Question

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,求详细过

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,求详细过程，谢谢，我换元用的是x=sint，真不知道，答案到底是怎么算的，求高人，详细过程，非常过程。

Answer 1

应该是书的答案错了,我都计算到1-π/4
过程如下:
∫<1/√2,1>√(1-x²)/x² dx,令x=sin(u),dx=cos(u)
x=1/√2,u=π/4,x=1,u=π/2
=∫<π/4,π/2>cos(u)√[1-sin²(u)]/sin²(u)du
=∫<π/4,π/2>cos²(u)/sin²(u)du
=∫<π/4,π/2>cot²(u)du
=∫<π/4,π/2>[csc²(u)-1]du
=[-cot(u)-u]<π/4,π/2>
=[-cot(π/2)-π/2]-[-cot(π/4)-π/4]
=-π/2+1+π/4
=1-π/4≈0.214602

Answer 2

怎么我用三角代换算的和你的答案是一样的啊