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Question Number 139353 by mohammad17 last updated on 26/Apr/21
∫0π2sin6θcos4θdθ
Answered by Dwaipayan Shikari last updated on 26/Apr/21
∫0π2sin2α−1θcos2β−1θdθ=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=∫0π/2sin6θcos4θdθ=∫0π/2cos6θsin4θdθ2I=∫0π/2sin4θcos4θdθ32I=∫0π/2sin42θdθ128I=∫0π/2(1−cos4θ)2dθ256I=∫0π/2(2−4cos4θ+1+cos8θ)dθ256I=3π2I=3π512
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