∫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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- March 8th 2009, 07:53 PMhlpplzEvaluate 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. - March 8th 2009, 08:01 PMChris L T521
In the first integral, make the substitution . The new lower limit is and the new upper limit is

So the integral

For the second integral, make the substitution . The new lower limit is and the new upper limit is

So the integral

Can you take it from here and solve each integral? - March 8th 2009, 08:12 PMhlpplz
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? - March 8th 2009, 08:27 PMGaloisTheory1
- March 8th 2009, 08:32 PMhlpplz
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?