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

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

Answers (1)

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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