0-pi-2-1-x-4-1-x-4-dx-

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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_{0}^{\frac{π}{2}} \frac{1 - x^{4}}{1 + x^{4}} d x

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))

{(1 + x^{4})}^{- 1}

= \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} . x^{4 n}

(1 - x^{4}) {(1 + x^{4})}^{- 1}

= (1 - x^{4}) . (\underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} . x^{4 n})

= \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} . x^{4 n} - \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} . x^{4 n + 4}

\int_{0}^{\frac{π}{2}} \frac{1 - x^{4}}{1 + x^{4}} d x

Missing \left or extra \right

= \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n} . {(\frac{π}{2})}^{4 n + 1}}{4 n + 1} - \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n} . {(\frac{π}{2})}^{4 n + 5}}{4 n + 5}

Commented by redmiiuser last updated on 03/Apr/20

pls pls pls check

p l s p l s p l s c h e c k

Commented by M±th+et£s last updated on 03/Apr/20

its right sir thank you

i t s r i g h t s i r t h a n k y o u

Commented by redmiiuser last updated on 03/Apr/20

welcome mister

w e l c o m e m i s t e r

Commented by MJS last updated on 03/Apr/20

what′s the value of your formula?

{what}^{'} s the value of your formula ?

Commented by redmiiuser last updated on 03/Apr/20

from the closed forms its much efficent to insert the borders

f r o m t h e c l o s e d f o r m s

i t s m u c h e f f i c e n t t o

i n s e r t t h e b o r d e r s

Commented by redmiiuser last updated on 03/Apr/20

Anyway Thanks Sir!

A n y w a y T h a n k s S i r!

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.

a s m u c h a s o n e c a n n o t g e t t h e v a l u e

o f t h e i n f i n i t e s e r i e s, i t h i n k s u c h a

s o l u t i o n i s n o t s o u s e f u l .

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.

T h e p r o c e s s i s n o t

a c l o s e d f o r m b u t y e t

i t i s e f f i c e n t .

A n y w a y T h a n k s f o r

y o u r c o m m e n t .

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{1 - x^{4}}{1 + x^{4}} d x = \int (\frac{2}{x^{4} + 1} - x) d x =

= - \int d x - \frac{\sqrt{2}}{2} \int \frac{x - \sqrt{2}}{x^{2} - \sqrt{2} x + 1} d x + \frac{\sqrt{2}}{2} \int \frac{x + \sqrt{2}}{x^{2} + \sqrt{2} x + 1} d x =

= - x -

- \frac{\sqrt{2}}{4} \ln (x^{2} - \sqrt{2} x + 1) + \frac{\sqrt{2}}{2} \arctan (\sqrt{2} x - 1) +

+ \frac{\sqrt{2}}{4} \ln (x^{2} + \sqrt{2} x + 1) + \frac{\sqrt{2}}{2} \arctan (\sqrt{2} x + 1) + C =

= - x + \frac{\sqrt{2}}{4} (\ln \frac{x^{2} + \sqrt{2} x + 1}{x^{2} - \sqrt{2} x + 1} + 2 (\arctan (\sqrt{2} x - 1) + \arctan (\sqrt{2} x + 1))) + C =

Commented by redmiiuser last updated on 03/Apr/20

sir its a definite integration.

s i r i t s a d e f i n i t e

i n t e g r a t i o n .

Commented by MJS last updated on 03/Apr/20

you can insert the borders

you can insert the borders

Commented by redmiiuser last updated on 03/Apr/20

sir can you make the process more smooth.

s i r c a n y o u m a k e t h e

p r o c e s s m o r e s m o o t h .

Commented by M±th+et£s last updated on 03/Apr/20

thank you its easy to insert the borders now

t h a n k y o u i t s e a s y t o i n s e r t t h e b o r d e r s n o w

Commented by redmiiuser last updated on 03/Apr/20

God bless you Sir. Have a nice day.

G o d b l e s s y o u S i r .

H a v e a n i c e d a y .

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))) ★

(- 1) \int_{0}^{\frac{π}{2}} \frac{x^{4} + 1 - 2}{1 + x^{4}} d x

\int_{0}^{\frac{π}{2}} \frac{2}{1 + x^{4}} - \int_{0}^{\frac{π}{2}} d x

\int_{0}^{\frac{π}{2}} \frac{1 + \frac{1}{x^{2}} - (1 - \frac{1}{x^{2}})}{x^{2} + \frac{1}{x^{2}}} - \int_{0}^{\frac{π}{2}} d x

\int_{0}^{\frac{π}{2}} \frac{d (x - \frac{1}{x})}{{(x - \frac{1}{x})}^{2} + 2} - \int_{0}^{\frac{π}{2}} \frac{d (x + \frac{1}{x})}{{(x + \frac{1}{x})}^{2} - 2} - \int_{0}^{\frac{π}{2}} d x

∣ \frac{1}{\sqrt{2}} {t a n}^{- 1} (\frac{x - \frac{1}{x}}{\sqrt{2}}) - \frac{1}{2 \sqrt{2}} l n (\frac{x + \frac{1}{x} - \sqrt{2}}{x + \frac{1}{x} + \sqrt{2}}) - x ∣_{0}^{\frac{π}{2}}

{\frac{1}{\sqrt{2}} {t a n}^{- 1} (\frac{\frac{π}{2} - \frac{2}{π}}{\sqrt{2}}) - \frac{1}{2 \sqrt{2}} l n (\frac{\frac{π}{2} + \frac{2}{π} - \sqrt{2}}{\frac{π}{2} + \frac{2}{π} + \sqrt{2}}) - \frac{π}{2}} -

{\frac{1}{\sqrt{2}} {t a n}^{- 1} (- \infty) - \frac{1}{2 \sqrt{2}} l n (\frac{0 + 1 - 0}{0 + 1 - 0}) - 0}

= \frac{1}{\sqrt{2}} {t a n}^{- 1} (\frac{π^{2} - 4}{2 \sqrt{2} π}) - \frac{1}{2 \sqrt{2}} l n (\frac{π^{2} + 2^{2} - 2 \sqrt{2} π}{π^{2} + 2^{2} + 2 \sqrt{2} π}) - \frac{π}{2} - \frac{1}{\sqrt{2}} (\frac{- π}{2})

= \frac{1}{\sqrt{2}} {t a n}^{- 1} (\frac{π^{2} - 4}{2 \sqrt{2 π}}) - \frac{1}{2 \sqrt{2}} l n (\frac{π^{2} + 4 - 2 \sqrt{2} π}{π^{2} + 4 + 2 \sqrt{2} π}) - \frac{π}{2} (1 + \frac{1}{\sqrt{2}})

★ l n (\frac{x + \frac{1}{x} - \sqrt{2}}{x + \frac{1}{x} + \sqrt{2}}) \to l n (\frac{x^{2} + 1 - \sqrt{2} x}{x^{2} + 1 + \sqrt{2} x}) ★

Commented by TANMAY PANACEA. last updated on 03/Apr/20

most welcome sir...

m o s t w e l c o m e s i r \dots

Commented by M±th+et£s last updated on 03/Apr/20

thank you sir

t h a n k y o u s i r

Commented by peter frank last updated on 03/Apr/20

thank you both

t h a n k y o u b o t h

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