Answer
$$\int_{0}^{\pi / 2} \frac{\cos t}{1+\sin t} d t =\ln 2$$
Work Step by Step
$$
\int_{0}^{\pi / 2} \frac{\cos t}{1+\sin t} d t
$$
Let $u =1+\sin t \ \ \to \ \ du =\cos t dt $ and
\begin{align*}
\text{at} \ t&=0\ \ \ \ \ \to \ u=1\\
\text{at} \ t&=\pi/2\ \to \ u=2
\end{align*}
Then
\begin{align*}
\int_{0}^{\pi / 2} \frac{\cos t}{1+\sin t} d t&=\int_{1}^{ 2} \frac{du}{u} \\
&=\ln u\bigg|_{1}^{2}\\
&=\ln 2
\end{align*}
