Answer
Convergent $\;,\;$ 2
Work Step by Step
Let \[I=\int_{0}^{\frac{π}{2}}\frac{\cos\theta}{\sqrt{\sin\theta}} \,d\theta\]
Since 0 is point of infinite discontinuity of integrand $\large\frac{\cos\theta}{\sqrt{\sin\theta}}$
\[\Rightarrow I=\lim_{t\rightarrow 0^+}\int_{t}^{\frac{π}{2}}\frac{\cos\theta}{\sqrt\sin{\theta}} \,d\theta\;\;\;\ldots (1)\]
Let \[I_1=\int\frac{\cos\theta}{\sqrt{\sin\theta}} \,d\theta\]
Substitute $\;r=\sin \theta\;\;\;\ldots (2)$
\[\Rightarrow dr=\cos \theta d\theta\]
\[I_1=\int\frac{dr}{r^{\frac{1}{2}}}=\int r^{-\frac{1}{2}}dr\]
\[I_1=2\sqrt r\]
From (2)
\[I_1=2\sqrt {\sin\theta}\;\;\;\ldots(3)\]
Using (3) in (1)
\[I=\lim_{t\rightarrow 0^+}\left[2\sqrt {\sin\theta}\right]_{t}^{\frac{π}{2}}\]
\[I=\lim_{t\rightarrow 0^+}\left[2-2\sqrt {\sin t}\right]=2\]
Since limit on R.H.S. of (1) exists so $I$ is convergent and $I=2$.
