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प्रश्न
cos y log (sec x + tan x) dx = cos x log (sec y + tan y) dy
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उत्तर
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
\[\int t dt = \int u du\]
\[ \Rightarrow \frac{t^2}{2} = \frac{u^2}{2} + C\]
\[ \Rightarrow \frac{\left[ \log\left( \sec y + \tan y \right) \right]^2}{2} = \frac{\left[ \log\left( \sec x + \tan x \right) \right]^2}{2} + C\]
\[ \Rightarrow \left[ \log\left( \sec y + \tan y \right) \right]^2 = \left[ \log\left( \sec x + \tan x \right) \right]^2 + 2C\]
\[ \Rightarrow \left[ \log\left( \sec y + \tan y \right) \right]^2 = \left[ \log\left( \sec x + \tan x \right) \right]^2 + k,\text{ where }k = 2C\]
