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Thread: integration by parts - proof

  1. #1
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    integration by parts - proof

    integration by parts - proof-capture.png

    So, i did the integration by parts. my u = sqrt(4-x^2) du = 1/(2sqrt(4-x^2)) , dv = dx , v= x. However, It does not like the proof on the right...
    What does the hint mean by clever zero?
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  2. #2
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    Re: integration by parts - proof

    To begin with, your "du" is incorrect. Taking $u= \sqrt{1- x^2}= (1- x^2)^{1/2}$, $du= (1/2)(1- x^2)^{-1/2}(-2x)= -\frac{x}{\sqrt{1- x^2}}$. You did not use the "chain rule".
    Last edited by topsquark; Feb 25th 2018 at 07:13 AM. Reason: Tweeked LaTeX
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  3. #3
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    Re: integration by parts - proof

    Quote Originally Posted by lc99 View Post
    Click image for larger version. 

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    So, i did the integration by parts. my u = sqrt(4-x^2) du = 1/(2sqrt(4-x^2)) , dv = dx , v= x. However, It does not like the proof on the right...
    What does the hint mean by clever zero?
    Addition by a "clever zero" means adding something like $1-1$. It adds to zero. So, for instance, inside the square root, you can add $-4x+4x$. Then you have:

    $\sqrt{4-x^2} = \sqrt{4-4x-x^2+4x} = \sqrt{(2-x)^2+4x}$

    This is not the "clever zero" the problem was referring to and in fact makes the problem harder, but it is an example of adding a zero to change the problem. You just need to figure out the zero needed to make the problem easier rather than harder.

    To check you work, let's solve the integral another way. Then, you can check to see if you get the same solution using integration by parts. How about the substitution $x=2\sin \theta$? Then you have $dx = 2\cos \theta d\theta$. Your integral becomes:

    $\displaystyle 2\int \sqrt{4(1-\sin^2 \theta)}\cos \theta d\theta = 2\int 2\cos^2\theta d\theta = 2\int \left(\cos(2\theta)+1\right)d\theta = \sin(2\theta)+2\theta + C = 2\arcsin\left(\dfrac{x}{2}\right)+\dfrac{x\sqrt{4-x^2}}{2}+C$
    Last edited by SlipEternal; Feb 25th 2018 at 06:49 AM.
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