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Need solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.7 question 11

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Answer: x \tan x-y \tan y=\log |\sec x|-\log |\sec y|+c

Hint: Separate the terms of x and y and then integrate them.

Given: x \cos ^{2} y d x=y \cos ^{2} x d y

Solution: x \cos ^{2} y d x=y \cos ^{2} x d y

        \begin{aligned} &\frac{x}{\cos ^{2} x} d x=\frac{y}{\cos ^{2} y} d y \\\\ &x \sec ^{2} x d x=y \sec ^{2} y d y \end{aligned}

        Integrating both sides

        \int x \sec ^{2} x d x=\int y \sec ^{2} y d y

        Using integration by parts

        \begin{aligned} &x \int \sec ^{2} x d x=y \int \sec ^{2} y d y \\\\ &x \tan x-\int \tan x d x=y \tan y-\int \tan y d y \end{aligned}

        Using identity, \int \tan x d x=\log |\sec x|

        \begin{aligned} &x \tan (x)-\log |\sec x|=y \tan y-\log |\sec y|+c \\\\ &x \tan x-y \tan y=\log |\sec x|-\log |\sec y|+c \end{aligned}

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