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Math Help - trig substitution

  1. #1
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    trig substitution

    trig substitution-trig-sub.jpg

    there is no numerator... now i'm confused. i know i use the trig sub is x=a tan theta
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    sqrt(x^2+2x)=sqrt(x^2+2x+1-1)=sqrt((x+1)^2-1)

    sub: t=x+1

    sqrt(t^2-1)

    sub: t=sec(k)
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  3. #3
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    ok so I get


    sqrt(sec^2(k)-1)d theta

    t=sec(k)
    so dt is the derivative of sec?
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  4. #4
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    Hyperbolic substitution is easier in this case than trigonometric...

    \int{\sqrt{x^2 + 2x}\,dx} = \int{\sqrt{x^2 + 2x + 1^2 - 1^2}\,dx}

     = \int{\sqrt{(x + 1)^2 - 1}\,dx}.

    Now let x + 1 = \cosh{t} so that dx = \sinh{t}\,dt and the integral becomes

    \int{\sqrt{\cosh^2{t} - 1}\,\sinh{t}\,dt}

     = \int{\sqrt{\sinh^2{t}}\,\sinh{t}\,dt}

     = \int{\sinh{t}\sinh{t}\,dt}

     = \int{\sinh^2{t}\,dt}

     = \int{\frac{\cosh{2t}}{2} - \frac{1}{2}\,dt}

     = \frac{\sinh{2t}}{4} - \frac{t}{2} + C

     = \frac{2\cosh{t}\sqrt{\cosh^2{t} - 1}}{2} - \frac{t}{2} + C

     = \frac{2(x + 1)\sqrt{(x + 1)^2 - 1}}{2} - \frac{\cosh^{-1}(x + 1)}{2} + C

     = \frac{(2x + 2)\sqrt{x^2 + 2x} - \cosh^{-1}(x + 1)}{2} + C.
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