Question

Solve:
$\int \frac{1}{2-3\cos 2x}dx$.

Answer 1

Now,

$\int \frac{1}{2-3\cos 2x}dx$

$=\int \frac{(1+{\tan }^{2}x)}{2(1+{\tan }^{2}x)-3(1-{\tan }^{2}x)}dx$ [ Since $\cos 2x=\frac{1-{\tan }^{2}x}{1+{\tan }^{2}x}$]

$=\int \frac{\left({\sec }^{2}x\right)}{-1+5{\tan }^{2}x}dx$

$=\frac{1}{5}\int \frac{\left({\sec }^{2}x\right)}{-\frac{1}{5}+{\tan }^{2}x}dx$

$=\frac{1}{5}\int \frac{d\left(\tan x\right)}{(\tan x{)}^2-{\left(\frac{1}{\sqrt{5}}\right)}^2}$

$=\frac{1}{5}.\frac{\sqrt{5}}{2}\log \left|\frac{\tan x-\frac{1}{\sqrt{5}}}{\tan x+\frac{1}{\sqrt{5}}}\right|+c$ [ Where $c$ being integrating constant]

$=\frac{1}{2\sqrt{5}}\log \left|\frac{\tan x-\frac{1}{\sqrt{5}}}{\tan x+\frac{1}{\sqrt{5}}}\right|+c$