Answer
$$\int^{\pi/6}_{-\pi/2}\frac{\cos t}{1-\sin t}dt=2\ln2$$
Work Step by Step
$$A=\int^{\pi/6}_{-\pi/2}\frac{\cos t}{1-\sin t}dt$$
Set $u=1-\sin t$, we have $$du=-\cos tdt$$ $$\cos tdt=-du$$
- For $t=\pi/6$, we have $$u=1-\sin\frac{\pi}{6}=1-\frac{1}{2}=\frac{1}{2}$$
- For $t=-\pi/2$, we have $$u=1-\sin\Big(-\frac{\pi}{2}\Big)=1-(-1)=2$$
Therefore, $$A=-\int^{1/2}_2\frac{1}{u}du$$ $$A=-\ln|u|\Big]^{1/2}_2$$ $$A=-(\ln\frac{1}{2}-\ln2)$$ $$A=\ln2-\ln\frac{1}{2}$$ $$A=\ln2-(\ln1-\ln2)=\ln2-(-\ln2)$$ $$A=2\ln2$$
