Answer
$-\dfrac{1}{2}\ln \space|\dfrac{2+\sqrt{4-\sin^2 (t)}}{\sin t}|+C$
Work Step by Step
Let $I=\int\dfrac{\cos t \space dt}{\sin t\sqrt{4-\sin^2t}}$
Suppose $u=\sin t \implies du=\cos t \space dt$
Therefore, $I=\int\dfrac{du}{u\sqrt{4-u^2}}du \space \\
=-\dfrac{1}{2}\ln|\dfrac{2+\sqrt{4-u^2}}{u}|+C \\=-\dfrac{1}{2}\ln \space|\dfrac{2+\sqrt{4-\sin^2 (t)}}{\sin t}|+C$