Question

$\underset{\mathrm{\pi }/6}{\overset{\mathrm{\pi }/3}{\int }}\frac{1}{\sin 2x}dx$ is equal to

(a) log_e 3

(b) ${\log }_e\sqrt{3}$

(c) $\frac{1}{2}\log \left(-1\right)$

(d) log (−1)

Answer 1

(b) ${\log }_e\sqrt{3}$

$$\begin{gathered} {\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{3}}\frac{1}{\sin 2x}dx \\ ={\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{3}}\cos ec2xdx \\ =\frac{1}{2}{\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{3}}2\cos ec2xdx \\ =\frac{-1}{2}{\left[\log \left(\cos ec2x+cot2x\right)\right]}_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{3}} \\ =\frac{-1}{2}\left[-2\log \sqrt{3}\right] \\ =\log \sqrt{3} \end{gathered}$$