\(\int_0^{\pi / 2} \frac{\cos x \sin x}{1+\sin x} d x\) is equal to

\[
\text { } \begin{aligned}
\text { Let } I & =\int_0^{\pi / 2} \frac{\cos x \sin x}{1+\sin x} d x \\
& =\int_0^{\pi / 2} \frac{\cos x(1+\sin x-1)}{1+\sin x} d x \\
I & =\int_0^{\pi / 2}\left(\frac{\cos x(1+\sin x)}{1+\sin x}-\frac{\cos x}{1+\sin x}\right) d x
\end{aligned}
\]

\[
\begin{aligned}
& =\int_0^{\pi / 2} \cos x d x-\int_0^{\pi / 2} \frac{\cos x}{1+\sin x} d x=[\sin x]_0^{\pi / 2}-I_1 \\
I_1 & =\int_0^{\pi / 2} \frac{\cos x}{1+\sin x} d x
\end{aligned}
\]
Let \(1+\sin x=t\)
\[
\begin{aligned}
\Rightarrow \cos x d x & =d t \\
I_1 & =\int_0^{\pi / 2} \frac{d t}{t}=[\log t]_0^{\pi / 2}=[\log (1+\sin x)]_0^{\pi / 2} \\
& =\log (1+1)-\log (1+0)=\log 2
\end{aligned}
\]
Now, from Eq. (i), we have
\[
I=[1-0]-\log 2=1-\log 2
\]