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Question Number 124957 by Study last updated on 07/Dec/20

Commented by Study last updated on 07/Dec/20

please help me??

pleasehelpme??

Answered by Dwaipayan Shikari last updated on 07/Dec/20

∫_0 ^(π/2) ((tanx))^(1/4)  dx  =∫_0 ^(π/2) sin^(1/4) x cos^(−(1/4)) x dx        sin^2 x=t⇒sin2x=(dt/dx)  =(1/2)∫_0 ^1 t^((1/8)−(1/2)) (1−t)^(−(1/8)−(1/2))  dt    =(1/2).((Γ((1/8)+(1/2))Γ((1/2)−(1/8)))/(Γ(1)))  =((Γ((5/8))Γ((3/8)))/2)=(π/(2sin(((3π)/8))))=(π/(2cos(π/8)))=(π/(((√(2+(√2))))))

0π2tanx4dx=0π2sin14xcos14xdxsin2x=tsin2x=dtdx=1201t1812(1t)1812dt=12.Γ(18+12)Γ(1218)Γ(1)=Γ(58)Γ(38)2=π2sin(3π8)=π2cosπ8=π(2+2)

Commented by mathmax by abdo last updated on 07/Dec/20

I=∫_0 ^(π/2)  ^4 (√(tanx))dx we do the changement (tanx)^(1/4) =t ⇒tanx=t^4  ⇒  x=arctan(t^4 ) ⇒ I =∫_0 ^∞    ((t×(4t^3 ))/(1+t^8 ))dt =4∫_0 ^∞   (t^4 /(1+t^8 ))dt  =_(t=u^(1/8) )    4 ∫_0 ^∞   (u^(1/2) /(1+u))×(1/8) u^((1/8)−1)  du =(1/2)∫_0 ^∞  (u^((1/2)+(1/8)−1) /(1+u))du  =(1/2)∫_0 ^∞   (u^((5/8)−1) /(1+u))du =(1/2)×(π/(sin(((5π)/8))))=(π/(2sin((π/2)+(π/8))))  =(π/(2cos((π/8))))=(π/(2×((√(2+(√2)))/2))) =(π/( (√(2+(√2))))) ⇒I=(π/( (√(2+(√2)))))

I=0π24tanxdxwedothechangement(tanx)14=ttanx=t4x=arctan(t4)I=0t×(4t3)1+t8dt=40t41+t8dt=t=u1840u121+u×18u181du=120u12+1811+udu=120u5811+udu=12×πsin(5π8)=π2sin(π2+π8)=π2cos(π8)=π2×2+22=π2+2I=π2+2

Commented by Study last updated on 07/Dec/20

why we chose border than 0 to ∞

whywechoseborderthan0to

Commented by Study last updated on 07/Dec/20

why (1/2)∫(u^((5/8)−1) /(1+u))du=(1/2)×(π/(sin(((5π)/8))))

why12u5811+udu=12×πsin(5π8)

Commented by Dwaipayan Shikari last updated on 07/Dec/20

∫_0 ^∞ (u^n /(1+u))du   take (u/(1+u))=t⇒(1/((1+u)^2 ))=(dt/du)  =∫_0 ^1 t^((1/n)−1) (1−t)^(−(1/n)) dt =Γ((1/n))Γ(1−(1/n))=(π/(sin((π/n))))

0un1+udutakeu1+u=t1(1+u)2=dtdu=01t1n1(1t)1ndt=Γ(1n)Γ(11n)=πsin(πn)

Commented by Bird last updated on 07/Dec/20

∫_0 ^∞  (t^(a−1) /(1+t))dt =(π/(sin(πa))) with0<a<1  this result is proved see the platform

0ta11+tdt=πsin(πa)with0<a<1thisresultisprovedseetheplatform

Commented by Study last updated on 08/Dec/20

ans me please???

ansmeplease???

Commented by Study last updated on 08/Dec/20

ans me please??

ansmeplease??

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