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Math Help - Integrate sin^2(4t) cos(4t)

  1. #1
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    Integrate sin^2(4t) cos(4t)

    Hi,

    I have the follwing that I am struggling with:

    \int sin^2(4t) cos(4t).dt

    so far I have got:

    u = sin^2(4t)

    du/dt = 4t

    du = 1/4

    Now I am not sure what to do with the cos(4t) can someone help explain and show me the answewer so I maybe able to work backwards?

    Thanks

    m
    Last edited by mr fantastic; May 30th 2011 at 01:18 PM. Reason: Re-titled.
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  2. #2
    Super Member Quacky's Avatar
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    Quote Originally Posted by mm874 View Post
    Hi,

    I have the follwing that I am struggling with:

    \int sin^2(4t) cos(4t).dt

    so far I have got:

    u = sin^2(4t)

    du/dt = 4t

    du = 1/4

    Now I am not sure what to do with the cos(4t) can someone help explain and show me the answewer so I maybe able to work backwards?

    Thanks

    m
    We had this discussion only recently. It is entirely unnecessary to use a u-substitution. You can just apply the general rule that:

    \int f'(x)[f(x)]^n dx=\dfrac{[f(x)]^{n+1}}{n+1}+C or you can just use the fact that this clearly resembles the derivative of sin^3(4t).

    If you absolutely have to use substitution (and you really don't) then your \frac{du}{dt} is incorrect. How do you differentiate Sin^2(4t)?
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  3. #3
    MHF Contributor
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    The substitution you are supposed to make is \displaystyle u = \sin{(4t)} \implies du = 4\cos{(4t)}\,dt.
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  4. #4
    Super Member Quacky's Avatar
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    Quote Originally Posted by Prove It View Post
    The substitution you are supposed to make is \displaystyle u = \sin{(4t)} \implies du = 4\cos{(4t)}\,dt.
    I agree that this is better, although the sin^2(x) substitution worked when I tried it just now, although it wasn't simple.
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