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Question Number 40131 by maxmathsup by imad last updated on 16/Jul/18
find the value of I = ∫_0 ^1 ((1+x^4 )/(1+x^6 ))dx
$${find}\:{the}\:{value}\:{of}\:{I}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{\mathrm{1}+{x}^{\mathrm{4}} }{\mathrm{1}+{x}^{\mathrm{6}} }{dx} \\ $$
Answered by maxmathsup by imad last updated on 24/Jul/18
we have ∫_0 ^1 (dx/(1+x^6 )) =_(x=(1/t)) ∫_1 ^(+∞) (1/(1+(1/t^6 ))) (dt/t^2 ) = ∫_1 ^(+∞) (dt/(t^2 +(1/t^4 ))) = ∫_1 ^(+∞) (t^4 /(1+t^6 ))dt = ∫_0 ^∞ (t^4 /(1+t^6 ))dt −∫_0 ^1 (t^4 /(1+t^6 ))dt ⇒ ∫_0 ^1 ((1+x^4 )/(1+x^6 ))dx = ∫_0 ^∞ (x^4 /(1+x^6 ))dx =_(x=u^(1/6) ) ∫_0 ^∞ (u^(4/6) /(1+u)) (1/6) u^((1/6)−1) du =(1/6) ∫_0 ^∞ (u^((2/3)+(1/6)−1) /(1+u)) du =(1/6) ∫_0 ^∞ (u^((5/6)−1) /(1+u)) du =(1/6) (π/(sin(((5π)/6)))) = (π/(6sin(π−(π/6)))) =(π/(6sin((π/6)))) =(π/(6.(1/2))) ⇒ I =(π/3) .
$${we}\:{have}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{6}} }\:\:=_{{x}=\frac{\mathrm{1}}{{t}}} \:\:\:\int_{\mathrm{1}} ^{+\infty} \:\:\:\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{6}} }}\:\frac{{dt}}{{t}^{\mathrm{2}} }\:=\:\int_{\mathrm{1}} ^{+\infty} \:\:\:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+\frac{\mathrm{1}}{{t}^{\mathrm{4}} }} \\ $$$$=\:\int_{\mathrm{1}} ^{+\infty} \:\:\frac{{t}^{\mathrm{4}} }{\mathrm{1}+{t}^{\mathrm{6}} }{dt}\:=\:\int_{\mathrm{0}} ^{\infty} \:\frac{{t}^{\mathrm{4}} }{\mathrm{1}+{t}^{\mathrm{6}} }{dt}\:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{t}^{\mathrm{4}} }{\mathrm{1}+{t}^{\mathrm{6}} }{dt}\:\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\mathrm{1}+{x}^{\mathrm{4}} }{\mathrm{1}+{x}^{\mathrm{6}} }{dx}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{x}^{\mathrm{4}} }{\mathrm{1}+{x}^{\mathrm{6}} }{dx}\:=_{{x}={u}^{\frac{\mathrm{1}}{\mathrm{6}}} } \:\:\:\:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{u}^{\frac{\mathrm{4}}{\mathrm{6}}} }{\mathrm{1}+{u}}\:\frac{\mathrm{1}}{\mathrm{6}}\:{u}^{\frac{\mathrm{1}}{\mathrm{6}}−\mathrm{1}} \:{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{u}^{\frac{\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{6}}−\mathrm{1}} }{\mathrm{1}+{u}}\:{du}\:=\frac{\mathrm{1}}{\mathrm{6}}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{u}^{\frac{\mathrm{5}}{\mathrm{6}}−\mathrm{1}} }{\mathrm{1}+{u}}\:{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\:\frac{\pi}{{sin}\left(\frac{\mathrm{5}\pi}{\mathrm{6}}\right)}\:=\:\frac{\pi}{\mathrm{6}{sin}\left(\pi−\frac{\pi}{\mathrm{6}}\right)}\:=\frac{\pi}{\mathrm{6}{sin}\left(\frac{\pi}{\mathrm{6}}\right)}\:=\frac{\pi}{\mathrm{6}.\frac{\mathrm{1}}{\mathrm{2}}}\:\Rightarrow\:{I}\:\:=\frac{\pi}{\mathrm{3}}\:. \\ $$

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