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Math Help - Trig substitution

  1. #1
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    Trig substitution

    <br /> <br />
\int\frac{dx}{4+x^2}\<br /> <br />

    I am supposed to use trig substitution here. I used a tangent sub but then everything seems to cancel itself out. Which sub would I need to use here?
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  2. #2
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    You're correct, use x=2tan{\theta}, \;\ dx=2sec^{2}{\theta}d{\theta}

    \int\frac{2sec^{2}{\theta}}{4+4tan^{2}{\theta}}d{\  theta}

    \int\frac{2sec^{2}{\theta}}{4(1+tan^{2}{\theta})}d  {\theta}

    \frac{1}{4}\int\frac{2sec^{2}{\theta}}{sec^{2}{\th  eta}}d{\theta}

    Yes, the sec^{2}{\theta} does cancel. Which makes it easy.

    \frac{1}{4}\int 2d{\theta}

    \frac{1}{2}\int d{\theta}

    \frac{\theta}{2}

    But, from our sub, {\theta}=tan^{-1}(\frac{x}{2})

    So, we get:

    \frac{tan^{-1}(\frac{x}{2})}{2}
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ur5pointos2slo View Post
    <br /> <br />
\int\frac{dx}{4+x^2}\<br /> <br />

    I am supposed to use trig substitution here. I used a tangent sub but then everything seems to cancel itself out. Which sub would I need to use here?
    there is a straight forward formula for this, but if you want to do the trig sub, that's fine. you have an expression of the form x^2 + a^2 thus your substitution must be x = a \tan \theta. that is, x = 2 \tan \theta
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  4. #4
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    Quote Originally Posted by ur5pointos2slo View Post
    <br /> <br />
\int\frac{dx}{4+x^2}\<br /> <br />

    I am supposed to use trig substitution here. I used a tangent sub but then everything seems to cancel itself out. Which sub would I need to use here?
    x = 2\tan{t}

    dx = 2\sec^2{t} \, dt

    \int \frac{1}{4 + x^2} \, dx

    \int \frac{1}{4(1 + \tan^2{t})} \, 2\sec^2{t} \, dt

    \frac{1}{2}\int \, dt

    \frac{1}{2} t + C

    \frac{1}{2} \arctan\left(\frac{x}{2}\right) + C
    Last edited by skeeter; April 5th 2009 at 09:48 AM. Reason: way too late ...
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  5. #5
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    Quote Originally Posted by Jhevon View Post
    there is a straight forward formula for this, but if you want to do the trig sub, that's fine. you have an expression of the form x^2 + a^2 thus your substitution must be x = a \tan \theta. that is, x = 2 \tan \theta
    I couldn't think of a formula that this integral could fit. Which formula are you referring to?
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  6. #6
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    Quote Originally Posted by galactus View Post
    You're correct, use x=2tan{\theta}, \;\ dx=2sec^{2}{\theta}d{\theta}

    \int\frac{2sec^{2}{\theta}}{4+4tan^{2}{\theta}}d{\  theta}

    \int\frac{2sec^{2}{\theta}}{4(1+tan^{2}{\theta})}d  {\theta}

    \frac{1}{4}\int\frac{2sec^{2}{\theta}}{sec^{2}{\th  eta}}d{\theta}

    Yes, the sec^{2}{\theta} does cancel. Which makes it easy.

    \frac{1}{4}\int 2d{\theta}

    \frac{1}{2}\int d{\theta}

    \frac{\theta}{2}

    But, from our sub, {\theta}=tan^{-1}(\frac{x}{2})

    So, we get:

    \frac{tan^{-1}(\frac{x}{2})}{2}
    I forgot all about the dtheta in there. I ended up with just the 1/2. Thanks a lot
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ur5pointos2slo View Post
    I couldn't think of a formula that this integral could fit. Which formula are you referring to?
    on the inside cover of your calc text, there should be a formula that looks a little like this: \int \frac {dx}{a^2 + x^2} = \frac 1a \tan^{-1} \frac xa + C
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  8. #8
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    Quote Originally Posted by Jhevon View Post
    on the inside cover of your calc text, there should be a formula that looks a little like this: \int \frac {dx}{a^2 + x^2} = \frac 1a \tan^{-1} \frac xa + C

    O yes I totally forgot all about that. thanks for the reminder!
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