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Thread: 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
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    Re: Integration by Parts

    Hello, sjmiller!

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

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

    Why is this wrong?

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


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


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

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