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Math Help - integration by parts question 2

  1. #1
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    integration by parts question 2

    Find the integral.

    im sure u = x^5 and v'= cos(x^3) but not sure where to go from there
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by gabet16941 View Post
    Find the integral.

    im sure u = x^5 and v'= cos(x^3) but not sure where to go from there
    well, from there you would do the integration by parts method, but as it stands, you'd have to do it 5 times!

    a preliminary substitution makes the problem easier:

    Let t = x^3

    \Rightarrow dt = 3x^2 ~dx

    \Rightarrow \frac 13 ~dt = x^2~dx

    So our integral becomes

    \frac 13 \int t \cos t~dt

    which is a relatively easy integration by parts problem
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  3. #3
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    Hello, gabet16941!

    \int x^5\cos(x^3)\,dx

    I'm sure u =  x^5\,\text{ and }\,v' = \cos(x^3) . . . . but you can't integrate {\color{blue}\cos(x^3)}
    Break it up like this:

    . . \begin{array}{ccccccc}u &=& x^3 & & dv &=& x^2\cos(x^3)\,dx \\ du &=& 3x^2\,dx & & v &=& \frac{1}{3}\sin(x^3) \end{array}


    Then we have: . \tfrac{1}{3}x^3\sin(x^3) - \tfrac{1}{3}\int x^2\sin(x^3)\,dx

    . . . . . . . . . = \;\tfrac{1}{3}x^3\sin(x^3) + \tfrac{1}{9}\cos(x^3) + C

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