Area of a Polar Region Practice Test

$\displaystyle \frac12\int_{0}^{\pi} (4\cos\theta)^2,d\theta$

$\displaystyle \int_{-\pi/2}^{\pi/2} (4\cos\theta)^2,d\theta$

$\displaystyle \frac12\int_{-\pi/2}^{\pi/2} (4\cos\theta)^2,d\theta$

$\displaystyle \frac12\int_{-\pi/2}^{\pi/2} 4\cos\theta,d\theta$

$\displaystyle \frac12\int_{-\pi}^{\pi} (4\cos\theta)^2,d\theta$