Question Number 112494 by bemath last updated on 08/Sep/20
$$\:\mathrm{y}\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:\mathrm{dx}\:+\:\mathrm{x}\sqrt{\mathrm{1}+\mathrm{y}^{\mathrm{2}} }\:\mathrm{dy}\:=\:\mathrm{0} \\ $$
Answered by ajfour last updated on 08/Sep/20
![((√(1+x^2 ))/x^2 )d(x^2 )+((√(1+y^2 ))/y^2 )d(y^2 )=0 let (√(1+x^2 ))=s ⇒ d(x^2 )=2sds ⇒ ∫ ((2s^2 ds)/(s^2 −1))+∫ ((2t^2 dt)/(t^2 −1)) = 0 ⇒ 2s+ln ∣((s−1)/(s+1))∣+2t+ln ∣((t−1)/(t+1))∣+c ⇒ 2((√(1+x^2 ))+(√(1+y^2 ))) +ln [((((√(1+x^2 ))−1)/( (√(1+x^2 ))+1)))((((√(1+y^2 ))−1)/( (√(1+y^2 ))+1))]+c](https://www.tinkutara.com/question/Q112496.png)
$$\frac{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{{x}^{\mathrm{2}} }{d}\left({x}^{\mathrm{2}} \right)+\frac{\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }}{{y}^{\mathrm{2}} }{d}\left({y}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$${let}\:\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }={s}\:\:\:\Rightarrow\:\:{d}\left({x}^{\mathrm{2}} \right)=\mathrm{2}{sds} \\ $$$$\Rightarrow\:\:\int\:\frac{\mathrm{2}{s}^{\mathrm{2}} {ds}}{{s}^{\mathrm{2}} −\mathrm{1}}+\int\:\frac{\mathrm{2}{t}^{\mathrm{2}} {dt}}{{t}^{\mathrm{2}} −\mathrm{1}}\:=\:\mathrm{0} \\ $$$$\Rightarrow\:\:\:\mathrm{2}{s}+\mathrm{ln}\:\mid\frac{{s}−\mathrm{1}}{{s}+\mathrm{1}}\mid+\mathrm{2}{t}+\mathrm{ln}\:\mid\frac{{t}−\mathrm{1}}{{t}+\mathrm{1}}\mid+{c} \\ $$$$\Rightarrow\:\mathrm{2}\left(\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }+\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{ln}\:\left[\left(\frac{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }−\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }+\mathrm{1}}\right)\left(\frac{\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }−\mathrm{1}}{\:\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }+\mathrm{1}}\right]+{c}\right. \\ $$
Answered by john santu last updated on 08/Sep/20
![y(√(1+x^2 )) dx = −x(√(1+y^2 )) dy (((√(1+x^2 )) dx)/(−x)) = (((√(1+y^2 )) dy)/y) −∫ (((√(1+x^2 )) dx)/x) = ∫ (((√(1+y^2 )) dy )/y) [ x = tan q ; y = tan z ] ⇔ −∫ ((sec q sec^2 q dq)/(tan q)) = ∫ ((sec^3 z dz)/(tan z)) − ∫ sec^2 q cosec q dq = ∫ sec^2 z cosec z dz](https://www.tinkutara.com/question/Q112495.png)
$$\:\:{y}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx}\:=\:−{x}\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }\:{dy} \\ $$$$\:\frac{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx}}{−{x}}\:=\:\frac{\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }\:{dy}}{{y}} \\ $$$$−\int\:\frac{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx}}{{x}}\:=\:\int\:\frac{\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }\:{dy}\:}{{y}} \\ $$$$\left[\:{x}\:=\:\mathrm{tan}\:{q}\:;\:{y}\:=\:\mathrm{tan}\:{z}\:\right] \\ $$$$\Leftrightarrow\:−\int\:\frac{\mathrm{sec}\:{q}\:\mathrm{sec}\:^{\mathrm{2}} {q}\:{dq}}{\mathrm{tan}\:{q}}\:=\:\int\:\frac{\mathrm{sec}\:^{\mathrm{3}} {z}\:{dz}}{\mathrm{tan}\:{z}} \\ $$$$−\:\int\:\mathrm{sec}\:^{\mathrm{2}} {q}\:\mathrm{cosec}\:{q}\:{dq}\:=\:\int\:\mathrm{sec}\:^{\mathrm{2}} {z}\:\mathrm{cosec}\:{z}\:{dz} \\ $$$$ \\ $$