What is the solution of the differential equation \[\cos y log\left( {\sec x + \tan x} \right)dx = \cos x log\left( {\sec y + \tan y} \right)dy\]?
A. \[\sec^{2}x + \sec^{2}y = c\]
B. \[\sec x + \sec y = c\]
C. \[\sec x - \sec y = c\]
D. None of these

Hint: Here, the first order differential equation is given. First, simplify the given equation by using the trigonometric formulas. Then, integrate both sides of the equation with respect to the corresponding variables. After that, solve the left-hand side integral by using the U-substitution method. In the end, solve the integrals to get the solution of the differential equation.

Formula Used: \[\dfrac{1}{{\cos x}} = \sec x\]
\[\dfrac{d}{{dx}}\left( {\sec x} \right) = log\left( {\sec x + \tan x} \right)\]
\[\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}}\]

Complete step by step solution: The given differential equation is \[\cos y log\left( {\sec x + \tan x} \right)dx = \cos x log\left( {\sec y + \tan y} \right)dy\].

Simplify the given equation.
\[ \dfrac{{log\left( {\sec y + \tan y} \right)}}{{\cos y}}dy = \dfrac{{log\left( {\sec x + \tan x} \right)}}{{\cos x}} dx\]
Apply the trigonometric ratio \[\dfrac{1}{{\cos x}} = \sec x\].
\[\sec y log\left( {\sec y + \tan y} \right)dy = \sec x log\left( {\sec x + \tan x} \right) dx\]
Now integrate both sides with respect to the corresponding variables.
\[\int {\sec y log\left( {\sec y + \tan y} \right)dy} = \int {\sec x log\left( {\sec x + \tan x} \right) dx} \] \[.....\left( 1 \right)\]
It is difficult to find the integral of the above equation.
So, apply the substitution method on both sides.

Substitute \[\sec y = u\] and \[\sec x = v\].
Differentiating the substitute equations, we get
\[log\left( {\sec y + \tan y} \right)dy = du\] and \[log\left( {\sec x + \tan x} \right)dx = dv\]

Then, we get the equation \[\left( 1 \right)\] as
\[\int {u du} = \int {v dv} \]
Apply the integration rule \[\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}}\].
\[\dfrac{{{u^2}}}{2} = \dfrac{{{v^2}}}{2}+\dfrac{{{c}}}{2}\]
Resubstitute the values of \[u\], and \[v\].
\[\dfrac{{\sec^{2}y}}{2} = \dfrac{{\sec^{2}x}}{2} + \dfrac{c}{2}\]
\[ \Rightarrow \sec^{2}y = \sec^{2}x + c\]

Option ‘D’ is correct

Note: Students often do mistake to integrating \[\int {{x^n}} dx\]. They apply the formula \[\int {{x^n}} dx = {x^{n + 1}} + c\], which is an incorrect formula. The correct formula is \[\int {{x^n}} dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c\].