Question

求这道积分题怎么做

Answer 1

详细的解答过程如下图片：前面有一截是奇函数，结果 为0

Answer 2

前面一坨等于零，后面一坨直接算

追答

原式=∫√(25-x^2)dx
设x=5sint,则dx=5cosdt…………………………变量代换
原式=∫√[25-25(sint)^2]·5costdt
=∫5cost·5costdt
=25∫(cost)^2dt
=25∫{[1+cos(2t)]/2}dt……………………………二倍角公式cos2t=2(cost)^2-1
=(25/2)∫[1+cos(2t)]dt
=(25/2)[t+(1/2)sin(2t)]+C
=(25/4)sin(2t)+(25/2)t+C
把t=arcsin(x/5)代入上式,得
原式=(25/4)sin[2arcsin(x/5)]+(25/2)arcsin(x/5)+C
=(x/2)√(25-x^2)+(25/2)arcsin(x/5)+C
其中,
sin[2arcsin(x/5)]
=2sin(arcsin(x/5))cos(arcsin(x/5))
=2×(x/5)×(√(25-x^2)/5)
=(2/25)x√(25-x^2)

∫√（a^2-x^2）dx
设x=asint
则dx=dasint=acostdt
a^2-x^2
=a^2-a^2sint^2
=a^2cost^2
∫√（a^2-x^2）dx
=∫acost*acostdt
=a^2∫cost^2dt
=a^2∫（cos2t+1）/2dt
=a^2/4∫(cos2t+1)d2t
=a^2/4*(sin2t+2t)
将x=asint代回
∫√（a^2-x^2）dx=x√(a^2-x^2)/2+a^2*arcsin(x/a)/2+C

后面一坨好像也是零

Answer 3

f(x) =x^3.(sinx)^2/(x^4+2x^2+1)
f(-x) = -f(x)
let
x=5sinu
dx=5cosu du
∫(-5->5) [x^3.(sinx)^2/(x^4+2x^2+1) +√(25-x^2) ] dx
=∫(-5->5) √(25-x^2) dx
=2∫(0->5) √(25-x^2) dx
=50∫(0->π/2) (cosu)^2 du
=25∫(0->π/2) (1+cos2u) du
=25[ u+(1/2)sin2u]|(0->π/2)
=(25/2)π