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Math Help - Integration by change of variable

  1. #1
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    Integration by change of variable

    I've been trying to show that
    I = \int^1_{-1}{\frac{\sqrt{1-x^2}}{1+x^2}dx} = -\pi + \int^\pi_{-\pi}{\frac{1}{1+\cos^2{\theta}}d\theta}

    (I've not had any problems with the integration though)

    I've tried x = \sin{\frac{\theta}{2}} and x = \tan{\frac{\theta}{4}} as substitutions, but I can only get so far before I get stuck.

    The most progress I've made is with x = \sin{\frac{\theta}{2}} as the substitution, where I get I = -\pi + \int^\pi_{-\pi}{\frac{2}{3-\cos{\theta}}d\theta} but then I can't see how to proceed further.

    Incidentally, I noticed that the graphs for \frac{1}{1+\cos^2{\theta}} and \frac{2}{3-\cos{\theta}} are different, so I think I'm probably using the wrong substitution.

    Can anyone point me in the right direction?
    (this is part of a question in complex analysis)
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  2. #2
    Senior Member DeMath's Avatar
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    Quote Originally Posted by peterje View Post
    I've been trying to show that
    I = \int^1_{-1}{\frac{\sqrt{1-x^2}}{1+x^2}dx} = -\pi + \int^\pi_{-\pi}{\frac{1}{1+\cos^2{\theta}}d\theta}

    (I've not had any problems with the integration though)

    I've tried x = \sin{\frac{\theta}{2}} and x = \tan{\frac{\theta}{4}} as substitutions, but I can only get so far before I get stuck.

    The most progress I've made is with x = \sin{\frac{\theta}{2}} as the substitution, where I get I = -\pi + \int^\pi_{-\pi}{\frac{2}{3-\cos{\theta}}d\theta} but then I can't see how to proceed further.

    Incidentally, I noticed that the graphs for \frac{1}{1+\cos^2{\theta}} and \frac{2}{3-\cos{\theta}} are different, so I think I'm probably using the wrong substitution.

    Can anyone point me in the right direction?
    (this is part of a question in complex analysis)
    Hint:

    \int\limits_{ - 1}^1 {\frac{{\sqrt {1 - {x^2}} }}<br />
{{1 + {x^2}}}dx}  = \int\limits_{ - 1}^1 {\frac{{1 - {x^2}}}<br />
{{\left( {1 + {x^2}} \right)\sqrt {1 - {x^2}} }}dx}  = \int\limits_{ - 1}^1 {\frac{{2 - \left( {1 + {x^2}} \right)}}<br />
{{\left( {1 + {x^2}} \right)\sqrt {1 - {x^2}} }}dx}  =

    =  - \int\limits_{ - 1}^1 {\frac{{dx}}{{\sqrt {1 - {x^2}} }}}  + 2\int\limits_{ - 1}^1 {\frac{{dx}}{{\left( {1 + {x^2}} \right)\sqrt {1 - {x^2}} }}}  =  - \pi  + 2\int\limits_{ - 1}^1 {\frac{{dx}}{{\left( {1 + {x^2}} \right)\sqrt {1 - {x^2}} }}} .

    Now try to use this substitution x = \cos \theta, \, dx =  - \sin \theta \, d\theta

    2\int\limits_{ - 1}^1 {\frac{{dx}}<br />
{{\left( {1 + {x^2}} \right)\sqrt {1 - {x^2}} }}}  = \left\{ \begin{gathered}<br />
  x = \cos \theta, \hfill \\<br />
  dx =  - \sin \theta\,d\theta \hfill \\ <br />
\end{gathered}  \right\} =  - 2\int\limits_\pi ^0 {\frac{{d\theta}}<br />
{{1 + {{\cos }^2}\theta}}}  = \int\limits_{ - \pi }^\pi  {\frac{{d\theta}}<br />
{{1 + {{\cos }^2}\theta}}} .
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  3. #3
    MHF Contributor Bruno J.'s Avatar
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    If you know the residue theorem from complex analysis, it is easy to solve integrals like I = -\pi + \int^\pi_{-\pi}{\frac{2}{3-\cos{\theta}}d\theta}. If you want I can do it.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Bruno J. View Post
    If you know the residue theorem from complex analysis, it is easy to solve integrals like I = -\pi + \int^\pi_{-\pi}{\frac{2}{3-\cos{\theta}}d\theta}. If you want I can do it.
    Also, the substitution \theta=2\arctan\vartheta may be useful...as an alternative to CA.
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  5. #5
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    DeMath: Thanks a lot. I didn't think to use the fact it's an even function.

    Bruno J.: I've had no problems solving the integral, but thanks for the offer of help anyway.
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by peterje View Post
    Incidentally, I noticed that the graphs for \frac{1}{1+\cos^2{\theta}} and \frac{2}{3-\cos{\theta}} are different, so I think I'm probably using the wrong substitution.
    Just a point:

    \int_0^1 x\text{ }dx=\int_0^1\frac{\pi\sin(\pi x)}{2}\text{ }dx
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  7. #7
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    Quote Originally Posted by DeMath View Post
    Hint:

    \int\limits_{ - 1}^1 {\frac{{\sqrt {1 - {x^2}} }}<br />
{{1 + {x^2}}}dx} = \int\limits_{ - 1}^1 {\frac{{1 - {x^2}}}<br />
{{\left( {1 + {x^2}} \right)\sqrt {1 - {x^2}} }}dx} = \int\limits_{ - 1}^1 {\frac{{2 - \left( {1 + {x^2}} \right)}}<br />
{{\left( {1 + {x^2}} \right)\sqrt {1 - {x^2}} }}dx} =

    = - \int\limits_{ - 1}^1 {\frac{{dx}}{{\sqrt {1 - {x^2}} }}} + 2\int\limits_{ - 1}^1 {\frac{{dx}}{{\left( {1 + {x^2}} \right)\sqrt {1 - {x^2}} }}} = - \pi + 2\int\limits_{ - 1}^1 {\frac{{dx}}{{\left( {1 + {x^2}} \right)\sqrt {1 - {x^2}} }}} .

    Now try to use this substitution x = \cos \theta, \, dx = - \sin \theta \, d\theta

    2\int\limits_{ - 1}^1 {\frac{{dx}}<br />
{{\left( {1 + {x^2}} \right)\sqrt {1 - {x^2}} }}} = \left\{ \begin{gathered}<br />
x = \cos \theta, \hfill \\<br />
dx = - \sin \theta\,d\theta \hfill \\ <br />
\end{gathered} \right\} = - 2\int\limits_\pi ^0 {\frac{{d\theta}}<br />
{{1 + {{\cos }^2}\theta}}} = \int\limits_{ - \pi }^\pi {\frac{{d\theta}}<br />
{{1 + {{\cos }^2}\theta}}} .
    Can you please explain

    Integration by change of variable-untitled.jpg

    I don't understand how the "2" disappears, and how the limits change.
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  8. #8
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    Quote Originally Posted by signature View Post
    Can you please explain

    Click image for larger version. 

Name:	Untitled.jpg 
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ID:	14230

    I don't understand how the "2" disappears, and how the limits change.
    \frac{1}{1 + \cos^2{\theta}} is an even function, so it is symmetric about the f(\theta) axis.

    That is, \int^0_{-\pi}{\frac{1}{1 + \cos^2{\theta}} d\theta} = \int^\pi_0{\frac{1}{1 + \cos^2{\theta}} d\theta}

    So I hope it's clear that -2\int^0_{\pi}{\frac{1}{1 + \cos^2{\theta}} d\theta} = \int^\pi_{-\pi}{\frac{1}{1 + \cos^2{\theta}} d\theta}
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