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Thread: Hard Integration Q

  1. #1
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    Hard Integration Q

    Find $\displaystyle \int \left ( \frac{x^2-1}{x^2+1}.\frac{1}{\sqrt{1+x^4}} \right )dx$

    I've tried all sorts of substituors for this one but i still cant seem to work it out. Can someone help out on this one?
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  2. #2
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    You could try saying $\displaystyle x^4$ = $\displaystyle tan^2 y$

    $\displaystyle dx = (sec^2 y)/[4(tan y)^{3/2}]$

    $\displaystyle x^2 = tan y$

    Which would give you 1/4 times the integral of (tan y - 1)/[(tan y)^(3/2)(tan y + 1)]

    I am not sure if that will lead any where and you may need another substitution.
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  3. #3
    Junior Member eddie2042's Avatar
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    not so sure about the final answer but if you set

    $\displaystyle

    u = x^2

    $

    you'll get

    $\displaystyle

    \frac {1} {\sqrt{1 + u^2}}

    $

    and i know the antiderivative of that is

    $\displaystyle

    \arcsin u

    $

    Good luck
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  4. #4
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    $\displaystyle \frac{x^{2}-1}{\left( x^{2}+1 \right)\sqrt{1+x^{4}}}=\frac{x^{2}-1}{x\left( x^{2}+1 \right)\sqrt{x^{2}+\frac{1}{x^{2}}}}=\frac{1-\frac{1}{x^{2}}}{\left( x+\frac{1}{x} \right)\sqrt{\left( x+\frac{1}{x} \right)^{2}-2}},$

    substitute $\displaystyle x+\frac1x=\frac{\sqrt2}t$ and your integral equals $\displaystyle -\frac1{\sqrt2}\arcsin\left(\frac{\sqrt2x}{1+x^2}\r ight)+k.$
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Krizalid View Post
    $\displaystyle \frac{x^{2}-1}{\left( x^{2}+1 \right)\sqrt{1+x^{4}}}=\frac{x^{2}-1}{x\left( x^{2}+1 \right)\sqrt{x^{2}+\frac{1}{x^{2}}}}=\frac{1-\frac{1}{x^{2}}}{\left( x+\frac{1}{x} \right)\sqrt{\left( x+\frac{1}{x} \right)^{2}-2}},$

    substitute $\displaystyle x+\frac1x=\frac{\sqrt2}t$ and your integral equals $\displaystyle -\frac1{\sqrt2}\arcsin\left(\frac{\sqrt2x}{1+x^2}\r ight)+k.$
    You show 'em son.
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  6. #6
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    Wow, just wow. You are unbelievable!

    Quote Originally Posted by Krizalid View Post
    $\displaystyle \frac{x^{2}-1}{\left( x^{2}+1 \right)\sqrt{1+x^{4}}}=\frac{x^{2}-1}{x\left( x^{2}+1 \right)\sqrt{x^{2}+\frac{1}{x^{2}}}}=\frac{1-\frac{1}{x^{2}}}{\left( x+\frac{1}{x} \right)\sqrt{\left( x+\frac{1}{x} \right)^{2}-2}},$

    substitute $\displaystyle x+\frac1x=\frac{\sqrt2}t$ and your integral equals $\displaystyle -\frac1{\sqrt2}\arcsin\left(\frac{\sqrt2x}{1+x^2}\r ight)+k.$
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  7. #7
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    Quote Originally Posted by vuze88 View Post
    Find $\displaystyle \int \left ( \frac{x^2-1}{x^2+1}.\frac{1}{\sqrt{1+x^4}} \right )dx$

    I've tried all sorts of substituors for this one but i still cant seem to work it out. Can someone help out on this one?

    Or if you never think of the wonderful substitution $\displaystyle t = x - 1/x $ , you can choose this substitution $\displaystyle x = \tan(y)$ which is familiar with all of us .


    $\displaystyle dx = \sec^2(y) dy $

    the integral becomes :

    $\displaystyle I = \int \frac{ \tan^2(y) - 1}{(\tan^2(y) + 1)\sqrt{1+ \tan^4(y)}} \sec^2(y)~ dy$

    $\displaystyle = \int \frac{ \tan^2(y) - 1}{\sqrt{1+ \tan^4(y)}}~ dy $

    $\displaystyle = \int \frac{ \cos^2(y) }{ \cos^2(y) }\cdot \frac{ \tan^2(y) - 1}{\sqrt{1+ \tan^4(y)}}~ dy $

    $\displaystyle = \int \frac{ \sin^2(y) - \cos^2(y) }{ \sqrt{ \sin^4(y) + \cos^4(y) }} ~dy $

    $\displaystyle = - \int \frac{ \cos(2y) }{ \sqrt{ (\sin^2(y) + \cos^2(y) )^2 - 2\sin^2(y) \cos^2(y) }}~dy $

    $\displaystyle = - \sqrt{2} \int \frac{ \cos(2y) }{ \sqrt{ 2 - \sin^2(2y) }}~dy$

    Haha , then substitute $\displaystyle \sin(2y) = t ~\implies 2\cos(2y)dy = dt $

    $\displaystyle I = - \frac{1}{\sqrt{2}} \int \frac{dt}{\sqrt{2 - t^2}}$

    $\displaystyle = - \frac{1}{\sqrt{2}}\arcsin(\frac{t}{\sqrt{2}}) + C $

    $\displaystyle t = \sin(2y) = \frac{2\tan(y)}{ \tan^2(y) + 1 } = \frac{2x}{x^2+1} $

    so

    $\displaystyle I = - \frac{1}{\sqrt{2}}\arcsin(\frac{\sqrt{2}x}{x^2+1}) + C $
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  8. #8
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    cool, thanks for that simplependulum
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  9. #9
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by simplependulum View Post
    Or if you never think of the wonderful substitution $\displaystyle t = x - 1/x $ , you can choose this substitution $\displaystyle x = \tan(y)$ which is familiar with all of us .


    $\displaystyle dx = \sec^2(y) dy $

    the integral becomes :

    $\displaystyle I = \int \frac{ \tan^2(y) - 1}{(\tan^2(y) + 1)\sqrt{1+ \tan^4(y)}} \sec^2(y)~ dy$

    $\displaystyle = \int \frac{ \tan^2(y) - 1}{\sqrt{1+ \tan^4(y)}}~ dy $

    $\displaystyle = \int \frac{ \cos^2(y) }{ \cos^2(y) }\cdot \frac{ \tan^2(y) - 1}{\sqrt{1+ \tan^4(y)}}~ dy $

    $\displaystyle = \int \frac{ \sin^2(y) - \cos^2(y) }{ \sqrt{ \sin^4(y) + \cos^4(y) }} ~dy $

    $\displaystyle = - \int \frac{ \cos(2y) }{ \sqrt{ (\sin^2(y) + \cos^2(y) )^2 - 2\sin^2(y) \cos^2(y) }}~dy $

    $\displaystyle = - \sqrt{2} \int \frac{ \cos(2y) }{ \sqrt{ 2 - \sin^2(2y) }}~dy$

    Haha , then substitute $\displaystyle \sin(2y) = t ~\implies 2\cos(2y)dy = dt $

    $\displaystyle I = - \frac{1}{\sqrt{2}} \int \frac{dt}{\sqrt{2 - t^2}}$

    $\displaystyle = - \frac{1}{\sqrt{2}}\arcsin(\frac{t}{\sqrt{2}}) + C $

    $\displaystyle t = \sin(2y) = \frac{2\tan(y)}{ \tan^2(y) + 1 } = \frac{2x}{x^2+1} $

    so

    $\displaystyle I = - \frac{1}{\sqrt{2}}\arcsin(\frac{\sqrt{2}x}{x^2+1}) + C $
    You're a trooper for writing that out.
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