4x- cos24x)][1/4(1-2cos 4x - cos 24x)]dx)
    1/8 integral ((1-2cos4x - cos 24x)(1-2cos 4x - cos 24x) dx)
    1/8 integral (1-2cos 4x - cos24x-2 cos 4x-4cos 24x - 2 cos 34x - cos24x - 2 cos34x - cos44x
    1/8 integral (1-4cos4x-6cos24x - 4 cos34x -cos44x dx)

    I stopped here because it wouldn't give me the answer I needed. The answer is
    1/64 (24x + 8 sin 4x + 8 sin x) + C

  2. #2
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    Look at this website.
    B
    e sure to click the show steps button.
    “A professor is someone who talks in someone else’s sleep”
    W.H. Auden

  3. #3
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    Hello, Ohoneo!

    [tex]\displaystyle \int \cos^42x\,dx[/tex]

    We have: .[tex]\cos^4\!2x \:=\:(\cos^2\!2x)^2 \:=\:\left(\frac{1+\cos4x}{2}\right)^2 \;=\;\frac{1}{4}(1 + 2\cos4x + \cos^2\!4x) [/tex]

    . . . . . . [tex]=\;\frac{1}{4}\left(1 + 2\cos4x + \frac{1+\cos8x}{2}\right) \;=\;\frac{1}{4}\left(\frac{3}{2} + 2\cos4x + \frac{1}{2}\cos8x\right) [/tex]

    Then: .[tex]\dfrac{1}{4}\displaystyle \int\left(\tfrac{3}{2} + 2\cos4x + \tfrac{1}{2}\cos8x\right)\,dx \;=\; \tfrac{1}{4}\left(\tfrac{3}{2}x + \tfrac{1}{2}\sin4x + \tfrac{1}{16}\cos8x\right) + C[/tex]
    Last edited by Subhotosh Khan; 02-22-2012 at 12:58 PM. Reason: fixed a small typo

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    Quote Originally Posted by soroban View Post
    Hello, Ohoneo!


    We have: .[tex]\cos^4\!2x \:=\\cos^2\!2x)^2 \:=\:\left(\frac{1+\cos4x}{2}\right)^2 \;=\;\frac{1}{4}(1 + 2\cos4x + \cos^2\!4x) [/tex]

    . . . . . . [tex]=\;\frac{1}{4}\left(1 + 2\cos4x + \frac{1+\cos8x}{2}\right) \;=\;\frac{1}{4}\left(\frac{3}{2} + 2\cos4x + \frac{1}{2}\cos8x\right) [/tex]

    Then: .[tex]\dfrac{1}{4}\displaystyle \int\left(\tfrac{3}{2} + 2\cos4x + \tfrac{1}{2}\cos8x\right)\,dx \;=\; \tfrac{1}{4}\left(\tfrac{3}{2}x + \tfrac{1}{2}\sin4x + \tfrac{1}{16}\cos8x\right) + C[/tex]
    Hey I was just wondering where you got the (3/2) from after you foil both half angle identities?

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    Quote Originally Posted by esahc View Post
    Hey I was just wondering where you got the (3/2) from after you foil both half angle identities?
    Not to be rude answering for Soroban, but the 3/2 comes from the sum of 1 and 1/2 within the first term. Remember that [tex]\frac{1+cos8x}{2}[/tex] can be written as [tex]\frac{1}{2}+\frac{cos8x}{2}[/tex]
    "There are 10 types of people in this world - those who understand binary and those who don't."

  6. #6
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    Oh i see now thanks for the help

  7. #7
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    Quote Originally Posted by Ohoneo View Post
    I tried integrating this on my own but couldn't get the correct answer. I'm not supposed to use any reduction formulas. Any help is much appreciated

    Integral (cos42x dx)
    Integral (1/2(1-cos 4x))2
    Okay to here.
    Integral ([1/4(1-2cos 4x- cos24x)][1/4(1-2cos 4x - cos 24x)]dx)
    ?? (1/4(1- 2cos(4x)- cos2(4x)) is (1/2(1- cos(4x))^2. Why do you have it twice?

    1/8 integral ((1-2cos4x - cos 24x)(1-2cos 4x - cos 24x) dx)
    1/8 integral (1-2cos 4x - cos24x-2 cos 4x-4cos 24x - 2 cos 34x - cos24x - 2 cos34x - cos44x
    1/8 integral (1-4cos4x-6cos24x - 4 cos34x -cos44x dx)

    I stopped here because it wouldn't give me the answer I needed. The answer is
    1/64 (24x + 8 sin 4x + 8 sin x) + C

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