dx-1-x-2-1-x-2-

Question Number 126314 by benjo_mathlover last updated on 19/Dec/20

$$\:\:\int\:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:? \\ $$

Answered by Dwaipayan Shikari last updated on 19/Dec/20

∫(dx/((1+x^2 )(√(1−x^2 )))) x=sinθ =∫(dθ/(1+sin^2 θ))=2∫(1/(3−cos2θ))dθ tanθ=t =2∫(1/(3−((1−t^2 )/(1+t^2 )))).(1/(1+t^2 ))dt= 2∫(1/(4t^2 +2))dt =(1/( (√2))) tan^(−1) (√2)t+C =(1/( (√2)))tan^(−1) (((√2)x)/( (√((1−x^2 )))))+C

$$\int\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:\:\:\:\:\:\:\:{x}={sin}\theta \\ $$$$=\int\frac{{d}\theta}{\mathrm{1}+{sin}^{\mathrm{2}} \theta}=\mathrm{2}\int\frac{\mathrm{1}}{\mathrm{3}−{cos}\mathrm{2}\theta}{d}\theta\:\:\:\:\:\:\:\:\:\:{tan}\theta={t} \\ $$$$=\mathrm{2}\int\frac{\mathrm{1}}{\mathrm{3}−\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}.\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}=\:\mathrm{2}\int\frac{\mathrm{1}}{\mathrm{4}{t}^{\mathrm{2}} +\mathrm{2}}{dt}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:{tan}^{−\mathrm{1}} \sqrt{\mathrm{2}}{t}+{C} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{tan}^{−\mathrm{1}} \frac{\sqrt{\mathrm{2}}{x}}{\:\sqrt{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}}+{C} \\ $$

Answered by mathmax by abdo last updated on 20/Dec/20

I =∫ (dx/((x^2 +1)(√(1−x^2 )))) changement x=sint give I =∫ ((costdt)/((sin^2 t+1)cost)) =∫ (dt/(1+((1−cos(2t))/2)))=∫ ((2dt)/(3−cos(2t))) =_(2t=u) ∫ (du/(3−cosu)) =_(tan((u/2))=y) ∫ (1/(3−((1−y^2 )/(1+y^2 ))))((2dy)/(1+y^2 )) =∫ ((2dy)/(3+3y^2 −1+y^2 )) =∫ ((2dy)/(2+4y^2 ))=∫ (dy/(1+2y^2 )) =_((√2)y=z) ∫ (1/(1+z^2 ))(dz/( (√2))) =(1/( (√2)))arctan(z)+C =(1/( (√2)))arctan((√2)y)+C =(1/( (√2)))arctan((√2)tan(t))+C =(1/( (√2)))arctan((√2)tan(arcsinx)) +C

$$\mathrm{I}\:=\int\:\:\frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }}\:\:\:\mathrm{changement}\:\mathrm{x}=\mathrm{sint}\:\mathrm{give} \\ $$$$\mathrm{I}\:=\int\:\:\frac{\mathrm{costdt}}{\left(\mathrm{sin}^{\mathrm{2}} \mathrm{t}+\mathrm{1}\right)\mathrm{cost}}\:=\int\:\:\frac{\mathrm{dt}}{\mathrm{1}+\frac{\mathrm{1}−\mathrm{cos}\left(\mathrm{2t}\right)}{\mathrm{2}}}=\int\:\frac{\mathrm{2dt}}{\mathrm{3}−\mathrm{cos}\left(\mathrm{2t}\right)} \\ $$$$=_{\mathrm{2t}=\mathrm{u}} \:\:\:\int\:\:\frac{\mathrm{du}}{\mathrm{3}−\mathrm{cosu}}\:=_{\mathrm{tan}\left(\frac{\mathrm{u}}{\mathrm{2}}\right)=\mathrm{y}} \:\:\:\int\:\:\frac{\mathrm{1}}{\mathrm{3}−\frac{\mathrm{1}−\mathrm{y}^{\mathrm{2}} }{\mathrm{1}+\mathrm{y}^{\mathrm{2}} }}\frac{\mathrm{2dy}}{\mathrm{1}+\mathrm{y}^{\mathrm{2}} } \\ $$$$=\int\:\:\frac{\mathrm{2dy}}{\mathrm{3}+\mathrm{3y}^{\mathrm{2}} −\mathrm{1}+\mathrm{y}^{\mathrm{2}} }\:=\int\:\frac{\mathrm{2dy}}{\mathrm{2}+\mathrm{4y}^{\mathrm{2}} }=\int\:\frac{\mathrm{dy}}{\mathrm{1}+\mathrm{2y}^{\mathrm{2}} }\:=_{\sqrt{\mathrm{2}}\mathrm{y}=\mathrm{z}} \:\:\int\:\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{z}^{\mathrm{2}} }\frac{\mathrm{dz}}{\:\sqrt{\mathrm{2}}} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{arctan}\left(\mathrm{z}\right)+\mathrm{C}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{arctan}\left(\sqrt{\mathrm{2}}\mathrm{y}\right)+\mathrm{C} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{arctan}\left(\sqrt{\mathrm{2}}\mathrm{tan}\left(\mathrm{t}\right)\right)+\mathrm{C}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{arctan}\left(\sqrt{\mathrm{2}}\mathrm{tan}\left(\mathrm{arcsinx}\right)\right)\:+\mathrm{C} \\ $$

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