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Math Help - Integral problem

  1. #1
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    Question Integral problem

    I'm at a loss with this question: Could I get some pointer please?

    By using substitution t = tanx evaluate the following definite integral:

    Limits zero to pi/4

    I = 1/(9((cosx)^2) - (sinx)^2) dx

    Apologies for the notation (or lack of)

    I'm stuck at getting this into a solveable integral

    D
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  2. #2
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    Quote Originally Posted by dojo View Post
    I'm at a loss with this question: Could I get some pointer please?

    By using substitution t = tanx evaluate the following definite integral:

    Limits zero to pi/4

    I = 1/(9((cosx)^2) - (sinx)^2) dx

    Apologies for the notation (or lack of)

    I'm stuck at getting this into a solveable integral

    D
    I guess I means integral.

    \int_{0}^{\frac{\pi}{4}}\left(\frac{1}{9cos^2(x)}-sin^2(x)\right)dx

    Is this what you mean?
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  3. #3
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    Well if that is what you meant, I would do this:

    \int_{0}^{\frac{\pi}{4}}\left(\frac{1}{9cos^2(x)}-sin^2(x)\right)dx=\frac{1}{9}\int_{0}^{\frac{\pi}{  4}}\left(\frac{2}{1+cos(2x)}\right )dx-\int_{0}^{\frac{\pi}{4}}\left(\frac{1-cos(2x)}{2}\right)

    =\frac{2}{9}\int_{0}^{\frac{\pi}{4}}dx+\frac{1}{9}  \int_{0}^{\frac{\pi}{4}}sec(2x)(2dx)-\frac{1}{2}\int_{0}^{\frac{\pi}{4}}dx+...
    Last edited by dwsmith; May 18th 2010 at 11:33 AM.
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  4. #4
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    No the parenthesis includes all the trig functions as a denominator.
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  5. #5
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    brackets indicate, rather...

    integrate 1/(9 cos^2 x - sin^2 x) - Wolfram|Alpha

    Click on 'show steps'. Basically, use pythag to write cos in terms of sin, then use the indentity

    \sin x = \frac{2 \tan(\frac{x}{2})}{\tan^2(\frac{x}{2}) + 1}

    then substitute x = 2 arctan u, which is called the Weierstrass sub
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  6. #6
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    Yes that is the correct integral. But how do you solve it with the tanx = t substitution???
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  7. #7
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    Quote Originally Posted by dojo View Post
    Yes that is the correct integral. But how do you solve it with the tanx = t substitution???
    Yes, they've asked for x = arctan t rather than x = 2 arctan t, so that may be an even better route
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  8. #8
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    Talking

    Superduper! Thanks
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  9. #9
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    Quote Originally Posted by dojo View Post
    Yes that is the correct integral. But how do you solve it with the tanx = t substitution???
    t = \tan{x}

    x = \arctan{t}

    dx = \frac{dt}{1+t^2}

    \cos^2{x} = \frac{1}{t^2+1}

    \sin^2{x} = \frac{t^2}{t^2+1}


    \int_0^{\frac{\pi}{4}} \frac{dx}{9\cos^2{x} - \sin^2{x}}<br />

    substitute ...

    \int_0^1 \frac{dt}{9 - t^2}

    use partial fractions to finish
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