# Area that lies inside both curves: $r=sin2\theta, r=cos2\theta$

My integral is setup as: $$A=8\int_0^\frac{\pi}8{\frac12sin^22\theta}\space d\theta - 8\int_{\frac{\pi}8}^0{\frac12cos^2 2\theta}\space d\theta$$

$$=8\int_0^\frac{\pi}8{\frac12sin^22\theta}\space d\theta + 8\int_0^{\frac{\pi}8}{\frac12cos^2 2\theta}\space d\theta$$

$$=4\int_0^\frac{\pi}8({sin^22\theta+cos^22\theta})\space d\theta$$

$$=4\int_0^\frac{\pi}8{\theta}\space d\theta$$

which gives me $\frac\pi2$.

My $8$ comes from symmetry.

How did the book get their first integral? Why is there only $sin^22\theta$? What happened to $cos2\theta$? And where did their $2$ after the $8$ come from?

Study the diagram carefully. When the angle of integration is from $0 \le \theta \le \pi/8$, we see that the $r = \sin 2\theta$ curve is inside the $r = \cos 2\theta$ curve, so the area of the shaded region is just limited to the integral of $\frac{1}{2}r^2(\theta) \, d\theta$. But this only takes care of $1/16$th of the total area--the 8 comes from the 8-fold symmetry, but the 2 comes from the fact that each "petal" is made up of 2 symmetric halves; for example, from $0 \le \theta \le \pi/8$ and then from $\pi/8 \le \theta \le \pi/4$.