#### Explain Solution R.D.Sharma Class 12 Chapter 21 Differential Equations Exercise Revision Exercise Question 29 Maths Textbook Solution.

Answer: $\left [ log\left ( sec\: y+\tan \: y \right ) \right ]^{2}=\left [ log\left ( \sec \: x+\tan \: x \right ) \right ]^{2}+C$

Hint: Apply substitution method.

Given:$\cos \: y\: log|\sec \: x+\tan \: x|dx=\cos \: x\: log|\sec \: y+\tan \: y|dy$

Solution:$\cos \: y\: log|\sec \: x+\tan \: x|dx=\cos \: x\: log|\sec \: y+\tan \: y|dy$

\begin{aligned} &\Rightarrow \frac{\log |\sec x+\tan x|}{\cos x} d x=\frac{\log |\sec y+\tan y|}{\cos y} d y\\ &\text { Integrating both sides }\\ &\Rightarrow \int \frac{\log |\sec y+\tan y|}{\cos y} d y=\int \frac{\log |\sec x+\tan x|}{\operatorname{Cos} x} d x \end{aligned}

Let  $\log |\sec y+\tan y|=t \text { and } \log |\sec x+\tan x|=u \\$

$\Rightarrow\left(\frac{\sec ^{2} y+s e c y \text { tany }}{s e c y+\tan y}\right) d y=d t \text { and }\left(\frac{\sec ^{2} x+\sec x \tan x}{\sec x+\tan x}\right) d x=d u \\$

$\Rightarrow \sec y d y=d t \quad \text { and } \sec x d x=d u \\$

$\Rightarrow\left(\frac{1}{c o s y}\right) d y=d t \text { and }\left(\frac{1}{\cos x}\right) d x=d u \\ \$

$\Rightarrow \int\: t\: d t=\int u d u \\$

$\Rightarrow \frac{t^{2}}{2}=\frac{u^{2}}{2}+c_{1} \\$

$\Rightarrow t^{2}=u^{2}+2 c_{1} \\$

$\Rightarrow t^{2}=u^{2}+c\left(\because 2 c_{1}=C\right) \\$

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

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