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Math Help - integral with cos

  1. #1
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    integral with cos

    Hi everyone,

    Can someone give me a hint on how to do this problem, please? I've tried it several different ways, but it doesn't seem to be working.

    Find the integral x to the 5thcos(x cubed)dx

    u=x to the 3rd
    dv=x to the 5th
    du=3x2
    v=x6/6

    (x to the 3rd)(x to the 5th)=(x to the 3rd)(x to the 6th)-int.(x to the 6th/6)3x sq.

    but this integral looks worse

    Any suggestions?

    Thank you
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by chocolatelover View Post
    Hi everyone,

    Can someone give me a hint on how to do this problem, please? I've tried it several different ways, but it doesn't seem to be working.

    Find the integral x to the 5thcos(x cubed)dx

    u=x to the 3rd
    dv=x to the 5th
    du=3x2
    v=x6/6

    (x to the 3rd)(x to the 5th)=(x to the 3rd)(x to the 6th)-int.(x to the 6th/6)3x sq.

    but this integral looks worse

    Any suggestions?

    Thank you
    do a regular substitution first. Let u = x^3

    then do by parts on the integral that results
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  3. #3
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    I put

    t=x3
    x2dx=dt/3

    int. x to the 5th cos (x3)dx=

    int. cosu3dx=

    1/3int. cos t3dx

    u=cosu3
    dv=dx/3
    du=-sinu3(3u2)
    v=

    Can you tell me if this right so far and if dv is 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
    I put

    t=x3
    x2dx=dt/3

    int. x to the 5th cos (x3)dx=

    int. cosu3dx=

    1/3int. cos t3dx

    u=cosu3
    dv=dx/3
    du=-sinu3(3u2)
    v=

    Can you tell me if this right so far and if dv is right?

    Thank you
    i have no idea what you just did. let me do the first substitution for you. i leave the integration by parts to you

    \int x^5 \cos \left( x^3 \right) ~dx

    We proceed by substitution
    Let t = x^3

    \Rightarrow dt = 3x^2~dx

    \Rightarrow \frac {1}{3}dt = x^2 ~dx

    Now, \int x^5 \cos \left( x^3 \right) ~dx = \int x^3 \cdot x^2 \cos \left( x^3 \right) ~dx

    = \frac {1}{3} \int t \cos t~dt

    Now we continue with integration by parts ...

    (Hint: obviously we use u = t and dv = \cos u ~du)
    Last edited by Jhevon; September 3rd 2007 at 06:13 PM.
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  5. #5
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    I kind of confused. How do you have t and u in the final interval? In this case I'm using t so that I don't get the u from parts mixed up with the u substitution right? but wouldn't there be two t's instead of the u?

    1/3int.tcostdt

    Thank you
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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
    I kind of confused. How do you have t and u in the final interval? In this case I'm using t so that I don't get the u from parts mixed up with the u substitution right? but wouldn't there be two t's instead of the u?

    1/3int.tcostdt

    Thank you
    yes, you are correct. i meant to type t, but i'm so used to typing u for substitutions that my hand did it against my will...bad hand! bad! *slap*
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  7. #7
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    Would my final answer be tsin-cost?

    u=t
    dv=costdt
    du=1
    v=sint

    tsin-int.sint=

    tsin-cost

    Thank you
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by chocolatelover View Post
    Would my final answer be tsin-cost?

    u=t
    dv=costdt
    du=1
    v=sint

    tsin-int.sint=

    tsin-cost

    Thank you
    it is + cos(t). and you forgot to multiply through by the 1/3

    and you also forgot to back substitute to get the x back

    we started with x remember, you can't give an answer in terms of t
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