# $\int_0^{2\pi}\sin\frac{x}{2}\cos^2\frac{x}{2}\,dx$

$$\int_0^{2\pi}\sin\frac{x}{2}\cos^2\frac{x}{2}\,dx$$

Tried substitution ($u = \cos\frac{x}{2}$), but I get $-\frac{\cos^3\frac{x}{2}}{3}$ ($-\frac{2}{3}$) instead of the correct answer, which is $1\frac{1}{3}$

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try $u = \cos(x/2)$ instead. – Dan Petersen Apr 23 '12 at 20:22
Sorry, I did try cos(x/2), slipped the ^2 in by mistake. – Lior Apr 23 '12 at 20:30
How did you get an answer that includes an $x$? The integration variable is meaningless without the $\int$. – Henning Makholm Apr 23 '12 at 20:43
@Lior : it looks as if you evaluated an indefinite integral and then didn't plug in the two numbers and subtract, to get the value of the definite integral. – Michael Hardy Apr 23 '12 at 23:04

It makes it a little easier to do this simple substitution for you to not mess up with constant multiple like you did (You got an extra $\frac{1}{3}$)

Substitute first $\frac{x}{2}=t$. To see the limits, when $x=2\pi, t=\pi$ and when $x=0, u=0$ and $\mathrm{d}x = 2\mathrm{d}t$

\begin{align*} \int_0^{2\pi}\sin\frac{x}{2}\cos^2\frac{x}{2} \mathrm{d}x &= 2 \int_0^{\pi}\sin t \hspace{3pt}\cos^2 t \hspace{3pt} \mathrm{d}t\\ &= 2 \left|\frac{-\cos^3 t}{3} \right|_0^{\pi}\\ &= 2 \left[\left(\frac{1}{3}\right) - \left( \frac{-1}{3} \right) \right]\\ &= 2 \left(\frac{2}{3}\right) = \frac{4}{3} \end{align*}

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$u = \cos \dfrac{x}{2} \Rightarrow du = - \dfrac{1}{2}\sin\dfrac{x}{2} \cdot dx$

Then your integrand becomes $-2 u^2 du$.

Can you take it from here?

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Don't forget that after finding an antiderivative ("$F(x)$"), you evaluate the definite integral as $F(b) - F(a)$, where $a,b$ are the limits of integration. – The Chaz 2.0 Apr 23 '12 at 20:54

You can successively reduce this to simple integrals using sum formulae, $$\int_0^{2\pi}\sin\frac{x}{2}\cos^2\frac{x}{2} dx = \frac{1}{2}\int_0^{2\pi}\sin(x)\cos\frac{x}{2} dx = \frac{2}{4}\int_0^{\pi}\left(\sin\frac{3x}{2} + \sin\frac{x}{2}\right) dx$$

This will give you, $\frac{1}{2} \times \frac{8}{3} = \frac{4}{3}$

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