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Math Help - definite integral

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    definite integral

    \int_{0}^{\frac{\pi}{4}}xtanxdx
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    Almost anytime you have x, or a power of x, multiplied by another function of x, you should think "integration by parts". If you let u= x, then du= dx so you have gotten rid of that x (if you have a power of x, setting u equal to that power of x reduces the power and repeated integration by parts will eliminate it altogether).
    Of course, now dv= tan(x)dx. Do you know the anti-derivative of tan(x)?

    If not, think of it as \int \frac{sin(x)}{cos(x)}dx and use the substitution u= cos(x).
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    Quote Originally Posted by HallsofIvy View Post
    Almost anytime you have x, or a power of x, multiplied by another function of x, you should think "integration by parts". If you let u= x, then du= dx so you have gotten rid of that x (if you have a power of x, setting u equal to that power of x reduces the power and repeated integration by parts will eliminate it altogether).
    Of course, now dv= tan(x)dx. Do you know the anti-derivative of tan(x)?

    If not, think of it as \int \frac{sin(x)}{cos(x)}dx and use the substitution u= cos(x).
    Except in this case IBP reduces to needing to evaluate the disgusting integral \displaystyle \int{-\ln{|\cos{x}|}\,dx}...
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    So by integration by parts, we have:

    \begin{aligned} \displaystyle \int_{0}^{\frac{\pi}{4}}x\tan{x}\;{dx} &= \biggl\|- x\ln(\cos{x})\biggl\|_{x = 0}^{x = \frac{\pi}{4}} +\int_{0}^{\frac{\pi}{4}}\ln\left(\cos{x}}\right)\  ;{dx}  = \frac{1}{8}\pi\ln(2)+\int_{0}^{\frac{\pi}{4}}\ln\l  eft(\cos{x}}\right)\;{dx}.\end{aligned}

    At this point, I recall that one of the definitions of Catalan's constant was \displaystyle \int_{0}^{\frac{\pi}{4}}\ln\left(2\cos{t}\right)\;  {dt} = \frac{1}{2}K.


    So I write \begin{aligned}\displaystyle \int_{0}^{\frac{\pi}{4}}\ln\left(\cos{x}}\right) \;{dx} & = \int_{0}^{\frac{\pi}{4}}\ln\left(2\cos{x}}\right) \;{dx}-\int_{0}^{\frac{\pi}{4}}\ln(2)\;{dx}  = \frac{1}{2}K -\frac{\pi}{4}\ln(2).\end{aligned}

    Thus \displaystyle \int_{0}^{\frac{\pi}{4}}x\tan{x}\;{dx} = \frac{\pi}{8}\ln(2)+\frac{1}{2}K -\frac{\pi}{4}\ln(2) =  \frac{1}{2}K-\frac{1}{8}\pi\ln(2).
    Last edited by TheCoffeeMachine; January 27th 2011 at 09:19 AM.
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