Answer
$$\int_{0}^{\pi} \sqrt{1-\sin ^{2} t} d t =2$$
Work Step by Step
Given $$\int_{0}^{\pi} \sqrt{1-\sin ^{2} t} d t $$
So, we have
\begin{aligned}
I&=\int_{0}^{\pi} \sqrt{1-\sin ^{2} t} d t\\
&\text{since} \ \ \cos^2 t= 1- \sin^2 t ,\text{ we get}\\
&=\int_{0}^{\pi}|\cos t| d t\\
&\text { since} \ \cos t \geq 0 \text { for } t \in[0, \pi / 2] \text { and } \cos t \leq 0 \text { for } t \in[\pi / 2, \pi] \text {, we can separate the integral as } \\
I&=\int_{0}^{\pi / 2} \cos t d t+\int_{\pi / 2}^{\pi} (-\cos t) d t\\
&=[\sin t]_{0}^{\pi / 2}-[\sin t]_{\pi / 2}^{\pi}\\
&=\sin \pi/2-\sin0-\sin \pi +\sin\pi/2\\
&=1-0-0+1\\
&=2\\
\end{aligned}
