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Math Help - integral of x/cos^2x dx

  1. #1
    Boo
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    integral of x/cos^2x dx

    Hello!
    Could someone help me with:

    \int\frac{x}{cos^2x}dx?

    I think it should be how to get rid of trigonometrics...
    Many thanks!!!
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  2. #2
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    Re: integral of x/cos^2x dx

    Quote Originally Posted by Boo View Post
    Hello!
    Could someone help me with:

    \int\frac{x}{cos^2x}dx?

    I think it should be how to get rid of trigonometrics...
    Many thanks!!!
    \displaystyle \begin{align*} \int{\frac{x}{ \cos^2{(x)} }\,dx} &= \int{x\sec^2{(x)}\,dx} \end{align*}

    Now apply integration by parts with \displaystyle \begin{align*} u = x \implies du = dx \end{align*} and \displaystyle \begin{align*} dv = \sec^2{(x)}\,dx \implies v = \tan{(x)} \end{align*} and the integral becomes

    \displaystyle \begin{align*} \int{x\sec^2{(x)}\,dx} &= x\tan{(x)} - \int{\tan{(x)}\,dx} \\ &= x\tan{(x)} + \int{\frac{-\sin{(x)}}{\cos{(x)}}\,dx} \\ &= x\tan{(x)} + \int{\frac{1}{w}\,dw}\textrm{ after making the substitution } w = \cos{(x)} \implies dw = -\sin{(x)}\,dx \\ &= x\tan{(x)} + \ln{|w|} + C \\ &= x\tan{(x)} + \ln{\left| \cos{(x)} \right| } + C \end{align*}
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