作者ek0519 (热情)
看板trans_math
标题Re: [积分]
时间Fri Feb 9 14:28:41 2007
※ 引述《king911015 (早已放弃爱上你)》之铭言：
: ∫x sin^-1(x)dx
= sin^-1 x* (1/2)x^2 -∫(1/2)x^2 * 1/(1-x^2)^(1/2)dx
set (1-x^2)^(1/2)=u 感谢纠正
1-u^2=x^2
2xdx = -2udu
= sin^-1 x* (1/2)x^2 -(1/2)∫(1-u^2)^(-1/2)du
set u=sinθ du=cosθ
= sin^-1 x* (1/2)x^2 -(1/2)∫ cos^2θdu
= sin^-1 x* (1/2)x^2 -(1/4)∫1+cos2θ du
= sin^-1 x* (1/2)x^2 -(1/4)(θ+sin2θ/2)
= sin^-1 x* (1/2)x^2 -(1/4)(sin^-1 u + 2u(1-u^2)^(1/2))
^^^^^^^^^^^^^ sin2θ=2 sinθ cosθ
= sin^-1 x* (1/2)x^2 -(1/4)(sin^-1[(1-x^2)^(1/2)]+x(1-x^2)^(1/2))+c
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用我看得见的指尖
将你一身的华丽褪去
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※ 编辑: ek0519 来自: 220.135.70.14 (02/13 23:49)