1
∫(0^1/2)[arcsinx /√(1+x^2)]*dx
这个没想好
若∫(0^1/2)[arcsinx /√(1-x^2)]*dx可以做
∫(0^1/2)[arcsinx /√(1-x^2)]*dx
=ʃ(0~1/2)arcsinxd(arcsinx)
=1/2(arcsinx)²|(0~1/2
=π²/72
2.∫-1^1[1/2xe^x2]*dx
=∫-1^1[e^(x^2)dx^2
=e^(x^2))|(-1~1)
=0
3.∫0^a/2[xdx / √(a^2-x^2)] ( a>0)
=1/2∫0^a/2[dx² / √(a^2-x^2)]
设√(a²-x²)=t,
那么a²-x²=t²
∴x²=a²-t²
∴dx²=-2tdt
∴ʃdx²/√(a²-x²)=ʃ-2dt=-2t=-2√(a²-x²)
∴
∫0^a/2[xdx / √(a^2-x^2)] ( a>0)
=1/2∫0^a/2[dx² / √(a^2-x^2)]
=-√(a²-x²)|(0~a/2)
=-√3/2a+a
=(2-√3)a/2