Answer
Divergent
Work Step by Step
\[I=\int_{0}^{\infty}\sin^2{\alpha}\:d\alpha\]
\[I=\lim_{t\rightarrow \infty}\int_{0}^{t}\sin^2{\alpha}\:d\alpha\;\;\;\ldots (1)\]
Let\[I_1=\int\sin^2{\alpha}\:d\alpha\]
\[\left[\sin^2 \theta=\frac{1-\cos 2\theta}{2}\right]\]
\[I_1=\int\left( \frac{1-\cos 2\alpha}{2} \right)\:d\alpha\]
\[I_1=\frac{\alpha}{2}-\frac{\sin 2\alpha}{4}\;\;\;\ldots(2)\]
Using (2) in (1)
\[I=\lim_{t\rightarrow \infty}\left[\frac{\alpha}{2}-\frac{\sin 2\alpha}{4}\right]_{0}^{t}\]
\[I=\lim_{t\rightarrow \infty}\left[\frac{t}{2}-\frac{\sin 2t}{4}-0\right]\]
$\;\;\;\;\;\;\;\;\; $ does not exist
Since limit on R.H.S. of (1) does not exist so $I$ is divergent.
