 | 残り
2
∫1/√(x(x+1)) dx は
t=x+√(x(x+1)) とおくと
t-x=√(x(x+1))
(t-x)^2=x(x+1)
t^2-2tx+x^2=x^2+x
t^2-2tx-x=0
(2t+1)x=t^2
x=t^2/(2t+1)
dx=2t(t+1)/(2t+1)^2
√(x(x+1))=t-x=t-t^2/(2t+1)=t(t+1)/(2t+1)
なので
∫1/√(x(x+1)) dx
=∫(2t+1)/(t(t+1))・2t(t+1)/(2t+1)^2 dt
=∫2/(2t+1) dt
=log|2t+1|+C
=log|2x+2√(x(x+1))+1|+C
3
∫[-π/2～π/2](x-cos2x)^2 dx
=∫[-π/2～π/2]x^2-2xcos2x+(cos2x)^2 dx
=[x^3/3-xsin2x-cos2x/2+x/2+sin4x/8][-π/2～π/2]
=(π^3/24+1/2+π/4)-(-π^3/24+1/2-π/4)
=π^3/12+π/2
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