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Question Number 64419 by turbo msup by abdo last updated on 17/Jul/19
1) find ∫ (dx/(x−(√(1−x^2 )))) 2) calculate ∫_0 ^1 (dx/(x−(√(1−x^2 ))))
$$\left.\mathrm{1}\right)\:{find}\:\int\:\:\frac{{dx}}{{x}−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dx}}{{x}−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$
Commented by mathmax by abdo last updated on 18/Jul/19
1) let A =∫ (dx/(x−(√(1−x^2 )))) changement x =sinθ give A =∫ ((cosθ)/(sinθ −cosθ))dθ =∫ ((cosθ)/(cosθ(tanθ −1)))dθ =∫ (dθ/(tanθ −1)) =_(tanθ =t) ∫ (dt/((1+t^2 )(t−1))) let decompose F(t) =(1/((t−1)(t^2 +1))) ⇒F(t)=(a/(t−1)) +((bt +c)/(t^2 +1)) a =lim_(t→1) (t−1)F(t) =(1/2) lim_(t→+∞) tF(t) =0 =a+b ⇒b=−(1/2) ⇒F(t)=(1/(2(t−1))) +((−(1/2)t +c)/(t^2 +1)) F(0) =−1 =−(1/2) +c ⇒c=−(1/2) ⇒F(t)=(1/(2(t−1))) −(1/2) ((t+1)/(t^2 +1)) ⇒ A =(1/2) ∫ (dt/(t−1)) −(1/2) ∫ ((t+1)/(t^2 +1))dt +c =(1/2)ln∣t−1∣−(1/4)ln(t^2 +1)−(1/2) arctan(t) +c we have t=tanθ and θ =arcsinx ⇒ A =(1/2)ln∣tan(arcsinx)−1∣−(1/4)ln(1+tan^2 (arcsinx))−(1/2)arcsinx +c
$$\left.\mathrm{1}\right)\:{let}\:{A}\:=\int\:\:\:\:\frac{{dx}}{{x}−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:\:{changement}\:{x}\:={sin}\theta\:{give} \\ $$$${A}\:=\int\:\:\:\frac{{cos}\theta}{{sin}\theta\:−{cos}\theta}{d}\theta\:\:=\int\:\:\frac{{cos}\theta}{{cos}\theta\left({tan}\theta\:−\mathrm{1}\right)}{d}\theta \\ $$$$=\int\:\:\:\frac{{d}\theta}{{tan}\theta\:−\mathrm{1}}\:\:=_{{tan}\theta\:={t}} \:\:\:\:\:\int\:\:\frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left({t}−\mathrm{1}\right)}\:\:{let}\:{decompose} \\ $$$${F}\left({t}\right)\:=\frac{\mathrm{1}}{\left({t}−\mathrm{1}\right)\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)}\:\Rightarrow{F}\left({t}\right)=\frac{{a}}{{t}−\mathrm{1}}\:+\frac{{bt}\:+{c}}{{t}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${a}\:={lim}_{{t}\rightarrow\mathrm{1}} \left({t}−\mathrm{1}\right){F}\left({t}\right)\:=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${lim}_{{t}\rightarrow+\infty} {tF}\left({t}\right)\:=\mathrm{0}\:={a}+{b}\:\Rightarrow{b}=−\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{F}\left({t}\right)=\frac{\mathrm{1}}{\mathrm{2}\left({t}−\mathrm{1}\right)}\:+\frac{−\frac{\mathrm{1}}{\mathrm{2}}{t}\:+{c}}{{t}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${F}\left(\mathrm{0}\right)\:=−\mathrm{1}\:=−\frac{\mathrm{1}}{\mathrm{2}}\:+{c}\:\Rightarrow{c}=−\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{F}\left({t}\right)=\frac{\mathrm{1}}{\mathrm{2}\left({t}−\mathrm{1}\right)}\:−\frac{\mathrm{1}}{\mathrm{2}}\:\frac{{t}+\mathrm{1}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow \\ $$$${A}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\:\frac{{dt}}{{t}−\mathrm{1}}\:−\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\:\frac{{t}+\mathrm{1}}{{t}^{\mathrm{2}} \:+\mathrm{1}}{dt}\:+{c} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid{t}−\mathrm{1}\mid−\frac{\mathrm{1}}{\mathrm{4}}{ln}\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{2}}\:{arctan}\left({t}\right)\:+{c} \\ $$$${we}\:{have}\:\:{t}={tan}\theta\:{and}\:\theta\:={arcsinx}\:\Rightarrow \\ $$$${A}\:=\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid{tan}\left({arcsinx}\right)−\mathrm{1}\mid−\frac{\mathrm{1}}{\mathrm{4}}{ln}\left(\mathrm{1}+{tan}^{\mathrm{2}} \left({arcsinx}\right)\right)−\frac{\mathrm{1}}{\mathrm{2}}{arcsinx}\:+{c} \\ $$

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