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Question Number 125416 by bemath last updated on 10/Dec/20

∫ (dx/( (√(x(√x)−x^2 )))) ?

$$\int\:\frac{{dx}}{\:\sqrt{{x}\sqrt{{x}}−{x}^{\mathrm{2}} }}\:? \\ $$

Answered by Dwaipayan Shikari last updated on 10/Dec/20

∫(dx/( (√x)(√((√x)−x))))          (√x)=t⇒(1/(2(√x)))=(dt/dx)  =2∫(dt/( (√(t−t^2 ))))=2∫(1/( (√t)((√(1−t)))))dt             t=u^2 ⇒1=2u(du/dt)       =4∫(du/( (√(1−u^2 ))))=4sin^(−1) u +C =4sin^(−1) (x)^(1/4)  dx

$$\int\frac{{dx}}{\:\sqrt{{x}}\sqrt{\sqrt{{x}}−{x}}}\:\:\:\:\:\:\:\:\:\:\sqrt{{x}}={t}\Rightarrow\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}}=\frac{{dt}}{{dx}} \\ $$$$=\mathrm{2}\int\frac{{dt}}{\:\sqrt{{t}−{t}^{\mathrm{2}} }}=\mathrm{2}\int\frac{\mathrm{1}}{\:\sqrt{{t}}\left(\sqrt{\mathrm{1}−{t}}\right)}{dt}\:\:\:\:\:\:\:\:\:\:\:\:\:{t}={u}^{\mathrm{2}} \Rightarrow\mathrm{1}=\mathrm{2}{u}\frac{{du}}{{dt}}\:\:\:\:\: \\ $$$$=\mathrm{4}\int\frac{{du}}{\:\sqrt{\mathrm{1}−{u}^{\mathrm{2}} }}=\mathrm{4}{sin}^{−\mathrm{1}} {u}\:+{C}\:=\mathrm{4}{sin}^{−\mathrm{1}} \sqrt[{\mathrm{4}}]{{x}}\:{dx} \\ $$

Answered by liberty last updated on 10/Dec/20

I=∫ (dx/(x(√((1/( (√x)))−1)))) ; let (1/( (√x)))−1 = w ; x=(1/((w+1)^2 ))  I= ∫ (((w+1)^2 )/( (√w))) (−(2/((1+w)^3 )))dw  I=−2∫ (dw/( (√w)(1+w))) ; let (√w) = z ; dw = 2z dz  I=−4 ∫ ((z dz)/(z(1+z^2 ))) = −4 tan^(−1) (z) + c  I= −4 tan^(−1) ((√w) )+ c   I=−4 tan^(−1) ((√(((1−(√x))/( (√x))) )) ) + c

$${I}=\int\:\frac{{dx}}{{x}\sqrt{\frac{\mathrm{1}}{\:\sqrt{{x}}}−\mathrm{1}}}\:;\:{let}\:\frac{\mathrm{1}}{\:\sqrt{{x}}}−\mathrm{1}\:=\:{w}\:;\:{x}=\frac{\mathrm{1}}{\left({w}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${I}=\:\int\:\frac{\left({w}+\mathrm{1}\right)^{\mathrm{2}} }{\:\sqrt{{w}}}\:\left(−\frac{\mathrm{2}}{\left(\mathrm{1}+{w}\right)^{\mathrm{3}} }\right){dw} \\ $$$${I}=−\mathrm{2}\int\:\frac{{dw}}{\:\sqrt{{w}}\left(\mathrm{1}+{w}\right)}\:;\:{let}\:\sqrt{{w}}\:=\:{z}\:;\:{dw}\:=\:\mathrm{2}{z}\:{dz} \\ $$$${I}=−\mathrm{4}\:\int\:\frac{{z}\:{dz}}{{z}\left(\mathrm{1}+{z}^{\mathrm{2}} \right)}\:=\:−\mathrm{4}\:\mathrm{tan}^{−\mathrm{1}} \left({z}\right)\:+\:{c} \\ $$$${I}=\:−\mathrm{4}\:\mathrm{tan}^{−\mathrm{1}} \left(\sqrt{{w}}\:\right)+\:{c}\: \\ $$$${I}=−\mathrm{4}\:\mathrm{tan}^{−\mathrm{1}} \left(\sqrt{\frac{\mathrm{1}−\sqrt{{x}}}{\:\sqrt{{x}}}\:}\:\right)\:+\:{c}\: \\ $$

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