Answer
$$\int^{\pi/2}_{0}\tan\frac{x}{2}dx=\ln2$$
Work Step by Step
$$A=\int^{\pi/2}_{0}\tan\frac{x}{2}dx=\int^{\pi/2}_{0}\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}dx$$
We set $u=\cos\frac{x}{2}$, which means $$du=-\frac{1}{2}\sin\frac{x}{2}dx$$ $$\sin\frac{x}{2}dx=-2du$$
For $x=\pi/2$, we have $$u=\cos\frac{\pi}{4}=\frac{\sqrt2}{2}$$
For $x=0$, we have $$u=\cos0=1$$
Therefore, $$A=-2\int^{\sqrt2/2}_1\frac{1}{u}du$$ $$A=-2\ln|u|\Big]^{\sqrt2/2}_1$$ $$A=-2(\ln\frac{\sqrt2}{2}-\ln1)$$ $$A=-2\ln\frac{\sqrt2}{2}=\ln\Big(\frac{\sqrt2}{2}\Big)^{-2}$$ $$A=\ln\frac{4}{2}=\ln2$$
