关于不定积分和定积分

Question

关于不定积分和定积分两题都不怎么会，求学霸帮帮忙

Answer 1

1、∫(0,π/2) cos^6xdx
=∫(0,π/2) cos^5xd(sinx)
=cos^5xsinx|(0,π/2)+∫(0,π/2) 5sin^2xcos^4xdx
=∫(0,π/2) 5(1-cos^2x)cos^4xdx
=5∫(0,π/2) cos^4xdx-5∫(0,π/2) cos^6xdx
∫(0,π/2) cos^6xdx=(5/6)*∫(0,π/2) cos^4xdx
=(5/6)*(3/4)*∫(0,π/2) cos^2xdx
=(5/6)*(3/4)*(1/2)*∫(0,π/2) dx
=(5/6)*(3/4)*(1/2)*(π/2)
=5π/32
2、∫(0,1) f(x)/√xdx
=∫(0,1) (1/√x)dx∫(1,√x)e^(-t^2)dt
=-∫(0,1) (1/√x)dx∫(√x,1)e^(-t^2)dt
=-∫(0,1) e^(-t^2)dt∫(0,t^2)1/√xdx
=-∫(0,1) e^(-t^2)dt*2√x|(0,t^2)
=-∫(0,1) 2te^(-t^2)dt
=∫(0,1) e^(-t^2)d(-t^2)
=e^(-t^2)|(0,1)
=1/e-1

Answer 2

(1)
I(2n) = ∫(0->π/2) (cosx)^(2n) dx
I6
=∫(0->π/2) (cosx)^6 dx
=∫(0->π/2) (cosx)^5 dsinx
=[sinx.(cosx)^5]|(0->π/2) +5∫(0->π/2) (sinx)^2.(cosx)^4 dx
=5∫(0->π/2) (sinx)^2.(cosx)^4 dx
=5∫(0->π/2) [ 1-(cosx)^2].(cosx)^4 dx
6I6 = 5I4
I6 = (5/6)I4
=(5/6)(3/4)(1/2)I0
=(5/16) ∫(0->π/2) dx
=(5/32)π
ie
∫(0->π/2) (cosx)^6 dx=(5/32)π
(2)
f(x) =∫(1->√x) e^(-t^2) dt
=>f(1) = 0
两边取导
f'(x) = [1/(2√x)] e^(-x)
--------
∫(0->1) f(x)/√x dx
=2∫(0->1) f(x)d√x
=2[√x.f(x) ] |(0->1) -2∫(0->1) √x.f'(x)dx
=0 -2∫(0->1) √x.f'(x)dx
=-2∫(0->1) √x.{ [1/(2√x)] e^(-x) }dx
=-∫(0->1) e^(-x) dx
=[e^(-x)]|(0->1)
=e^(-1) -1

Answer 3

第一题点火公式