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[1+根号下(1-x的平方)]分之一的不定积分
1个回答
展开全部
令x=sint -π\2<=t<=π\2
则dx=costdt
∫1\[1+(1-x^2)^(1\2)]dx
=∫costdt\(1+cost)
=∫[1-1\(1+cost)]dt
=∫{1-(1\2)*[sec(t\2)]^2}dt
=t-tan(t\2)+C
=arcsinx-x\[1+(1-x^2)^(1\2)]+C
则dx=costdt
∫1\[1+(1-x^2)^(1\2)]dx
=∫costdt\(1+cost)
=∫[1-1\(1+cost)]dt
=∫{1-(1\2)*[sec(t\2)]^2}dt
=t-tan(t\2)+C
=arcsinx-x\[1+(1-x^2)^(1\2)]+C
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