Question

Solve $\underset{\pi /6}{\overset{\pi /3}{\int }}\frac{1}{\sin 2x}dx$

Answer 1

${\int }_{\pi /6}^{\pi /3}\frac{1}{\sin 2x}dx$

$={\int }_{\pi /6}^{\pi /3}\frac{1}{\left(\frac{2\tan x}{1+{\tan }^{2}x}\right)}dx$ $\{\therefore \sin 2x=\frac{2\tan x}{1+{\tan }^{2}x}\}$

$={\int }_{\pi /6}^{\pi /3}\frac{(1+{\tan }^{2}x)}{2\tan x}dx$

$={\int }_{\pi /6}^{\pi /3}\frac{{\sec }^{2}xdx}{2\tan x}$

$\tan x=t$

${\sec }^{2}xdx=dt$

$={\int }_{1/\sqrt{3}}^{\sqrt{3}}\frac{dt}{2t}$

$=\frac{1}{2}ln|t|{)}_{1/\sqrt{3}}^{\sqrt{3}}$

$=\frac{1}{2}(ln\sqrt{3}-ln\left(\frac{1}{\sqrt{3}}\right))$

$=\frac{1}{2}(ln\sqrt{3}+ln\sqrt{3})=ln\sqrt{3}=\frac{1}{2}ln3$.