Question

Solve $\frac{\ln (\sec x+\tan x)}{\cos x}dx=\frac{\ln (\sec y+\tan y)}{\cos y}dy$

Answer 1

Given differential equation is

$\frac{\ln (\sec x+\tan x)}{\cos x}dx=\frac{\ln (\sec y+\tan y)}{\cos y}dy$

Integrating we get,

$\int \frac{\ln (\sec x+\tan x)}{\cos x}dx=\int \frac{\ln (\sec y+\tan y)}{\cos y}dy$

Put $z=\ln (\sec x+\tan x)$ in LHS

$\Rightarrow \frac{dz}{dx}=\frac{\sec x.\tan x+{\sec }^{2}x}{\sec x+\tan x}$

$\Rightarrow \frac{dz}{dx}=\frac{\sec x(\tan x+\sec x)}{\sec x+\tan x}$

$\Rightarrow dz=\frac{dx}{\cos x}$

Put $t=\ln (\sec y+\tan y)$ in RHS

$\Rightarrow \frac{dt}{dx}=\frac{\sec y.\tan y+{\sec }^{2}y}{\sec x+\tan x}$

$\Rightarrow \frac{dt}{dy}=\frac{\sec y(\tan y+\sec y)}{\sec y+\tan y}$

$\Rightarrow dt=\frac{dy}{\cos y}$

So, the differential equation becomes

$\int zdz=\int tdt$

$\Rightarrow \frac{z^2}{2}=\frac{t^2}{2}+C$

$\Rightarrow z^2=t^2+c$

$\Rightarrow {\ln }^2(\sec x+\tan x)={\ln }^2(\sec y+\tan y)+c$