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Question Number 107923 by bemath last updated on 13/Aug/20

     ((⊚BeMath⊚)/)      ∫_0 ^1  x^(9/2)  (1−x)^(5/2)  dx ?

BeMath10x92(1x)52dx?

Answered by 1549442205PVT last updated on 13/Aug/20

Set (1/x)−1=t^2 ⇒−(1/x^2 )dx=2tdt⇒dx=−2tx^2 dt  t(x=1)=0,t(x=0)=∞  x=(1/( (√(1+t^2 ))))⇒I=∫_0 ^∞ x^(9/2) (xt^2 )^(1/2) 2tx^2 dt  =2∫_0 ^∞ x^7 t^2 dt=2∫_0 ^∞ ((t^2 dt)/( (√((1+t^2 )^7 ))))  =2∫_1 ^∞  ((t^2 +1−1)/((1+t^2 )^3 (√(1+t^2 ))))dt=2∫(dt/((1+t^2 )^2 (√(1+t^2 ))))  −2∫_0 ^∞ (dt/((1+t^2 )^3 (√(1+t^2 ))))=  Set t=tanϕ⇒dt=(1+t^2 )dϕ  I=2∫_0 ^(π/2) (dϕ/((1+tan^2 ϕ)(√(1+tan^2 ϕ)))))  −2∫_0 ^(π/2) (dϕ/((1+tan^2 ϕ)^2 (√(1+tan^2 ϕ)))))  =2∫_0 ^(π/2)  cos^3 ϕdϕ−2∫_0 ^(π/2) cos^5 dϕ  =2∫_0 ^(π/2) (1−sin^2 ϕ)dsinϕ−2∫_0 ^(π/2) (1−sin^2 ϕ)^2 dsinϕ  =(2sinϕ−((2sin^3 ϕ)/3))∣_0 ^(π/2) −2∫_0 ^(π/2) [1−2sin^2 ϕ+sin^4 ϕ)dsinϕ  =2−(2/3)−2[sinϕ−(2/3)sin^3 ϕ+((sin^5 ϕ)/5)]_0 ^(π/2)   =2−(2/3)−2(1−(2/3)+(1/5))=(4/(15))

Set1x1=t21x2dx=2tdtdx=2tx2dtt(x=1)=0,t(x=0)=x=11+t2I=0x92(xt2)122tx2dt=20x7t2dt=20t2dt(1+t2)7=21t2+11(1+t2)31+t2dt=2dt(1+t2)21+t220dt(1+t2)31+t2=Sett=tanφdt=(1+t2)dφI=20π2dφ(1+tan2φ)1+tan2φ)20π2dφ(1+tan2φ)21+tan2φ)=20π2cos3φdφ20π2cos5dφ=20π2(1sin2φ)dsinφ20π2(1sin2φ)2dsinφ=(2sinφ2sin3φ3)0π220π2[12sin2φ+sin4φ)dsinφ=2232[sinφ23sin3φ+sin5φ5]0π2=2232(123+15)=415

Answered by bobhans last updated on 13/Aug/20

    ((BobHans)/Π)  I=−(2/7)∫_0 ^1  x^(9/2)  d(1−x)^(7/2)  = −(2/7)[ x^(9/2) (1−x)^(7/2) ]_0 ^1 +  (9/7)∫_0 ^1  x^(7/2) (1−x)^(7/2)  dx   let J = ∫_0 ^1  x^(7/2)  (1−x)^(7/2)  dx=(9/7)∫_0 ^1  (x−x^2 )^(7/2)  dx  J=(9/7)∫_0 ^1  [ (1/4)−((1/2)−x)^2  ]^(7/2) dx  set (1/2)−x = z   J=(9/(7×2^8 )) ∫_(−π/2) ^(π/2)  cos^8  z dz    Reduction formula   J_n  = ((n−1)/n).J_(n−2)  ⇒J_8 = (7/8).(5/6).(3/4).(1/2).J_0   J_8 = ((35)/(128)).[(π/2)−(−(π/2))]= ((35π)/(128))  conclusion   I= (9/(7×2^8 )) ×((35π)/(128)) = ((45π)/(32768)).

BobHansΠI=2710x9/2d(1x)7/2=27[x9/2(1x)7/2]01+9710x7/2(1x)7/2dxletJ=10x7/2(1x)7/2dx=9710(xx2)7/2dxJ=9710[14(12x)2]7/2dxset12x=zJ=97×28π/2π/2cos8zdzReductionformulaJn=n1n.Jn2J8=78.56.34.12.J0J8=35128.[π2(π2)]=35π128conclusionI=97×28×35π128=45π32768.

Commented by bemath last updated on 14/Aug/20

correct....thank you

correct....thankyou

Answered by mathmax by abdo last updated on 13/Aug/20

I =∫_0 ^1  x^(9/2) (1−x)^(5/2)  dx  we know B(p,q) =∫_0 ^1  x^(p−1) (1−x)^(q−1)  dx  =((Γ(p).Γ(q))/(Γ(p+q)))   here p=(9/2)+1 and q=(5/2)+1 ⇒  I =((Γ((9/2)+1)Γ((5/2)+1))/(Γ((9/2)+(5/2)+2))) =(9/2).(5/2).((Γ((9/2))Γ((5/2)))/(Γ(9))) =((45)/(4.8!)) Γ((9/2))Γ((5/2))  Γ((9/2)) =Γ(1+(7/2)) =(7/2)Γ((7/2))=(7/2)Γ(1+(5/2))=(7/2).(5/2)Γ((5/2))  =((35)/4)Γ(1+(3/2)) =((35)/4).(3/2)Γ((3/2)) =((3.35)/8)Γ((1/2)+1) =((3.35)/(16))Γ((1/2))  we follow the same way to find Γ((5/2))

I=01x92(1x)52dxweknowB(p,q)=01xp1(1x)q1dx=Γ(p).Γ(q)Γ(p+q)herep=92+1andq=52+1I=Γ(92+1)Γ(52+1)Γ(92+52+2)=92.52.Γ(92)Γ(52)Γ(9)=454.8!Γ(92)Γ(52)Γ(92)=Γ(1+72)=72Γ(72)=72Γ(1+52)=72.52Γ(52)=354Γ(1+32)=354.32Γ(32)=3.358Γ(12+1)=3.3516Γ(12)wefollowthesamewaytofindΓ(52)

Commented by bemath last updated on 14/Aug/20

typo it should be 8! in denumerator

typoitshouldbe8!indenumerator

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