#### Explain Solution R.D.Sharma Class 12 Chapter 21 Differential Equations Exercise Revision Exercise Question 28 Maths Textbook Solution.

Answer: $x\tan x-y\tan |\frac{\sec x}{\sec y}|+C$

Hint: Use the formula of $\int u\: v\: dx$

Given: $x\cos ^{2}y\: dx=y\: \cos ^{2}x\: dy$

Solution:$x\cos ^{2}y\: dx=y\: \cos ^{2}x\: dy$

\begin{aligned} &\Rightarrow \quad \frac{x}{\cos ^{2} x} d x=\frac{y}{\cos ^{2} y} d y \\ &\Rightarrow \quad y \sec ^{2} y d y=x \sec ^{2} x d x \quad\left(\because \frac{1}{\cos x}=\sec x\right) \\ &\Rightarrow \int y \sec ^{2} y d y=\int x \sec ^{2} x d x \quad(\because \text { integrating both sides }) \\ &\Rightarrow \quad y \int \sec ^{2} y d y-\int \tan y d y=x \int \sec ^{2} x d x-\int \tan x d x \\ &\quad\left[\because \int u v d x=u \int v d x-\int\left(\frac{d}{d x} u \int v d x\right) d x\right. \\ &\Rightarrow \quad y \tan y-\log |\sec y|=x \tan x-\log |\sec x|+C \\ &\Rightarrow \quad x \tan x-y \tan y=\log |\sec x|-\log |\sec y|+C \\ &\Rightarrow \quad x \tan x-y \tan y=\log \left|\frac{\sec x}{\sec y}\right|+C \quad[\because \log a-\log b=\log (a / b)] \end{aligned}