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Question Number 63065 by Enebeli Chinedu Vitalis last updated on 28/Jun/19
If I=∫_( 0) ^1 (dx/( (√(1+x^4 )))) , then
$$\mathrm{If}\:{I}=\underset{\:\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{{dx}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{4}} }}\:,\:\mathrm{then} \\ $$
Commented by Enebeli Chinedu Vitalis last updated on 28/Jun/19
thanks boss★★
$${thanks}\:{boss}\bigstar\bigstar \\ $$
Commented by mathmax by abdo last updated on 28/Jun/19
changement x^2 =tanθ give x =(tanθ)^(1/2) ⇒I =∫_0 ^(π/4) (1/(2(√(1+tan^2 θ))))(1+tan^2 θ)(tanθ)^(−(1/2)) dθ = (1/2)∫_0 ^(π/4) (√(1+tan^2 θ))(tanθ)^(−(1/2)) =(1/2) ∫_0 ^(π/4) (1/(cosθ (√((sinθ)/(cosθ))))) =(1/2) ∫_0 ^(π/2) (dθ/( (√(cosθsinθ)))) =(1/( (√2))) ∫_0 ^(π/2) (dθ/( (√(sin(2θ))))) =(1/( (√2)))∫_0 ^π (dt/(2(√(sint)))) (2θ=t) =(1/(2(√2))) ∫_0 ^π (dt/( (√(sint)))) changement tan((t/2)) =u give ∫_0 ^π (dt/( (√(sint)))) =∫_0 ^∞ ((2du)/((1+u^2 )(√((2u)/(1+u^2 ))))) =(√2) ∫_0 ^∞ (du/( (√u)(√(1+u^2 )))) =_((√u)=α) (√2)∫_0 ^∞ ((2αdα)/(α(√(1+α^4 )))) =2(√2)∫_0 ^∞ (dα/( (√(1+α^4 )))) ⇒I =∫_0 ^∞ (dx/( (√(1+x^4 )))) I =∫_0 ^1 (dx/( (√(1+x^4 )))) + ∫_1 ^(+∞) (dx/( (√(1+x^4 )))) ⇒ ∫_1 ^(+∞) (dx/( (√(1+x^4 )))) =0....be contnued....
$${changement}\:{x}^{\mathrm{2}} ={tan}\theta\:{give}\:{x}\:=\left({tan}\theta\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:\Rightarrow{I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{1}+{tan}^{\mathrm{2}} \theta}}\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right)\left({tan}\theta\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} {d}\theta \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\:\:\sqrt{\mathrm{1}+{tan}^{\mathrm{2}} \theta}\left({tan}\theta\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \:=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{1}}{{cos}\theta\:\sqrt{\frac{{sin}\theta}{{cos}\theta}}}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{d}\theta}{\:\sqrt{{cos}\theta{sin}\theta}} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{d}\theta}{\:\sqrt{{sin}\left(\mathrm{2}\theta\right)}}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{dt}}{\mathrm{2}\sqrt{{sint}}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{2}\theta={t}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\int_{\mathrm{0}} ^{\pi} \:\:\:\:\:\frac{{dt}}{\:\sqrt{{sint}}}\:\:{changement}\:\:{tan}\left(\frac{{t}}{\mathrm{2}}\right)\:={u}\:{give} \\ $$$$\int_{\mathrm{0}} ^{\pi} \:\:\frac{{dt}}{\:\sqrt{{sint}}}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\:\:\frac{\mathrm{2}{du}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\sqrt{\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }}}\:=\sqrt{\mathrm{2}}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\frac{{du}}{\:\sqrt{{u}}\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }} \\ $$$$=_{\sqrt{{u}}=\alpha} \:\:\:\:\sqrt{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{2}\alpha{d}\alpha}{\alpha\sqrt{\mathrm{1}+\alpha^{\mathrm{4}} }}\:\:=\mathrm{2}\sqrt{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{d}\alpha}{\:\sqrt{\mathrm{1}+\alpha^{\mathrm{4}} }}\:\Rightarrow{I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dx}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{4}} }} \\ $$$${I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dx}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{4}} }}\:+\:\int_{\mathrm{1}} ^{+\infty} \:\:\:\frac{{dx}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{4}} }}\:\Rightarrow\:\int_{\mathrm{1}} ^{+\infty} \:\:\frac{{dx}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{4}} }}\:=\mathrm{0}….{be}\:{contnued}…. \\ $$
Commented by mathmax by abdo last updated on 28/Jun/19
you are welcome.
$${you}\:{are}\:{welcome}. \\ $$

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