Answer
$$\int^{\pi/2}_{-\pi/2}\frac{2\cos\theta d\theta}{1+(\sin\theta)^2}=\pi$$
Work Step by Step
$$A=\int^{\pi/2}_{-\pi/2}\frac{2\cos\theta d\theta}{1+(\sin\theta)^2}$$
We set $u=\sin\theta$, which means $$du=\cos\theta d\theta$$
For $x=\pi/2$, we have $$u=\sin\frac{\pi}{2}=1$$
For $x=-\pi/2$, we have $$u=\sin\Big(-\frac{\pi}{2}\Big)=-1$$
Therefore, $$A=2\int^1_{-1}\frac{1}{1+u^2}du=2\tan^{-1}u\Big]^1_{-1}$$ $$A=2(\tan^{-1}1-\tan^{-1}(-1))$$ $$A=2\Big(\frac{\pi}{4}-\Big(-\frac{\pi}{4}\Big)\Big)$$ $$A=2\Big(\frac{\pi}{2}\Big)=\pi$$
