Question

cos y log (sec x + tan x) dx = cos x log (sec y + tan y) dy

Answer 1

We have,

cos y log (sec x + tan x) dx = cos x log (sec y + tan y) dy

\[\Rightarrow \frac{\log\left( \sec y + \tan y \right)}{\cos y}dy = \frac{\log\left( \sec x + \tan x \right)}{\cos x}dx\]

Integrating both sides, we get

\[\int\frac{\log\left( \sec y + \tan y \right)}{\cos y}dy = \int\frac{\log\left( \sec x + \tan x \right)}{\cos x}dx . . . . . . . . . \left( 1 \right)\]

\[\text{Putting }\log\left( \sec y + \tan y \right) = t\text{ and }\log\left( \sec x + \tan x \right) = u\]

\[ \Rightarrow \frac{\sec^2 y + \sec y \tan y}{\sec y + \tan y}dy = dt and \frac{\sec^2 x + \sec x \tan x}{\sec x + \tan x}dx = du\]

\[ \Rightarrow \sec y dy = dt\text{ and }\sec x dx = du\]

Therefore, (1) becomes

\[\int t dt = \int u du\]

\[ \Rightarrow \frac{t^2}{2} = \frac{u^2}{2} + C\]

\[ \Rightarrow \frac{\left[ \log\left( \sec y + \tan y \right) \right]^2}{2} = \frac{\left[ \log\left( \sec x + \tan x \right) \right]^2}{2} + C\]

\[ \Rightarrow \left[ \log\left( \sec y + \tan y \right) \right]^2 = \left[ \log\left( \sec x + \tan x \right) \right]^2 + 2C\]

\[ \Rightarrow \left[ \log\left( \sec y + \tan y \right) \right]^2 = \left[ \log\left( \sec x + \tan x \right) \right]^2 + k,\text{ where }k = 2C\]