Question

Solve $\int \frac{1}{2-3\cos 2x}dx=\frac{1}{5}.\frac{1}{\frac{2}{\sqrt{\left(5\right)}}}\log \frac{\tan x-\frac{1}{\sqrt{\left(5\right)}}}{\tan x+\frac{1}{\sqrt{\left(5\right)}}}\log \frac{\sqrt{5}\tan x-1}{\sqrt{\left(5\right)\tan x+1}}.$

Answer 1

Let $I=\int \frac{dx}{2-3\cos 2x}=\int \frac{dx}{2-3(2{\cos }^{2}x-1)}$
$=\int \frac{dx}{5-6{\cos }^{2}x}$

Dividing numerator and denominator by ${\cos }^{2}x$

$=\int \frac{{\sec }^{2}xdx}{5(1+{\tan }^{2}x)-6}$
Put $\tan x=t$

$\Rightarrow {\sec }^{2}xdx=dt$
Therefore
$I=\int \frac{dt}{5t^2-1}=\frac{1}{5}\int \frac{dt}{t^2-{\left[1/\sqrt{\left(5\right)}\right]}^2}$]
$=\frac{1}{5}.\frac{1}{2\frac{1}{\sqrt{\left(5\right)}}}\log \frac{t-\frac{1}{\sqrt{\left(5\right)}}}{t+\frac{1}{\sqrt{\left(5\right)}}}+C$
$=\frac{1}{5}.\frac{1}{2\frac{1}{\sqrt{\left(5\right)}}}\log \frac{\tan x-\frac{1}{\sqrt{\left(5\right)}}}{\tan x+\frac{1}{\sqrt{\left(5\right)}}}+C$

$I=\frac{1}{5}.\frac{1}{2\frac{1}{\sqrt{\left(5\right)}}}\log \frac{\sqrt{5}\tan x-1}{\sqrt{5}\tan x+1}+C$