Question

微积分题目定积分

微积分题目定积分利用华里士公式

Answer 1

(1)
∫(-π/2->π/2) (cosx)^5 dx
=2∫(0->π/2) (cosx)^5 dx
=2∫(0->π/2) (cosx)^4 dsinx
=2∫(0->π/2) [1-(sinx)^2]^2 dsinx
=2∫(0->π/2) [1-2(sinx)^2 + (sinx)^4] dsinx
=2[ sinx - (2/3)(sinx)^3 + (1/5)(sinx)^5] |(0->π/2)
=2(1 -2/3 +1/5)
=16/15
(2)
5+4x-x^2 = 9-(x-2)^2
let
x-2 = 3sinu
dx = 3cosu du
x=-1, u=-π/2
x=5, u=π/2
∫(-1->5) x(5+4x-x^2)^(3/2) dx
=-(1/2)∫(-1->5) (4-2x)(5+4x-x^2)^(3/2) dx + 2∫(-1->5) (5+4x-x^2)^(3/2) dx
=-(1/5)[(5+4x-x^2)^(5/2)]|(-1->5) +2∫(-1->5) (5+4x-x^2)^(3/2) dx
=(2^(5/2)/5) +2∫(-π/2->π/2) (3cosu)^3 (3cosu du)
=(2^(5/2)/5) +324∫(0->π/2) (cosu)^4 du
=(2^(5/2)/5) +81∫(0->π/2) (1+cos2u)^2 du
=(2^(5/2)/5) +81∫(0->π/2) [1+2cos2u+ (cos2u)^2] du
=(2^(5/2)/5) +(81/2)∫(0->π/2) [3+4cos2u+ cos4u ] du
=(2^(5/2)/5) +(81/2) [3u+2sin2u+ (1/4)sin4u ]|(0->π/2)
=(2^(5/2)/5) +(243/4)π

Answer 2

∫1/(1+根号(1+x))dx
设 根号(1+x)=t 所以 t^2-1=x dx/dt=2t代入
= ∫2t/(1+t)dx =∫2t+2-2/(1+t)dx=∫2dx-∫2/(1+t)dx
=2t-2ln|1+t|+c 代入 根号(1+x)=t
=2根号(1+x)-2ln|1+根号(1+x)|+C