Answer
$$\int^{\pi}_0\sqrt{1-\sin^2t}dt=2$$
Work Step by Step
$$A=\int^{\pi}_0\sqrt{1-\sin^2t}dt$$
Use the identity $$1-\sin^2t=\cos^2t$$
That means $$A=\int^{\pi}_0\sqrt{\cos^2t}dt$$ $$A=\int^{\pi}_0|\cos t|dt$$
We have $\cos t\ge0$ on $[0,\pi/2]$ and $\lt0$ on $(\pi/2,\pi]$. Therefore,
$$A=\int_0^{\pi/2}\cos tdt-\int^\pi_{\pi/2}\cos tdt$$ $$A=\sin t\Big]^{\pi/2}_0-\sin t\Big]^\pi_{\pi/2}$$ $$A=(1-0)-(0-1)$$ $$A=1+1=2$$
