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