Answer
$$\int\cot t\sqrt{1-\sin^2t}dt=\sqrt{1-\sin^2t}-\ln\Big|\frac{1+\sqrt{1-\sin^2t}}{\sin t}\Big|+C$$
Work Step by Step
$$A=\int\cot t\sqrt{1-\sin^2t}dt$$ ($0\lt t\lt\pi/2$)
$$A=\int\frac{\cos t}{\sin t}\sqrt{1-\sin^2t}dt$$
We set $u=\sin t$, which means $$du=\cos tdt$$
Therefore, $$A=\int\frac{\sqrt{1-u^2}}{u}du$$
Here apply Formula 47, which states that
$$\int\frac{\sqrt{a^2-x^2}}{x}dx=\sqrt{a^2-x^2}-a\ln\Big|\frac{a+\sqrt{a^2-x^2}}{x}\Big|+C$$
for $a=1$.
$$A=\sqrt{1-u^2}-\ln\Big|\frac{1+\sqrt{1-u^2}}{u}\Big|+C$$ $$A=\sqrt{1-\sin^2t}-\ln\Big|\frac{1+\sqrt{1-\sin^2t}}{\sin t}\Big|+C$$
