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Use-gamma-function-to-prove-i-0-8-cos-3-4x-dx-1-6-ii-0-6-cos-4-3-sin-2-6-d-5-192-




Question Number 90574 by niroj last updated on 24/Apr/20
 Use gamma function to prove    (i) .  ∫_0 ^(  (𝛑/8))  cos^3 4x dx= (1/6).    (ii). ∫_0 ^( (𝛑/6))  cos^4 3𝛉 sin^2 6𝛉 d𝛉 = ((5𝛑)/(192)).
Usegammafunctiontoprove(i).0π8cos34xdx=16.(ii).0π6cos43θsin26θdθ=5π192.
Commented by mathmax by abdo last updated on 24/Apr/20
I =∫_0 ^(π/8)  cos^3 (4x)dx ⇒ I =_(4x=t)    (1/4)∫_0 ^(π/2)  cos^3 (t) dt  =(1/4) ∫_0 ^(π/2)   cos^2 t  sint dt  B(p,q) =∫_0 ^(π/2)  cos^(2p−1) x sin^(2q−1) xdx   ⇒∫_0 ^(π/2)  cos^2 t sin^2 t dt =∫_0 ^(π/2)  cos^(2((3/2))−1) t sin^(2(1)−1) tdt  =B((3/2),1) =((Γ((3/2)).Γ(1))/(Γ((3/2)+1))) we have    Γ(n)=(n−1)! ⇒Γ(1)=1  Γ(x+1)=xΓ(x) ⇒  Γ((3/2)+1) =(3/2)Γ((3/2)) ⇒B((3/2),1) =((Γ((3/2)))/((3/2)Γ((3/2)))) =(2/3) ⇒  I =(1/4)×(2/3) =(1/6)
I=0π8cos3(4x)dxI=4x=t140π2cos3(t)dt=140π2cos2tsintdtB(p,q)=0π2cos2p1xsin2q1xdx0π2cos2tsin2tdt=0π2cos2(32)1tsin2(1)1tdt=B(32,1)=Γ(32).Γ(1)Γ(32+1)wehaveΓ(n)=(n1)!Γ(1)=1Γ(x+1)=xΓ(x)Γ(32+1)=32Γ(32)B(32,1)=Γ(32)32Γ(32)=23I=14×23=16
Commented by niroj last updated on 25/Apr/20
thank you .
thankyou.

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