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Math Help - integral with cos

  1. #1
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    integral with cos

    Hi,

    Can someone help me with the interval 0 to pie/2 cos to the 5thxdx, please?

    This is what I did:

    cos2x=1/2(1+cos2x)

    int. 0 to pie/2 (cos3x)cos2xdx=

    1/2 int. (1+cos2x)cos3dx=

    Then would I do a u substitution?

    u=cos2x
    du=-sin2x(2)dx

    Thank you
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  2. #2
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    \cos^5 x = \cos^4 x \cos x = (1-\sin^2 x)^2 \cos x.

    Let t=\sin x now continue.
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  3. #3
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    Hi,

    Can you explain to me how you got this, please?.

    I understand how you went to the second step, but I don't understand how you got the third equation.

    After that I need to do u substitution, right?

    t=sinx
    dt=cosxdx

    int. 1 to pie/1 (1-sin2x) squared cosx
    =integ. -sin2xcosx

    Now, I just need to plug in pie/2 and 1, right?

    Thank you
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by chocolatelover View Post
    Hi,

    Can you explain to me how you got this, please?.

    I understand how you went to the second step, but I don't understand how you got the third equation.
    recall that \cos^2 x = 1 - \sin^2 x

    therefore, \cos^4 x = \left( \cos^2 x \right)^2 = \left( 1 - \sin^2 x \right)^2

    After that I need to do u substitution, right?

    t=sinx
    dt=cosxdx
    yes

    int. 1 to pie/1 (1-sin2x) squared cosx
    =integ. -sin2xcosx

    Now, I just need to plug in pie/2 and 1, right?

    Thank you
    what are you doing here? if you made the t-substitution, why do you still have x's in the expressions??
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  5. #5
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    Would it be (-sin2t)sq. dt?

    then do I need to do integration by parts?
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by chocolatelover View Post
    Would it be (-sin2t)sq. dt?

    then do I need to do integration by parts?
    no integration by parts necessary (i see that you have a problem knowing when to use by parts and when to use regular substitution. we must address this).

    anyway, as TPH showed us:

    \int \cos^5 x dx = \int (1-\sin^2 x)^2 \cos x~dx

    We proceed by substitution.

    Let t=\sin x

    \Rightarrow dt = \cos x ~dx

    So now, we look on our integral. wherever we see \sin x, we replace it with t. if we see \cos x ~dx, we replace it with dt

    So our integral becomes:

    \int \left( 1 - t^2 \right)^2~dt

    Hopefully, you can take it from here

    Quote Originally Posted by chocolatelover View Post

    1/2 int. (1+cos2x)cos3dx=

    Then would I do a u substitution?

    u=cos2x
    du=-sin2x(2)dx

    Thank you
    by the way. please use "^" when you want to show that something is a power. so to type \cos^5 x you should type cos^5(x) or (cos(x))^5 etc.
    Last edited by ThePerfectHacker; September 5th 2007 at 09:19 AM.
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