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Question Number 122157 by mathmax by abdo last updated on 14/Nov/20

find ∫_(−1) ^1 (√(1+x^4 ))dx

$$\mathrm{find}\:\int_{−\mathrm{1}} ^{\mathrm{1}} \sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{4}} }\mathrm{dx} \\ $$

Commented by peter frank last updated on 14/Nov/20

Qn 121384

$$\mathrm{Qn}\:\mathrm{121384} \\ $$

Answered by mindispower last updated on 15/Nov/20

=2∫_0 ^1 (√(1+x^4 ))dx x^4 =u⇒dx=u^(−(3/4)) (du/4) =(1/2)∫_0 ^1 (√(1+u)).u^(−(3/4)) du=I β(b,c−b)_2 F_1 (a,b,c;z)=∫_0 ^1 x^(b−1) (1−x)^(c−b−1) (1−zx)^(−a) dx β betta function,_2 F_1 (a,b,c;z) Hypergeometric function I⇔(1/2)∫_0 ^1 (1+u)^(1/2) u^(−(3/4)) (1−u)^0 du ⇒b−1=−(3/4),c−b−1=0,−a=(1/2),−z=1 ⇒b=(1/4),c=(5/4),a=−(1/2),z=−1 I=(1/2)β((1/4),1) _2 F_1 (−(1/2),(1/4),(5/4);−1) ((β((1/4),1))/2)=(1/2)((Γ((1/4)))/(Γ(1+(1/4)))) =2 we get ∫_(−1) ^1 (√(1+x^4 ))dx=2∫_0 ^1 (√(1+x^4 ))dx=2._2 F_1 (−(1/2),(1/4),(5/4);−1)

$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}+{x}^{\mathrm{4}} }{dx} \\ $$$${x}^{\mathrm{4}} ={u}\Rightarrow{dx}={u}^{−\frac{\mathrm{3}}{\mathrm{4}}} \frac{{du}}{\mathrm{4}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}+{u}}.{u}^{−\frac{\mathrm{3}}{\mathrm{4}}} {du}={I} \\ $$$$\beta\left({b},{c}−{b}\right)_{\mathrm{2}} {F}_{\mathrm{1}} \left({a},{b},{c};{z}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{b}−\mathrm{1}} \left(\mathrm{1}−{x}\right)^{{c}−{b}−\mathrm{1}} \left(\mathrm{1}−{zx}\right)^{−{a}} {dx} \\ $$$$\beta\:{betta}\:{function},_{\mathrm{2}} {F}_{\mathrm{1}} \left({a},{b},{c};{z}\right)\:{Hypergeometric} \\ $$$${function} \\ $$$${I}\Leftrightarrow\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+{u}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} {u}^{−\frac{\mathrm{3}}{\mathrm{4}}} \left(\mathrm{1}−{u}\right)^{\mathrm{0}} {du} \\ $$$$\Rightarrow{b}−\mathrm{1}=−\frac{\mathrm{3}}{\mathrm{4}},{c}−{b}−\mathrm{1}=\mathrm{0},−{a}=\frac{\mathrm{1}}{\mathrm{2}},−{z}=\mathrm{1} \\ $$$$\Rightarrow{b}=\frac{\mathrm{1}}{\mathrm{4}},{c}=\frac{\mathrm{5}}{\mathrm{4}},{a}=−\frac{\mathrm{1}}{\mathrm{2}},{z}=−\mathrm{1} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}}\beta\left(\frac{\mathrm{1}}{\mathrm{4}},\mathrm{1}\right)\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(−\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{4}},\frac{\mathrm{5}}{\mathrm{4}};−\mathrm{1}\right) \\ $$$$\frac{\beta\left(\frac{\mathrm{1}}{\mathrm{4}},\mathrm{1}\right)}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\Gamma\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}\right)}\:=\mathrm{2} \\ $$$${we}\:{get}\:\int_{−\mathrm{1}} ^{\mathrm{1}} \sqrt{\mathrm{1}+{x}^{\mathrm{4}} }{dx}=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}+{x}^{\mathrm{4}} }{dx}=\mathrm{2}._{\mathrm{2}} {F}_{\mathrm{1}} \left(−\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{4}},\frac{\mathrm{5}}{\mathrm{4}};−\mathrm{1}\right) \\ $$

Answered by Dwaipayan Shikari last updated on 15/Nov/20

∫_(−1) ^1 (√(1+x^4 ))dx =2∫_0 ^1 (√(1−t)) dt x^4 =−t⇒4x^3 =−(dt/dx) =(1/2)∫_0 ^1 x^(−3) (√(1−t)) dt =(1/2)∫_0 ^1 (−1)^(−(3/4)) t^(−(3/4)) (1−t)^(1/2) dt =((𝚪((1/4))𝚪((3/2)))/(2(√i)))

$$\int_{−\mathrm{1}} ^{\mathrm{1}} \sqrt{\mathrm{1}+{x}^{\mathrm{4}} }{dx} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}−{t}}\:{dt}\:\:\:\:\:\:{x}^{\mathrm{4}} =−{t}\Rightarrow\mathrm{4}{x}^{\mathrm{3}} =−\frac{{dt}}{{dx}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{−\mathrm{3}} \sqrt{\mathrm{1}−{t}}\:{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \left(−\mathrm{1}\right)^{−\frac{\mathrm{3}}{\mathrm{4}}} {t}^{−\frac{\mathrm{3}}{\mathrm{4}}} \left(\mathrm{1}−{t}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} {dt} \\ $$$$=\frac{\boldsymbol{\Gamma}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\boldsymbol{\Gamma}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)}{\mathrm{2}\sqrt{{i}}} \\ $$

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