Question

The solution of $\cos y\log (\sec x+\tan x)dx=\cos x\log (\sec y+\tan y)dy$ is

• A. $\left[\log \right(\sec x+\tan x){]}^2-[\log (\sec y+\tan y){]}^2=c$
• B. $\left[\log \right(secx+\tan x){]}^2+[\log (\sec y+\tan y){]}^2=c$
• C. $\left[\log \right(\sec x-\tan x){]}^2-[\log (\sec y-\tan y){]}^2=c$
• D. $\left[\log \right(\sec x-\tan x){]}^2+[\log (\sec y-\tan y){]}^2=c$

Answer 1

The correct option is A $\left[\log \right(\sec x+\tan x){]}^2-[\log (\sec y+\tan y){]}^2=c$

Given,

$cosylog(secx+tanx)dx=cosxlog(secy+tany)dy$

or

$\frac{log(secy+tany)dy}{cosy}=\frac{log(secx+tanx)dx}{cosx}$

Integrate Both the sides

$\int \frac{log(secy+tany)dy}{cosy}=\int \frac{log(secx+tanx)dx}{cosx}$$-\left(1\right)$

Let u=$log(sey+tany)$

$du=\frac{1}{cosy}dy$

and $log(secx+tanx)=v$

$\frac{1}{cosx}=dv$

$\left(1\right)\to \int udu=\int vdv$

$\to \frac{u^2}{2}=\frac{v^2}{2}+c$

$u^2=v^2+2c$

$\left[log\right(secy+tany){]}^2-[log(secx+tanx){]}^2=c$