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Math Help - Evaluate the Integral

  1. #1
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    Evaluate the Integral

    ∫sin³(x/2) from 0 to pi/3.

    I have..

    ∫sin²(x/2)sin(x/2)
    ∫[1-cos²(x/2)]sin(x/2)
    ∫sin(x/2) - cos²(x/2)sin(x/2)
    ∫sin(x/2) - ∫cos²(x/2)sin(x/2)

    I don't know where to go from there... Please help me.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by hlpplz View Post
    ∫sin³(x/2) from 0 to pi/3.

    I have..

    ∫sin²(x/2)sin(x/2)
    ∫[1-cos²(x/2)]sin(x/2)
    ∫sin(x/2) - cos²(x/2)sin(x/2)
    ∫sin(x/2) - ∫cos²(x/2)sin(x/2)

    I don't know where to go from there... Please help me.
    In the first integral, make the substitution u=\frac{x}{2}\implies \,du=\tfrac{1}{2}\,dx\implies 2\,du=\,dx. The new lower limit is u\!\left(0\right)=\frac{0}{2}=0 and the new upper limit is u\!\left(\frac{\pi}{3}\right)=\frac{\pi}{6}

    So the integral \int_0^{\frac{\pi}{3}}\sin\!\left(\frac{x}{2}\righ  t)\,dx\xrightarrow{u=\frac{x}{2}}{}2\int_0^{\frac{  \pi}{6}}\sin\!\left(u\right)\,du

    For the second integral, make the substitution u=\cos\!\left(\frac{x}{2}\right)\implies\,du=-\tfrac{1}{2}\sin\!\left(\frac{x}{2}\right)\,dx\imp  lies 2\,du=-\sin\!\left(\frac{x}{2}\right)\,dx. The new lower limit is u\!\left(0\right)=\cos\!\left(\frac{0}{2}\right)=\  cos\!\left(0\right)=1 and the new upper limit is u\!\left(\frac{\pi}{3}\right)=\cos\!\left(\frac{\p  i}{6}\right)=\frac{\sqrt{3}}{2}

    So the integral -\int_0^{\frac{\pi}{3}}\cos^2\!\left(\frac{x}{2}\ri  ght)\sin\!\left(\frac{x}{2}\right)\,dx\xrightarrow  {u=\cos\!\left(\frac{x}{2}\right)}{}2\int_0^{\frac  {\sqrt{3}}{2}} u^2\,du

    Can you take it from here and solve each integral?
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  3. #3
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    from that i got

    2cosu + 2/3 u³

    2cos x/2 + 2/3cos³(x/2)

    Evaluating: 0-pi/3

    [2cos pi/6 + 2/3cos³ pi/6] - [2cos0 + 2/3cos0]

    am i right so far?
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  4. #4
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    Quote Originally Posted by hlpplz View Post
    from that i got

    2cosu + 2/3 u³

    2cos x/2 + 2/3cos³(x/2)

    Evaluating: 0-pi/3

    [2cos pi/6 + 2/3cos³ pi/6] - [2cos0 + 2/3cos0]

    am i right so far?
    yes, but you can simplify further.

    what is cos pi/6
    and what is cos 0
    ??
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  5. #5
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    i think it's

    [2cos√3/2 + 2/3cos³√3/2] - [2 + 2/3]

    Should i just plug it in the calculator and get a numerical answer?
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