Question

If $\int \frac{1}{{\cos }^{2}x(1-4{\tan }^{2}x)}dx=K\cdot \ln \left|\frac{1+2\tan x}{1-2\tan x}\right|+C,$ then $\frac{1}{K}$ is
(where $C$ is integration constant)

Answer 1

$I=\int \frac{1}{{\cos }^{2}x(1-4{\tan }^{2}x)}dx$
Put $\tan x=t$
$\Rightarrow {\sec }^{2}xdx=dt\Rightarrow I=\int \frac{dt}{1-(2t{)}^2}=\int \frac{1}{2}\cdot \frac{(1+2t)+(1-2t)}{(1-2t)(1+2t)}dt=\frac{1}{2}[\frac{-\ln |1-2t|}{2}+\frac{\ln |1+2t|}{2}]=\frac{1}{4}\ln \left|\frac{1+2t}{1-2t}\right|+C=\frac{1}{4}\ln \left|\frac{1+2\tan x}{1-2\tan x}\right|+C$
$\Rightarrow K=\frac{1}{4}$