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Question Number 87121 by M±th+et£s last updated on 03/Apr/20
∫_0 ^(π/2) ((1−x^4 )/(1+x^4 ))dx
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}−{x}^{\mathrm{4}} }{\mathrm{1}+{x}^{\mathrm{4}} }{dx} \\ $$
Answered by redmiiuser last updated on 03/Apr/20
(1+x^4 )^(−1) =Σ_(n=0 ) ^∞ (−1)^n .x^(4n) (1−x^4 )(1+x^4 )^(−1) =(1−x^4 ).(Σ_(n=0) ^∞ (−1)^n .x^(4n) ) =Σ_(n=0) ^∞ (−1)^n .x^(4n) −Σ_(n=0) ^∞ (−1)^n .x^(4n+4) ∫_0 ^(π/2) ((1−x^4 )/(1+x^4 ))dx =[Σ_(n=0) ^∞ (((−1)^n .x^(4n+1) )/(4n+1))]_0 ^(π/2) −[Σ_(n=0) ^∞ (((−1)^n .x^(4n+5) )/(4n+5))]_0 ^(π/2) =Σ_(n=0) ^∞ (((−1)^n .((π/2))^(4n+1) )/(4n+1)) − Σ_(n=0) ^∞ (((−1)^n .((π/2))^(4n+5) )/(4n+5))
$$\left(\mathrm{1}+{x}^{\mathrm{4}} \right)^{−\mathrm{1}} \\ $$$$=\underset{{n}=\mathrm{0}\:\:} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} .{x}^{\mathrm{4}{n}} \\ $$$$\left(\mathrm{1}−{x}^{\mathrm{4}} \right)\left(\mathrm{1}+{x}^{\mathrm{4}} \right)^{−\mathrm{1}} \\ $$$$=\left(\mathrm{1}−{x}^{\mathrm{4}} \right).\left(\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} .{x}^{\mathrm{4}{n}} \right) \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} .{x}^{\mathrm{4}{n}} −\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} .{x}^{\mathrm{4}{n}+\mathrm{4}} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}−{x}^{\mathrm{4}} }{\mathrm{1}+{x}^{\mathrm{4}} }{dx} \\ $$$$=\left[\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} .{x}^{\mathrm{4}{n}+\mathrm{1}} }{\mathrm{4}{n}+\mathrm{1}}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\right]}}−\left[\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} .{x}^{\mathrm{4}{n}+\mathrm{5}} }{\mathrm{4}{n}+\mathrm{5}}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\right]}} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} .\left(\frac{\pi}{\mathrm{2}}\right)^{\mathrm{4}{n}+\mathrm{1}} }{\mathrm{4}{n}+\mathrm{1}}\:\:−\:\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} .\left(\frac{\pi}{\mathrm{2}}\right)^{\mathrm{4}{n}+\mathrm{5}} }{\mathrm{4}{n}+\mathrm{5}} \\ $$
Commented by redmiiuser last updated on 03/Apr/20
pls pls pls check
$${pls}\:{pls}\:{pls}\:{check} \\ $$
Commented by M±th+et£s last updated on 03/Apr/20
its right sir thank you
$${its}\:{right}\:{sir}\:{thank}\:{you} \\ $$
Commented by redmiiuser last updated on 03/Apr/20
welcome mister
$${welcome}\:{mister} \\ $$
Commented by MJS last updated on 03/Apr/20
what′s the value of your formula?
$$\mathrm{what}'\mathrm{s}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{your}\:\mathrm{formula}? \\ $$
Commented by redmiiuser last updated on 03/Apr/20
from the closed forms its much efficent to insert the borders
$${from}\:{the}\:{closed}\:{forms} \\ $$$${its}\:{much}\:{efficent}\:{to} \\ $$$${insert}\:\:{the}\:{borders} \\ $$
Commented by redmiiuser last updated on 03/Apr/20
Anyway Thanks Sir!
$${Anyway}\:{Thanks}\:{Sir}! \\ $$
Commented by mr W last updated on 03/Apr/20
as much as one can not get the value of the infinite series, i think such a solution is not so useful.
$${as}\:{much}\:{as}\:{one}\:{can}\:{not}\:{get}\:{the}\:{value} \\ $$$${of}\:{the}\:{infinite}\:{series},\:{i}\:{think}\:{such}\:{a} \\ $$$${solution}\:{is}\:{not}\:{so}\:{useful}. \\ $$
Commented by redmiiuser last updated on 03/Apr/20
The process is not a closed form but yet it is efficent. Anyway Thanks for your comment.
$${The}\:{process}\:{is}\:{not}\: \\ $$$${a}\:{closed}\:{form}\:{but}\:{yet} \\ $$$${it}\:{is}\:{efficent}. \\ $$$${Anyway}\:{Thanks}\:{for} \\ $$$${your}\:{comment}. \\ $$
Answered by MJS last updated on 03/Apr/20
∫((1−x^4 )/(1+x^4 ))dx=∫((2/(x^4 +1))−x)dx= =−∫dx−((√2)/2)∫((x−(√2))/(x^2 −(√2)x+1))dx+((√2)/2)∫((x+(√2))/(x^2 +(√2)x+1))dx= =−x− −((√2)/4)ln (x^2 −(√2)x+1) +((√2)/2)arctan ((√2)x−1) + +((√2)/4)ln (x^2 +(√2)x+1) +((√2)/2)arctan ((√2)x+1) +C= =−x+((√2)/4)(ln ((x^2 +(√2)x+1)/(x^2 −(√2)x+1)) +2(arctan ((√2)x−1) +arctan ((√2)x+1))) +C=
$$\int\frac{\mathrm{1}−{x}^{\mathrm{4}} }{\mathrm{1}+{x}^{\mathrm{4}} }{dx}=\int\left(\frac{\mathrm{2}}{{x}^{\mathrm{4}} +\mathrm{1}}−{x}\right){dx}= \\ $$$$=−\int{dx}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\int\frac{{x}−\sqrt{\mathrm{2}}}{{x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{x}+\mathrm{1}}{dx}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\int\frac{{x}+\sqrt{\mathrm{2}}}{{x}^{\mathrm{2}} +\sqrt{\mathrm{2}}{x}+\mathrm{1}}{dx}= \\ $$$$=−{x}− \\ $$$$\:\:\:\:\:−\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\mathrm{ln}\:\left({x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{x}+\mathrm{1}\right)\:+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{arctan}\:\left(\sqrt{\mathrm{2}}{x}−\mathrm{1}\right)\:+ \\ $$$$\:\:\:\:\:+\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\mathrm{ln}\:\left({x}^{\mathrm{2}} +\sqrt{\mathrm{2}}{x}+\mathrm{1}\right)\:+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{arctan}\:\left(\sqrt{\mathrm{2}}{x}+\mathrm{1}\right)\:+{C}= \\ $$$$=−{x}+\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\left(\mathrm{ln}\:\frac{{x}^{\mathrm{2}} +\sqrt{\mathrm{2}}{x}+\mathrm{1}}{{x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{x}+\mathrm{1}}\:+\mathrm{2}\left(\mathrm{arctan}\:\left(\sqrt{\mathrm{2}}{x}−\mathrm{1}\right)\:+\mathrm{arctan}\:\left(\sqrt{\mathrm{2}}{x}+\mathrm{1}\right)\right)\right)\:+{C}= \\ $$
Commented by redmiiuser last updated on 03/Apr/20
sir its a definite integration.
$${sir}\:{its}\:{a}\:{definite} \\ $$$${integration}. \\ $$
Commented by MJS last updated on 03/Apr/20
you can insert the borders
$$\mathrm{you}\:\mathrm{can}\:\mathrm{insert}\:\mathrm{the}\:\mathrm{borders} \\ $$
Commented by redmiiuser last updated on 03/Apr/20
sir can you make the process more smooth.
$${sir}\:{can}\:{you}\:{make}\:{the} \\ $$$${process}\:{more}\:{smooth}. \\ $$
Commented by M±th+et£s last updated on 03/Apr/20
thank you its easy to insert the borders now
$${thank}\:{you}\:{its}\:{easy}\:{to}\:{insert}\:{the}\:{borders}\:{now} \\ $$
Commented by redmiiuser last updated on 03/Apr/20
God bless you Sir. Have a nice day.
$${God}\:{bless}\:{you}\:\:{Sir}. \\ $$$${Have}\:{a}\:{nice}\:{day}. \\ $$
Answered by TANMAY PANACEA. last updated on 03/Apr/20
(−1)∫_0 ^(π/2) ((x^4 +1−2)/(1+x^4 ))dx ∫_0 ^(π/2) (2/(1+x^4 ))−∫_0 ^(π/2) dx ∫_0 ^(π/2) ((1+(1/x^2 )−(1−(1/x^2 )))/(x^2 +(1/x^2 )))−∫_0 ^(π/2) dx ∫_0 ^(π/2) ((d(x−(1/x)))/((x−(1/x))^2 +2))−∫_0 ^(π/2) ((d(x+(1/x)))/((x+(1/x))^2 −2))−∫_0 ^(π/2) dx ∣(1/( (√2)))tan^(−1) (((x−(1/x))/( (√2))))−(1/(2(√2)))ln(((x+(1/x)−(√2))/(x+(1/x)+(√2))))−x∣_0 ^(π/2) {(1/( (√2)))tan^(−1) ((((π/2)−(2/π))/( (√2))))−(1/(2(√2)))ln((((π/2)+(2/π)−(√2))/((π/2)+(2/π)+(√2))))−(π/2)}− {(1/( (√2)))tan^(−1) (−∞)−(1/(2(√2)))ln(((0+1−0)/(0+1−0)))−0} =(1/( (√2)))tan^(−1) (((π^2 −4)/(2(√2) π)))−(1/(2(√2)))ln(((π^2 +2^2 −2(√2) π)/(π^2 +2^2 +2(√2)π)))−(π/2)−(1/( (√2)))(((−π)/2)) =(1/( (√2)))tan^(−1) (((π^2 −4)/(2(√(2π)))))−(1/(2(√2)))ln(((π^2 +4−2(√2) π)/(π^2 +4+2(√2) π)))−(π/2)(1+(1/( (√2)))) ★ln(((x+(1/x)−(√2))/(x+(1/x)+(√2))))→ln(((x^2 +1−(√2) x)/(x^2 +1+(√2) x))) ★
$$\left(−\mathrm{1}\right)\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{x}^{\mathrm{4}} +\mathrm{1}−\mathrm{2}}{\mathrm{1}+{x}^{\mathrm{4}} }{dx} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{2}}{\mathrm{1}+{x}^{\mathrm{4}} }−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {dx} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)}{{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {dx} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{d}\left({x}−\frac{\mathrm{1}}{{x}}\right)}{\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} +\mathrm{2}}−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{d}\left({x}+\frac{\mathrm{1}}{{x}}\right)}{\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} −\mathrm{2}}−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {dx} \\ $$$$\mid\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{tan}^{−\mathrm{1}} \left(\frac{{x}−\frac{\mathrm{1}}{{x}}}{\:\sqrt{\mathrm{2}}}\right)−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{ln}\left(\frac{{x}+\frac{\mathrm{1}}{{x}}−\sqrt{\mathrm{2}}}{{x}+\frac{\mathrm{1}}{{x}}+\sqrt{\mathrm{2}}}\right)−{x}\mid_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$\left\{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{tan}^{−\mathrm{1}} \left(\frac{\frac{\pi}{\mathrm{2}}−\frac{\mathrm{2}}{\pi}}{\:\sqrt{\mathrm{2}}}\right)−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{ln}\left(\frac{\frac{\pi}{\mathrm{2}}+\frac{\mathrm{2}}{\pi}−\sqrt{\mathrm{2}}}{\frac{\pi}{\mathrm{2}}+\frac{\mathrm{2}}{\pi}+\sqrt{\mathrm{2}}}\right)−\frac{\pi}{\mathrm{2}}\right\}− \\ $$$$ \\ $$$$\left\{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{tan}^{−\mathrm{1}} \left(−\infty\right)−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{ln}\left(\frac{\mathrm{0}+\mathrm{1}−\mathrm{0}}{\mathrm{0}+\mathrm{1}−\mathrm{0}}\right)−\mathrm{0}\right\} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{tan}^{−\mathrm{1}} \left(\frac{\pi^{\mathrm{2}} −\mathrm{4}}{\mathrm{2}\sqrt{\mathrm{2}}\:\pi}\right)−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{ln}\left(\frac{\pi^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{2}}\:\pi}{\pi^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{2}}\pi}\right)−\frac{\pi}{\mathrm{2}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left(\frac{−\pi}{\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{tan}^{−\mathrm{1}} \left(\frac{\pi^{\mathrm{2}} −\mathrm{4}}{\mathrm{2}\sqrt{\mathrm{2}\pi}}\right)−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{ln}\left(\frac{\pi^{\mathrm{2}} +\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}\:\pi}{\pi^{\mathrm{2}} +\mathrm{4}+\mathrm{2}\sqrt{\mathrm{2}}\:\pi}\right)−\frac{\pi}{\mathrm{2}}\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right) \\ $$$$\bigstar{ln}\left(\frac{{x}+\frac{\mathrm{1}}{{x}}−\sqrt{\mathrm{2}}}{{x}+\frac{\mathrm{1}}{{x}}+\sqrt{\mathrm{2}}}\right)\rightarrow{ln}\left(\frac{{x}^{\mathrm{2}} +\mathrm{1}−\sqrt{\mathrm{2}}\:{x}}{{x}^{\mathrm{2}} +\mathrm{1}+\sqrt{\mathrm{2}}\:{x}}\right)\:\bigstar \\ $$$$ \\ $$
Commented by TANMAY PANACEA. last updated on 03/Apr/20
most welcome sir...
$${most}\:{welcome}\:{sir}… \\ $$
Commented by M±th+et£s last updated on 03/Apr/20
thank you sir
$${thank}\:{you}\:{sir} \\ $$
Commented by peter frank last updated on 03/Apr/20
thank you both
$${thank}\:{you}\:{both} \\ $$

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