Answer
$$\int^\pi_0\tan\frac{x}{3}dx=3\ln2$$
Work Step by Step
$$A=\int^\pi_0\tan\frac{x}{3}dx=\int^\pi_0\frac{\sin\frac{x}{3}}{\cos\frac{x}{3}}dx$$
Set $u=\cos\frac{x}{3}$, which means $$du=-\frac{1}{3}\sin\frac{x}{3}dx$$ $$\sin\frac{x}{3}dx=-3du$$
- For $x=\pi$, we have $u=\cos\frac{\pi}{3}=\frac{1}{2}$
- For $x=0$, we have $u=\cos0=1$
Therefore, $$A=-3\int^{1/2}_1\frac{1}{u}du$$ $$A=-3\ln|u|\Big]^{1/2}_1$$ $$A=-3(\ln(1/2)-\ln1)$$ $$A=-3\ln(1/2)$$ $$A=-3(\ln1-\ln2)$$ $$A=-3(-\ln2)=3\ln2$$
