Answer
$$\int^{\pi}_{0}\frac{\sin t}{2-\cos t}dt=\ln3$$
Work Step by Step
$$A=\int^{\pi}_{0}\frac{\sin t}{2-\cos t}dt$$
We set $u=2-\cos t$, which means $$du=-(-\sin t)dt=\sin tdt$$
For $t=\pi$, we have $$u=2-\cos\pi=2-(-1)=3$$
For $t=0$, we have $$u=2-\cos0=2-1=1$$
Therefore, $$A=\int^3_1\frac{1}{u}du=\ln|u|\Big]^3_1$$ $$A=\ln3-\ln1=\ln3-0$$ $$A=\ln3$$
