∫0^1/2[arcsinx / (根号下1+x^2)]*dx 2.∫-1^1[xe^x2/2]*dx 3.∫0^a/2[xdx / (根号下a^2-x^2)]*dx a>0
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![∫0^1/2[arcsinx / (根号下1+x^2)]*dx 2.∫-1^1[xe^x2/2]*dx 3.∫0^a/2[xdx / (根号下a^2-x^2)]*dx a>0](/uploads/image/z/5391700-52-0.jpg?t=%E2%88%AB0%5E1%2F2%5Barcsinx+%2F+%28%E6%A0%B9%E5%8F%B7%E4%B8%8B1%2Bx%5E2%29%5D%2Adx+2.%E2%88%AB-1%5E1%5Bxe%5Ex2%2F2%5D%2Adx+3.%E2%88%AB0%5Ea%2F2%5Bxdx+%2F+%28%E6%A0%B9%E5%8F%B7%E4%B8%8Ba%5E2-x%5E2%29%5D%2Adx+a%3E0)
∫0^1/2[arcsinx / (根号下1+x^2)]*dx 2.∫-1^1[xe^x2/2]*dx 3.∫0^a/2[xdx / (根号下a^2-x^2)]*dx a>0
∫0^1/2[arcsinx / (根号下1+x^2)]*dx 2.∫-1^1[xe^x2/2]*dx 3.∫0^a/2[xdx / (根号下a^2-x^2)]*dx a>0
∫0^1/2[arcsinx / (根号下1+x^2)]*dx 2.∫-1^1[xe^x2/2]*dx 3.∫0^a/2[xdx / (根号下a^2-x^2)]*dx a>0
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