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Math Help - What is wrong with my solution?

  1. #1
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    What is wrong with my solution?

    Given:

    \int sin \frac{x}{2}cos \frac{x}{2}

    -2 \int udu

    -2\frac{u^{2}}{2}+c

    -u^{2}+c

    -cos\frac{x}{2}^{2}+c


    the answer should be:

    -\frac{1}{2}cos[x]



    What's wrong with my solution?

    any help will be appreciated, thanks
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  2. #2
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    your solution is actually correct with the method you applied


    to get the other answer make use of the identity \sin 2x=2\sin x\cos x to subsitute and then integrate
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  3. #3
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    Quote Originally Posted by honestliar View Post
    Given:

    \int sin \frac{x}{2}cos \frac{x}{2}

    -2 \int udu

    -2\frac{u^{2}}{2}+c

    -u^{2}+c

    -cos\frac{x}{2}^{2}+c


    the answer should be:

    -\frac{1}{2}cos[x]



    What's wrong with my solution?

    any help will be appreciated, thanks
    The simplest approach is to note that the integrand is the same as \frac{1}{2} \sin (x) so you get -\frac{1}{2} \cos (x) + K as the answer.

    Now you should note that the difference between this and your answer of - \cos \left( \frac{x}{2}\right) ^{2}+c (note the brackets by the way) is merely a constant ....
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  4. #4
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    Quote Originally Posted by mr fantastic View Post
    The simplest approach is to note that the integrand is the same as \frac{1}{2} \sin (x) so you get -\frac{1}{2} \cos (x) + K as the answer.

    Now you should note that the difference between this and your answer of - \cos \left( \frac{x}{2}\right) ^{2}+c (note the brackets by the way) is merely a constant ....

    But the given is not just \frac{1}{2} \sin (x)
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  5. #5
    Super Member Aryth's Avatar
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    Recall the identity:

    sin(x) = 2sin\left(\frac{x}{2}\right)cos\left(\frac{x}{2}\r  ight)
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  6. #6
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    Thank you, I got it.
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