>>22

(1)
　∫arctan(x)dx = x･arctan(x) - ∫{x/(1+x^2)}dx = x･arctan(x) - (1/2)log(1+x^2) +c,
　(与式) = π/(√3) - log(2) = 1.12065218367427254117684613618398 …

(2)
　∫[√(1-x^2), x+1] y･dy = [ (1/2)y^2 ](y=√(1-x^2)→x+1) = x(x+1),　　(0<x<1)
　(与式) = ∫[0,1] (x^2)(x+1)dx = [ (1/4)x^4 + (1/3)x^3 ](x=0→1) = (1/4) + (1/3) = 7/12.

(3)
　∫[0,x^2] e^(y/x)dy = [ x･e^(y/x) ](y=0→x^2) = x･(e^x),
　(与式) = ∫[0,1] x･(e^x)dx = [ (x-1)(e^x) ](x=0→1) = 1.
　
(4)
　∫[0,sin(x)] (y^3)dy = [ (1/4)y^4 ](y=0→sin(x)) = (1/4){sin(x)}^4,
まづ、sin(x)^2 = {1-cos(2x)}/2,　cos(x)^2 = {1+cos(2x)}/2　を使って cos(2x)の多項式にする。
　sin(x)^4･cos(x)^2 = {1-cos(x)^2}^2･cos(x)^2
　= (1/8){1-cos(2x)}^2･(1+cos(2x)) = (1/8){1 -cos(2x) -cos(2x)^2 +cos(2x)^3}
次に、cos(2x)^2 = {1+cos(4x)}/2,　cos(2x)^3 = {3cos(2x)+cos(6x)}/4　を使ってフーリエ展開。
　sin(x)^4･cos(x)^2 = (1/32){2 -cos(2x) -2cos(4x) +cos(6x) },
[0,π/2] で積分すると、定数項だけ残る.
　(1/32)[2x -(1/2)sin(2x) -(1/2)sin(4x) +(1/6)sin(6x) ](x=0→π/2) = π/32.
　(与式) = (1/4)･(π/32) = π/128.

(5)
　∫[0,y] x･dx = [ (1/2)x^2 ](x=0→y) = (1/2)y^2,
　(与式) = ∫[0,π] {∫[0,y] x･dx}{sin(y)/y}dy = (1/2)∫[0,π] y･sin(y)dy
　= (1/2)[ sin(y) - y･cos(y) ](y=0→π) = π/2.