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Thread: Re-substituion problem

  1. #1
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    Re-substituion problem

    this is part of an integration, thats why i posted it here.
    the needed substitution here is $\displaystyle u=sin(x)$

    i end up with this sometime, which is correct:

    $\displaystyle 2\sqrt{sin(x)}-\frac{2*\sqrt{sin(x^5)}}{5}+C$

    now my book says the result should be:

    $\displaystyle \frac{2}{5}\sqrt{sin(x)}[4+cos^2(x)]+C$

    and wofram alpha says it should be:

    $\displaystyle \frac{1}{5}\sqrt{sin(x)}[cos(2x)+9]+C$

    to be honest... i tried and i tried, i dont get it how i will get any of those 2 solutions. would be so nice if someone could explain it me reeeeal slow

    thx in advance
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  2. #2
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    I think there's a typo in there. Did you mean $\displaystyle \sin^5( x)$ instead of $\displaystyle \sin (x^5)$?

    In that case:


    $\displaystyle 2\sqrt{\sin x}-\frac{2\sqrt{\sin^5 x}}{5}+C=2\cdot \frac{5}{5}\sqrt{\sin x}-\frac{2\sqrt{\sin x}\sqrt{\sin^4 x}}{5}+C=$ $\displaystyle \frac{2\sqrt{\sin x}}{5}(5-\sqrt{\sin^4 x})+C=\frac{2\sqrt{\sin x}}{5}(5-\sin^2 x)+C=$ $\displaystyle \frac{2\sqrt{\sin x}}{5}(4 + (1-\sin^2 x))+C=$ $\displaystyle \frac{2\sqrt{\sin x}}{5}(4+\cos^2 x)+C$
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  3. #3
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    oh yes that was a typo... thank you very much, all clear now!

    need to improve my algebra skills
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  4. #4
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    Quote Originally Posted by coobe View Post
    this is part of an integration, thats why i posted it here.
    the needed substitution here is $\displaystyle u=sin(x)$

    i end up with this sometime, which is correct:

    $\displaystyle 2\sqrt{sin(x)}-\frac{2*\sqrt{sin(x^5)}}{5}+C$ $\displaystyle {\color{red} 2\sqrt{sin(x)}-\frac{2*\sqrt{(sin(x))^5}}{5}+CMr F says: You mean }$.

    now my book says the result should be:

    $\displaystyle \frac{2}{5}\sqrt{sin(x)}[4+cos^2(x)]+C$

    and wofram alpha says it should be:

    $\displaystyle \frac{1}{5}\sqrt{sin(x)}[cos(2x)+9]+C$

    to be honest... i tried and i tried, i dont get it how i will get any of those 2 solutions. would be so nice if someone could explain it me reeeeal slow

    thx in advance
    $\displaystyle 2\sqrt{\sin(x)}-\frac{2 \sqrt{(\sin(x))^5}}{5}+C = 2\sqrt{\sin(x)}-\frac{2 \sqrt{(\sin(x))^4 \sin x}}{5}+C$

    $\displaystyle = 2\sqrt{\sin(x)}-\frac{2 \sin^2 x\sqrt{\sin x}}{5}+C $.

    Now take out a factor of $\displaystyle \frac{2 \sqrt{\sin x}}{5}$ and substitute $\displaystyle \sin^2 x = 1 - \cos^2 x$ to get your book's answer.

    To show that Wolfram's answer is equivalent to the book's answer you need to recall that $\displaystyle \cos (2x) = 2 \cos^2 (x) - 1$ ....
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