Question

求∫xsin²xdx

Answer 1

显然sin²x= -0.5cos2x+0.5
所以得到
原积分
=∫ x *(-0.5cos2x+0.5) dx
=0.25x² - 0.25∫ x d(sin2x)
=0.25x² - 0.25x *sin2x + ∫0.25sin2x dx
=0.25x² - 0.25x *sin2x -0.125cos2x +C，C为常数

追问

那形如∫√[a²-（x-b）²]dx怎么求，a，b都是常数 复制去Google翻译翻译结果

追答

令x-b=a*sinθ
那么
∫√[a²-(x-b)²]dx

=∫ a cosθ d(a*sinθ)
=∫ a² cos²θ dθ
=∫ a² (0.5cos2θ+0.5) dθ
=∫ a² (0.25cos2θ+0.25) d2θ
=0.25a² *sin2θ +0.5a²θ +C
=0.5a² *sinθ*cosθ +0.5a² *arcsin[(x-b)/a] +C
=0.5(x-b) *√[a²-(x-b)²] +0.5a² *arcsin[(x-b)/a] +C，C为常数