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Question Number 112494 by bemath last updated on 08/Sep/20

$$\mathrm{y}\sqrt{1+{\mathrm{x}}^2}\mathrm{dx}+\mathrm{x}\sqrt{1+{\mathrm{y}}^2}\mathrm{dy}=0$$

Answered by ajfour last updated on 08/Sep/20

$$\frac{\sqrt{1+x^2}}{x^2}d\left(x^2\right)+\frac{\sqrt{1+y^2}}{y^2}d\left(y^2\right)=0$$$$let\sqrt{1+x^2}=s\Rightarrow d\left(x^2\right)=2sds$$$$\Rightarrow \int \frac{2s^{2}ds}{s^2-1}+\int \frac{2t^{2}dt}{t^2-1}=0$$$$\Rightarrow 2s+\ln \mid \frac{s-1}{s+1}\mid +2t+\ln \mid \frac{t-1}{t+1}\mid +c$$$$\Rightarrow 2(\sqrt{1+x^2}+\sqrt{1+y^2})$$$$+\ln [\left(\frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}\right)\left(\frac{\sqrt{1+y^2}-1}{\sqrt{1+y^2}+1}\right]+c$$

Answered by john santu last updated on 08/Sep/20

$$y\sqrt{1+x^2}dx=-x\sqrt{1+y^2}dy$$$$\frac{\sqrt{1+x^2}dx}{-x}=\frac{\sqrt{1+y^2}dy}{y}$$$$-\int \frac{\sqrt{1+x^2}dx}{x}=\int \frac{\sqrt{1+y^2}dy}{y}$$$$[x=\tan q;y=\tan z]$$$$\iff -\int \frac{\sec q\sec {}^{2}qdq}{\tan q}=\int \frac{\sec {}^{3}zdz}{\tan z}$$$$-\int \sec {}^{2}q\mathrm{cosec}qdq=\int \sec {}^{2}z\mathrm{cosec}zdz$$$$$$

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