解数学题，求不定积分、凑微分、换元、分部积分，要正确答案

Question

如题所述

Answer 1

1
=xarcsinx-∫x/[(1-x^2)^1/2]dx=xarcsinx+1/2*∫d(1-x^2)/[(1-x^2)^1/2]=xarcsinx+(1-x^2)^1/2+c
2
∫e^xsin^2xdx=∫(1-cos2x)e^x/2dx=1/2[∫e^xdx-∫e^xcos2xdx]
下面着重求出第二项
∫e^xcos2xdx=∫cos2xd(e^x)=e^xcos2x+2∫e^xsin2xdx=e^xcos2x+2∫sin2xde^x
=e^xcos2x+2e^xsin2x-4∫e^xcos2xdx
移项得到
5∫e^xcos2xdx=e^xcos2x+2e^xsin2x
所以∫e^xcos2xdx=1/5(e^xcos2x+2e^xsin2x)
代入原式得到
∫e^xsin^2xdx=1/2[e^x-1/5(e^xcos2x+2e^xsin2x)]=e^x(1/2-1/10cos2x-1/5sin2x)+c
3
原式=∫{-无穷到+无穷}d(x+1)/[1+(x+1)^2]=arctan(x+1)|{-无穷到+无穷}=π/2-(-π/2）=π
4
原式=∫e^(-5/2)d[e^(x-1/2)]/[1+[e^(x-1/2)]^2]=e^(-5/2)arctan[e^(x-1/2)]
|{负无穷到正无穷}=π/2*(e^(-5/2))
5
原式=∫√sin^(3)x
(1-sin^(2)x)
dx=∫sin^(3/2)x
|cosx|dx
=∫{0到π/2}sin^(3/2)x
cosxdx-∫{π/2到π}sin^(3/2)x
cosxdx
=∫{0到π/2}sin^(3/2)xdsinx-∫{π/2到π}sin^(3/2)xdsinx
=2/5(sin^(5/2)x)|
{0到π/2}-2/5(sin^(5/2)x)|
{π/2到π}
=4/5
6
设t=1+√3x+1
，2<t<5
那么x=1/3
[(t-1)^2-1]
所以dx=2/3
(t-1)
dt
那么
原式=2/3
∫{t从2到5}[(t-1)/t]dt
=2/3
∫{t从2到5}[(1-1/t)]dt
=2/3(t-lnt)
|
{t从2到5}
=2-2/3
ln(5/2)