计算定积分∫下限1/√2上限1 [√(1-x^2)/x^2]dx=?我用换元法算得1-π/4,答案给的是π/4+√2/2
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计算定积分∫下限1/√2上限1 [√(1-x^2)/x^2]dx=?我用换元法算得1-π/4,答案给的是π/4+√2/2,我换元用的是x=sint,答案到底是怎么算的,
应该是书的答案错了,我都计算到1-π/4
过程如下:
∫√(1-x²)/x² dx,令x=sin(u),dx=cos(u)
x=1/√2,u=π/4,x=1,u=π/2
=∫cos(u)√[1-sin²(u)]/sin²(u)du
=∫cos²(u)/sin²(u)du
=∫cot²(u)du
=∫[csc²(u)-1]du
=[-cot(u)-u]
=[-cot(π/2)-π/2]-[-cot(π/4)-π/4]
=-π/2+1+π/4
=1-π/4≈0.214602
过程如下:
∫√(1-x²)/x² dx,令x=sin(u),dx=cos(u)
x=1/√2,u=π/4,x=1,u=π/2
=∫cos(u)√[1-sin²(u)]/sin²(u)du
=∫cos²(u)/sin²(u)du
=∫cot²(u)du
=∫[csc²(u)-1]du
=[-cot(u)-u]
=[-cot(π/2)-π/2]-[-cot(π/4)-π/4]
=-π/2+1+π/4
=1-π/4≈0.214602
计算定积分∫下限1/√2上限1 [√(1-x^2)/x^2]dx=?我用换元法算得1-π/4,答案给的是π/4+√2/2
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