Answer
$$\int\frac{dt}{\tan t\sqrt{4-\sin^2t}}=-\frac{1}{2}\ln\Big|\frac{2+\sqrt{4-\sin^2t}}{\sin t}\Big|+C$$
Work Step by Step
$$A=\int\frac{dt}{\tan t\sqrt{4-\sin^2t}}$$ $$A=\int\frac{\cos tdt}{\sin t\sqrt{4-\sin^2t}}$$
We set $u=\sin t$, which means $$du=\cos tdt$$
Therefore, $$A=\int\frac{du}{u\sqrt{4-u^2}}du$$
Here apply Formula 50, which states that
$$\int\frac{dx}{x\sqrt{a^2-x^2}}=-\frac{1}{a}\ln\Big|\frac{a+\sqrt{a^2-x^2}}{x}\Big|+C$$
for $a=2$.
$$A=-\frac{1}{2}\ln\Big|\frac{2+\sqrt{4-u^2}}{u}\Big|+C$$ $$A=-\frac{1}{2}\ln\Big|\frac{2+\sqrt{4-\sin^2t}}{\sin t}\Big|+C$$
