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Question Number 139353 by mohammad17 last updated on 26/Apr/21

∫_0 ^( (π/2)) sin^6 θ cos^4 θ dθ

0π2sin6θcos4θdθ

Answered by Dwaipayan Shikari last updated on 26/Apr/21

∫_0 ^(π/2) sin^(2α−1) θ cos^(2β−1) θ dθ=((Γ(α)Γ(β))/(2Γ(α+β)))  ∫_0 ^(π/2) sin^6 θ cos^4 θ dθ=((Γ((7/2))Γ((5/2)))/(2Γ(6)))=(((5/2).(3/2).(1/2).(3/2).(1/2)Γ^2 ((1/2)))/(240))  =(3/2^9 )π=((3π)/(512))

0π2sin2α1θcos2β1θdθ=Γ(α)Γ(β)2Γ(α+β)0π2sin6θcos4θdθ=Γ(72)Γ(52)2Γ(6)=52.32.12.32.12Γ2(12)240=329π=3π512

Answered by ajfour last updated on 26/Apr/21

I=∫_0 ^( π/2) sin^6 θcos^4 θdθ    =∫_0 ^( π/2) cos^6 θsin^4 θdθ  2I=∫_0 ^( π/2) sin^4 θcos^4 θdθ  32I=∫_0 ^( π/2) sin^4 2θdθ  128I=∫_0 ^( π/2) (1−cos 4θ)^2 dθ  256I=∫_0 ^( π/2) (2−4cos 4θ+1+cos 8θ)dθ  256I=((3π)/2)  I=((3π)/(512))

I=0π/2sin6θcos4θdθ=0π/2cos6θsin4θdθ2I=0π/2sin4θcos4θdθ32I=0π/2sin42θdθ128I=0π/2(1cos4θ)2dθ256I=0π/2(24cos4θ+1+cos8θ)dθ256I=3π2I=3π512

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