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let-A-n-0-pi-2-cos-n-xdx-1-calculate-A-0-A-2-and-A-3-2-calculate-A-n-interms-of-n-3-find-0-pi-2-cos-8-xdx-




Question Number 65770 by mathmax by abdo last updated on 03/Aug/19
let A_n =∫_0 ^(π/2)  cos^n xdx  1) calculate A_0 ,A_2  and A_3   2)calculate A_n interms of n  3) find ∫_0 ^(π/2)  cos^8 xdx
letAn=0π2cosnxdx1)calculateA0,A2andA32)calculateAnintermsofn3)find0π2cos8xdx
Commented by ~ À ® @ 237 ~ last updated on 04/Aug/19
 A_0 =[x]_0_  ^(π/2) =(π/2)  2A_n  = 2∫_0 ^(π/2) (sinx)^(2((1/2))−1) (cosx)^(2(((n+1)/2))−1) dx=B((1/2).  ((n+1)/2))  Then A_n  = (1/2). ((Γ((1/2))Γ(((n+1)/2)))/(Γ(((n+2)/2))))  when using the Lagrange formula Π_(k=1) ^(p−1) Γ(z+(k/p))=(2π)^((p−1)/2) p^((1/2)−pz) Γ(pz)  Γ(n+(1/2))=(√(π )) ((Γ(2n))/2^(2n−1) )  ∀ n>1  so we can find   A_(2n) = ((√π)/2).((((√π) Γ(2n))/2^(2n−1) )/(Γ(n+1)))= (((2n−1)!)/(2^(2n)  n!)) π  A_(2n+1) = ((√π)/2). ((Γ(n+1))/( (√π) .((Γ(2n+2))/2^(2n+1) )))=((2^(2n) .n!)/((2n+1)!))  So   A_1 =1      A_3 = ((2^2 .1!)/(3!))      A_8 =((7!π)/(2^8 4!))
A0=[x]0π2=π22An=20π2(sinx)2(12)1(cosx)2(n+12)1dx=B(12.n+12)ThenAn=12.Γ(12)Γ(n+12)Γ(n+22)whenusingtheLagrangeformulap1k=1Γ(z+kp)=(2π)p12p12pzΓ(pz)Γ(n+12)=πΓ(2n)22n1n>1sowecanfindA2n=π2.πΓ(2n)22n1Γ(n+1)=(2n1)!22nn!πA2n+1=π2.Γ(n+1)π.Γ(2n+2)22n+1=22n.n!(2n+1)!SoA1=1A3=22.1!3!A8=7!π284!
Commented by mathmax by abdo last updated on 04/Aug/19
thank you sir.
thankyousir.