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Math Help - 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 u=sin(x)

    i end up with this sometime, which is correct:

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

    now my book says the result should be:

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

    and wofram alpha says it should be:

    \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 \sin^5( x) instead of \sin (x^5)?

    In that case:


    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= \frac{2\sqrt{\sin x}}{5}(5-\sqrt{\sin^4 x})+C=\frac{2\sqrt{\sin x}}{5}(5-\sin^2 x)+C= \frac{2\sqrt{\sin x}}{5}(4 + (1-\sin^2 x))+C= \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 u=sin(x)

    i end up with this sometime, which is correct:

    2\sqrt{sin(x)}-\frac{2*\sqrt{sin(x^5)}}{5}+C 2\sqrt{sin(x)}-\frac{2*\sqrt{(sin(x))^5}}{5}+CMr F says: You mean }" alt="{\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:

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

    and wofram alpha says it should be:

    \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
    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

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

    Now take out a factor of \frac{2 \sqrt{\sin x}}{5} and substitute \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 \cos (2x) = 2 \cos^2 (x) - 1 ....
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