Find the area inside one leaf of the rose:
r = 4 sin(5θ)
Please help!
the sine functin is equal to zero at points of the form $\displaystyle \pi \cdot k \mbox{ where} k \in \mathbb{Z}$
so setting the argument equal to the above we get
$\displaystyle 5 \theta = \pi \cdot k \iff \theta =\frac{\pi \cdot k}{5}$
so our limits of integration are k=0 and k=1$\displaystyle 0 \mbox{and } \frac{\pi}{5}$
$\displaystyle A=\frac{1}{2}\int_{0}^{\pi/5}(4\sin(5 \theta))^2 d\theta$
$\displaystyle =8\int_{0}^{\pi/5}\sin^{2}(5\theta)d\theta$
I think you can get it from here.
Good luck