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Math Help - Solving an integral using ash substitution

  1. #1
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    Solving an integral using ash substitution

    Hi there, I must solve this:

    \displaystyle\int_{}^{}x^2\sqrt[ ]{x^2+3}dx

    The statement says that I must solve it using an adequate trigonometric substitution.

    I followed this way:

    x=\sqrt[ ]{3}sh(t)
    dx=\sqrt[ ]{3}ch(t)dt

    I've used the identity: ch^2(t)-sh^2(t)=1\Rightarrow{ch(t)=\sqrt[ ]{1+sh^2(t)}}


    u=ch(t)
    du=sh(t)dt

    dv=ch(t)dt
    v=sh(t)

    \displaystyle\int_{}^{}ch^2(t)dt=ch(t)sh(t)-\displaystyle\int_{}^{}sh(t)sh(t)dt

    \displaystyle\int_{}^{}ch^2(t)dt=ch(t)sh(t)-\displaystyle\int_{}^{}sh^2(t)dt=ch(t)sh(t)-\displaystyle\int_{}^{}(ch^2(t)-1)dt

    * \displaystyle\int_{}^{}ch^2(t)dt=\displaystyle\fra  c{1}{2}(ch(t)sh(t)+t)


    9\displaystyle\int_{}^{}sh^2(t)ch^2(t)dt=9(\displa  ystyle\frac{sh^2(t)}{2}ch(t)sh(t)+t)-\displaystyle\int_{}^{}(ch(t)sh(t)+t)sh(t)ch(t)dt=  9(\displaystyle\frac{sh^2(t)}{2}ch(t)sh(t)+t)-\displaystyle\int_{}^{}ch(t)^2sh(t)^2dt-\displaystyle\int_{}^{}ch(t)sh(t)tdt

    There must be an easier way yo solve this.
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  2. #2
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    sh and ch are the hyperbolic function, sinh(t) and cosh(t)?

    One other method would be to use trig functions rather than hyperbolic functions. Since sin^2(t)+ cos^2(t)=1, dividing both sides by cos^2(t), tan^2(t)+ 1= sec^2(t). That suggests using the substitution x= \sqrt{3}tan(t). I doubt that that is any simpler but most people are more familiar with trig functions than with hyperbolic functions.
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  3. #3
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    Thanks HallsofIvy. I thought of that too, but it was easier to memorize the hyperbolic substitutions to me, and I did all those kind of integrations using hyperbolic substitutions when I could, and a\sin(t)=x in the others. I think you're right, but I'm not much familiarized with trigonometric identities, unless not much more than what I am with hyperbolic identities. But I'll try that way.

    Bye there.

    Oh, BTW

    Quote Originally Posted by HallsofIvy
    sh and ch are the hyperbolic function, sinh(t) and cosh(t)?
    The answer is yes :P
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  4. #4
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    Quote Originally Posted by Ulysses View Post
    Hi there, I must solve this:

    \displaystyle\int_{}^{}x^2\sqrt[ ]{x^2+3}dx

    The statement says that I must solve it using an adequate trigonometric substitution.

    I followed this way:

    x=\sqrt[ ]{3}sh(t)
    dx=\sqrt[ ]{3}ch(t)dt

    I've used the identity: ch^2(t)-sh^2(t)=1\Rightarrow{ch(t)=\sqrt[ ]{1+sh^2(t)}}


    u=ch(t)
    du=sh(t)dt

    dv=ch(t)dt
    v=sh(t)

    \displaystyle\int_{}^{}ch^2(t)dt=ch(t)sh(t)-\displaystyle\int_{}^{}sh(t)sh(t)dt

    \displaystyle\int_{}^{}ch^2(t)dt=ch(t)sh(t)-\displaystyle\int_{}^{}sh^2(t)dt=ch(t)sh(t)-\displaystyle\int_{}^{}(ch^2(t)-1)dt

    * \displaystyle\int_{}^{}ch^2(t)dt=\displaystyle\fra  c{1}{2}(ch(t)sh(t)+t)


    9\displaystyle\int_{}^{}sh^2(t)ch^2(t)dt=9(\displa  ystyle\frac{sh^2(t)}{2}ch(t)sh(t)+t)-\displaystyle\int_{}^{}(ch(t)sh(t)+t)sh(t)ch(t)dt=  9(\displaystyle\frac{sh^2(t)}{2}ch(t)sh(t)+t)-\displaystyle\int_{}^{}ch(t)^2sh(t)^2dt-\displaystyle\int_{}^{}ch(t)sh(t)tdt

    There must be an easier way yo solve this.
    \int{x^2\sqrt{x^2 + 3}\,dx}.

    Let x = \sqrt{3}\sinh{t} so that dx = \sqrt{3}\cosh{t}\,dt.

    The integral becomes

    \int{(\sqrt{3}\sinh{t})^2\sqrt{(\sqrt{3}\sinh{t})^  2 + 3}\,\sqrt{3}\cosh{t}\,dt}

     = \int{3\sinh^2{t}\sqrt{3\sinh^2{t} + 3}\,\sqrt{3}\cosh{t}\,dt}

     = \int{3\sqrt{3}\cosh{t}\sinh^2{t}\sqrt{3(\sinh^2{t} + 1)}\,dt}

     = \int{3\sqrt{3}\cosh{t}\sinh^2{t}\sqrt{3\cosh^2{t}}  \,dt}

     = \int{3\sqrt{3}\cosh{t}\sinh^2{t}\sqrt{3}\cosh{t}\,  dt}

     = \int{9\cosh^2{t}\sinh^2{t}\,dt}

     = 9\int{\left(\frac{1}{2}\cosh{2t} + \frac{1}{2}\right)\left(\frac{1}{2}\cosh{2t} - \frac{1}{2}\right)\,dt}

     = 9\int{\frac{1}{4}\cosh^2{2t} - \frac{1}{4}\,dt}

     = \frac{9}{4}\int{\cosh^2{2t} - 1\,dt}

     = \frac{9}{4}\int{\sinh^2{2t}\,dt}

     = \frac{9}{4}\int{\frac{1}{2}\cosh{4t} - \frac{1}{2}\,dt}

     = \frac{9}{8}\int{\cosh{4t} - 1\,dt}

     = \frac{9}{8}\left[\frac{1}{4}\sinh{4t} - t\right] + C

     = \frac{9}{32}\sinh{4t} - \frac{9}{8}t + C

     = \frac{9}{32}(2\sinh{2t}\cosh{2t}) - \frac{9}{8}t + C

     = \frac{9}{16}(\sinh{2t}\cosh{2t}) - \frac{9}{8}t + C

     = \frac{9}{16}[2\sinh{t}\cosh{t}(2\sinh^2{t} + 1)] - \frac{9}{8}t + C

     = \frac{9}{8}[\sinh{t}\sqrt{\sinh^2{t} + 1}(2\sinh^2{t} + 1)] - \frac{9}{8}t + C

     = \frac{9}{8}\left[\frac{x}{\sqrt{3}}\sqrt{\frac{x^2}{3} + 1}\left(\frac{2x^2}{3} + 1\right)\right] - \frac{9}{8}\sinh^{-1}\left(\frac{x}{\sqrt{3}}\right) + C

     = \frac{9}{8}\left[\frac{x}{\sqrt{3}}\sqrt{\frac{x^2 + 3}{3}}\left(\frac{2x^2 + 3}{3}\right)\right] - \frac{9\sinh^{-1}\left(\frac{\sqrt{3}x}{3}\right)}{8} + C

     = \frac{9}{8}\left[\frac{x}{\sqrt{3}}\frac{\sqrt{x^2 + 3}}{\sqrt{3}}\left(\frac{2x^2 + 3}{3}\right)\right] - \frac{9\sinh^{-1}\left(\frac{\sqrt{3}x}{3}\right)}{8} + C

     = \frac{9}{8}\left[\frac{x(2x^2 + 3)\sqrt{x^2 + 3}}{9}\right] - \frac{9\sinh^{-1}\left(\frac{\sqrt{3}x}{3}\right)}{8} +C

     = \frac{x(2x^2 + 3)\sqrt{x^2 + 3} - 9\sinh^{-1}\left(\frac{\sqrt{3}x}{3}\right)}{8} + C
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  5. #5
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    But notice in the original post, "The statement says that I must solve it using an adequate trigonometric substitution". (Emphasis mine.)
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  6. #6
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    Quote Originally Posted by HallsofIvy View Post
    But notice in the original post, "The statement says that I must solve it using an adequate trigonometric substitution". (Emphasis mine.)
    Yet the title of the thread says "using ash substitution", in other words, a "sinh" subsitution...
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