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Math Help - Calculus Integral

  1. #1
    HMV
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    Calculus Integral

    Int(x/sqr(x^2+4x+8))dx
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by HMV View Post
    Int(x/sqr(x^2+4x+8))dx
    This is the same as \int\frac{x\,dx}{\sqrt{x^2+4x+4+4}}=\int\frac{x\,d  x}{\sqrt{(x+2)^2+4}}=\int\frac{x\,dx}{\sqrt{(x+2)^  2+2^2}}

    Now apply the trigonometric substitution x+2=2\tan\theta

    Can you try to take it from here?
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  3. #3
    HMV
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    Is int (secx) the right answer?
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by HMV View Post
    Is int (secx) the right answer?
    Not quite...

    You should simplify the integral down to 2\int \left(\sec\theta\tan\theta-\sec \theta\right)\,d\theta after applying the substitution. Keep in mind that \int \sec\theta\,d\theta=\ln\left|\sec\theta+\tan\theta  \right|+C

    Can you continue on?
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  5. #5
    HMV
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    I can certainly work the integral from there. Would you mind explainimg how you got to that simplyfied form of the integral?
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  6. #6
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by HMV View Post
    I can certainly work the integral from there. Would you mind explainimg how you got to that simplyfied form of the integral?
    Sure

    Our integral was \int\frac{x\,dx}{\sqrt{(x+2)^2+2^2}}

    Applying the trig substitution x+2=2\tan\theta\implies x=2\left(\tan\theta-1\right), we see that \,dx=2\sec^2\theta\,d\theta.

    Substituting all this into the integral, we have \int\frac{2\left(\tan\theta-1\right)2\sec^2\theta\,d\theta}{\sqrt{4\left(\tan^  2\theta+1\right)}}=4\int\frac{\left(\tan\theta-1\right)\sec^2\theta\,d\theta}{2\sec\theta} =2\int\left(\tan\theta-1\right)\sec\theta\,d\theta=2\int\left(\sec\theta\  tan\theta-\sec\theta\right)\,d\theta

    Does this make sense?
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  7. #7
    Newbie madmartigano's Avatar
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    Here's an alternate way of solving this problem. It might be easier or harder for you, depending on what you're used to.

    Setup the integral as \int {\frac{{x dx}}{{\sqrt {{{(x + 2)}^2} + 4} }}} <br />

    Let x + 2 = 2\sinh \theta, since {\sinh ^2}\theta  + 1 = {\cosh ^2}\theta

    so dx = 2\cosh \theta  d\theta and x = 2\sinh \theta  - 2<br />

    \int {\frac{{(2\sinh \theta - 2)(2\cosh \theta )d\theta }}{{\sqrt {4{{\sinh }^2}\theta + 4} }}} = \frac{4}{2}\int {\frac{{\sinh \theta \cosh \theta - {{\cosh }}\theta }}{{\sqrt {{{\sinh }^2}\theta + 1} }} d\theta }

     = 2\int {\frac{{\sinh \theta \cosh \theta - \cosh \theta }}{{\sqrt {{{\cosh }^2}\theta } }}} d\theta = 2\int {\frac{{\sinh \theta \cosh \theta }}{{\cosh \theta }} d\theta - 2\int {\frac{{\cosh \theta }}{{\cosh \theta }} d\theta } } <br />

    = 2\cosh \theta  - 2\theta  + c

    remember that x + 2 = 2\sinh \theta , so \theta  = {\sinh ^{ - 1}}\left( {\frac{{x + 2}}{2}} \right)<br />

    use the identity {\sinh ^2}\theta  + 1 = {\cosh ^2}\theta <br />
to show:

    \cosh \theta = \sqrt {1 + {{\sinh }^2}\theta } = \sqrt {1 + {{\left( {\frac{{x + 2}}{2}} \right)}^2}} = \sqrt {\frac{{{x^2}}}{4} + x + 2} = \frac{1}{2}\sqrt {{x^2} + 4x + 8} <br />

    and we have our answer:

    2\cosh \theta  - 2\theta  + c = \sqrt {{x^2} + 4x + 8}  - 2{\sinh ^{ - 1}}\left( {\frac{{x + 2}}{2}} \right) + c<br />
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