Answer
$$
\int_{0}^{\pi / 2} \frac{\cos \theta}{\sqrt{\sin \theta}} d \theta
$$
Observe that the given integral is improper because $ f(\theta)= \frac{\cos \theta}{\sqrt{\sin \theta}} $ has the vertical asymptote $\theta =0 $. The infinite discontinuity occurs at the left endpoint of $ [0, \pi / 2] $
The given improper integral is convergent and its value is
$$
\int_{0}^{\pi / 2} \frac{\cos \theta}{\sqrt{\sin \theta}} d \theta = 2
$$
Work Step by Step
$$
\int_{0}^{\pi / 2} \frac{\cos \theta}{\sqrt{\sin \theta}} d \theta
$$
Observe that the given integral is improper because $ f(\theta)= \frac{\cos \theta}{\sqrt{\sin \theta}} $ has the vertical asymptote $\theta =0 $. Since the infinite discontinuity occurs at the left endpoint of $ [0, \pi / 2] $ , we must use part (b) of Definition 3 :
$$
\begin{split}
\int_{0}^{\pi / 2} \frac{\cos \theta}{\sqrt{\sin \theta}} d \theta & = \lim _{t \rightarrow 0 ^{+}} \int_{t}^{\pi / 2} \sin ^{-\frac{1}{2 }}\theta (\cos \theta d \theta ) \\
& =\lim _{t \rightarrow 0 ^{+}} \int_{t}^{\pi / 2} \sin ^{-\frac{1}{2 }}\theta ( d \sin \theta ) \\
& =\lim _{t \rightarrow 0^{+}}\left[2 \sin ^{\frac{1}{2 }}\theta \right]_{t}^{\pi / 2} \\
& =2 \lim _{t \rightarrow 0^{+}}\left[ \sin ^{\frac{1}{2 }}(\pi / 2) -\sin ^{\frac{1}{2 }} t \right] \\
& =2 \lim _{t \rightarrow 0^{+}}\left[ 1 -\sin ^{\frac{1}{2 }} t \right]\\
& = 2
\end{split}
$$
Thus the given improper integral
$$
\int_{0}^{\pi / 2} \frac{\cos \theta}{\sqrt{\sin \theta}} d \theta = 2
$$
is convergent.
