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Math Help - Integration by Parts

  1. #1
    Junior Member
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    Integration by Parts

    Hi

    Evaluate the integral of t(sec^2(6t))dt

    My answer is: (t/6)tan(6t) - (1/12)Ln|sec(6t)| + C

    why is this wrong?

    Thanks
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  2. #2
    Super Member

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    Lexington, MA (USA)
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    Re: Integration by Parts

    Hello, sjmiller!

    Integrate: . I \;=\;\int t\sec^2(6t)\,dt

    My answer is: . \tfrac{1}{6}t\tan(6t) - \tfrac{1}{12}\ln|\sec(6t)| + C

    Why is this wrong?

    Did you do something silly? . . . .Like: \tfrac{1}{6}\!\cdot\!\tfrac{1}{6} \,=\,\tfrac{1}{12} ??


    By parts: . \begin{Bmatrix}u &=& t && dv &=& \sec^2(6t)\,dt \\ du &=& dt && v &=& \frac{1}{6}\tan(6t) \end{Bmatrix}


    We have: . I \;=\;\tfrac{1}{6}t\tan(6t) - \tfrac{1}{6}\int\tan(6t)\,dt

    . . . . . . . . I \;=\;\tfrac{1}{6}t\tan(6t) - {\bf\tfrac{1}{36}}\ln|\sec(6t)| + C
    . . . . . . . . . . . . . . . . . . . . . \uparrow
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