Question: I'm trying to find the area of the region bounded by $r =|\sin \theta | $ in the $xy$-plane using the formula: 
\begin{align\*}
A &= \frac{1}{2}\int\_{0}^{\pi} (|\sin \theta|)^2 d\theta \\
&=\frac{1}{2}\int\_{0}^{\pi} (1-\cos2\theta)^2 d\theta \\
&= \frac{1}{2}\int\_{0}^{\pi} (1-2\cos\theta + \cos^22\theta)^2 d\theta \\
\end{align\*}

which I get $\frac{\pi}{4}$ as the answer but I got it wrong on a test.

The other options are:

1. $\frac{\pi}{2}$
2. $\frac{\pi}{4}$
3. $\pi$
4. $1$
5. $2$

Could there have been a mistake on the test or did I miss something?

Answer: 
Hint: square it, then make use of the fact that the 5 fifth roots of unity sum to 0.
Let $w=e^{2\pi i/5}$. Then $w^5=1$ and $w\ne1$ so $w^4+w^3+w^2+w+1=0$, $w^4=w^{-1},$ and $w^3=w^{-2}$.

Also, $\cos(2\pi i/5)=\dfrac{w+w^{-1}}2$, 

so $\cos^2(2\pi i/5)=\dfrac{w^2+w^{-2}+2}4=\dfrac{w^2+w^3+2}4=\dfrac{-1-w-w^{-1}+2}4=\dfrac{1-2\cos(2\pi i/5)}{4}.$

Can you take it from here?