Answer
$$\int^{\pi/3}_{0}\frac{4\sin\theta}{1-4\cos\theta}d\theta=-\ln3$$
Work Step by Step
$$A=\int^{\pi/3}_{0}\frac{4\sin\theta}{1-4\cos\theta}d\theta$$
We set $u=1-4\cos\theta$, which means $$du=-(-4\sin\theta)d\theta=4\sin\theta d\theta$$
For $\theta=\pi/3$, we have $$u=1-4\cos(\pi/3)=1-4\times\frac{1}{2}=-1$$
For $\theta=0$, we have $$u=1-4\cos0=1-4\times1=-3$$
Therefore, $$A=\int^{-1}_{-3}\frac{1}{u}du=\ln|u|\Big]^{-1}_{-3}$$ $$A=\ln1-\ln3=0-\ln3$$ $$A=-\ln3$$
