# Math Help - Evaluate the Integral

1. ## 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)

2. Originally Posted by hlpplz
∫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)

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?

3. 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?

4. Originally Posted by hlpplz
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
??

5. 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?