Question Number 149739 by iloveisrael last updated on 07/Aug/21

Answered by Olaf_Thorendsen last updated on 07/Aug/21

λ = ∫_0 ^π xsin^9 x dx (1) λ = ∫_π ^0 (π−u)sin^9 (π−u) (−du) λ = ∫_0 ^π (π−u)sin^9 u du (2) (1)+(2) : 2λ = π∫_0 ^π sin^9 x dx λ = (π/2)∫_0 ^(π/2) sin^9 x dx+(π/2)∫_(π/2) ^π sin^9 x dx λ = (π/2)∫_0 ^(π/2) sin^9 x dx+(π/2)∫_0 ^(π/2) sin^9 (t+(π/2)) dt λ = (π/2)∫_0 ^(π/2) sin^9 x dx+(π/2)∫_0 ^(π/2) cos^9 t dt λ = (π/2)W_9 +(π/2)W_9 = πW_9 (Wallis) W_(2p+1) = (((2^p p!)^2 )/((2p+1)!)) W_9 = W_(2×4+1) = (((2^4 4!)^2 )/(9!)) = ((128)/(315)) ⇒ λ = ((128π)/(315))

Answered by mathmax by abdo last updated on 07/Aug/21

Ψ=∫_0 ^π x sin^9 x dx ⇒Ψ=∫_0 ^π x(((e^(ix) −e^(−ix) )/(2i)))^9 dx =(1/((2i)^9 ))∫_0 ^π x(Σ_(k=0) ^9 C_9 ^k (e^(ix) )^k (−e^(−ix) )^(9−k) )dx =(1/((2i)^9 ))Σ_(k=0) ^9 C_9 ^k (−1)^(9−k) ∫_0 ^π x e^(ikx) e^((k−9)ix) dx =−(1/((2i)^9 ))Σ_(k=0) ^9 (−1)^k C_9 ^k u_k u_k =∫_0 ^π x e^((2k−9)ix) dx =[(x/((2k−9)i))e^((2k−9)ix) ]_0 ^π −∫_0 ^π (1/((2k−9)i))e^((2k−9)ix) dx =(π/((2k−9)i))e^((2k−9)iπ) −(1/(((2k−9)i)^2 ))[e^((2k−9)ix) ]_0 ^π =−(π/((2k−9)i))+(1/((2k−9)^2 ))(−2) ⇒ Ψ=−(1/((2i)^9 ))Σ_(k=0) ^9 (−1)^k C_9 ^k (−(π/((2k−9)i))−(2/((2k−9)^2 )))....

Answered by gsk2684 last updated on 07/Aug/21

λ+λ=∫_0 ^π xsin ^9 x dx+∫_0 ^π (π−x)sin ^9 (π−x) dx 2λ=∫_0 ^π xsin ^9 x dx+∫_0 ^π (π−x)sin ^9 x dx 2λ=∫_0 ^π xsin ^9 x dx+π∫_0 ^π sin ^9 x dx−∫_0 ^π xsin ^9 x dx 2λ=π∫_0 ^π sin ^9 xdx 2λ=2π∫_0 ^(π/2) sin ^9 xdx λ=π (8/9) (6/7) (4/5) (2/3) 1