Question

高等数学，数学。分部积分做，怎么搞？？？

Answer 1

1、∫(π/4,3π/4) x/sin^2xdx
=∫(π/4,3π/4) xcsc^2xdx
=-∫(π/4,3π/4) xd(cotx)
=-xcotx|(π/4,3π/4)+∫(π/4,3π/4) cotxdx
=π+ln|sinx||(π/4,3π/4)
=π
2、∫(0,π/2) e^x*cos2xdx
=∫(0,π/2) cos2xd(e^x)
=cos2x*e^x|(0,π/2) +2*∫(0,π/2) e^x*sin2xdx
=-e^(π/2)-1+2*∫(0,π/2) sin2xd(e^x)
=-e^(π/2)-1+2sin2x*e^x|(0,π/2)-4*∫(0,π/2) e^x*cos2xdx
=-e^(π/2)-1-4*∫(0,π/2) e^x*cos2xdx
所以∫(0,π/2) e^x*cos2xdx=[-e^(π/2)-1]/5
3、原式=(1/2)*∫x(1-cosx)dx
=(1/2)*∫xdx-(1/2)*∫xcosxdx
=(x^2)/4-(1/2)*∫xd(sinx)
=(x^2)/4-(1/2)*(xsinx-∫sinxdx)
=(x^2)/4-(xsinx)/2-(cosx)/2+C，其中C是任意常数
4、原式=-∫(x^2+1)d[e^(-x)]
=-(x^2+1)e^(-x)+∫2xe^(-x)dx
=-(x^2+1)e^(-x)-∫2xd[e^(-x)]
=-(x^2+1)e^(-x)-[2xe^(-x)-∫2e^(-x)dx]
=-(x^2+1)e^(-x)-2xe^(-x)-2e^(-x)+C
=(-x^2-2x-3)e^(-x)+C，其中C是任意常数
5、原式=(1/2)*∫sinxd[e^(2x)]
=(1/2)*[sinx*e^(2x)-∫cosx*e^(2x)dx]
=(sinx)/2*e^(2x)-(1/4)*∫cosxd[e^(2x)]
=(sinx)/2*e^(2x)-(1/4)*[cosx*e^(2x)+∫sinx*e^(2x)dx]
=(sinx)/2*e^(2x)-(cosx)/4*e^(2x)-(1/4)*∫sinx*e^(2x)dx
所以∫sinx*e^(2x)dx=(1/5)*[(2sinx)-(cosx)]*e^(2x)+C，其中C是任意常数

Answer 2

(1)
∫x/(sinx)^2 dx
=∫x(cscx)^2 dx
=-∫xdcotx
=-xcotx+ ∫cotx dx
=-xcotx + ln|sinx| + C

∫(π/4->3π/4) x/(sinx)^2 dx
=[-xcotx + ln|sinx|]|(π/4->3π/4)
=[ 3π/4 -(1/2)ln2 ] -[-π/4 - (1/2)ln2 ]
=π

(2)
∫e^x.cos2x dx
=∫cos2x de^x
=e^x.cos2x +2∫e^x.sin2x dx
=e^x.cos2x +2∫sin2x de^x
=e^x.cos2x +2sin2x.e^x -4∫e^x.cos2x dx
5∫e^x.cos2x dx = e^x.cos2x +2sin2x.e^x
∫e^x.cos2x dx = (1/5) [e^x.cos2x +2sin2x.e^x ] + C

∫(0->π/2) e^x.cos2x dx
=(1/5) [e^x.cos2x +2sin2x.e^x ]| (0->π/2)
=(1/5) [ -e^(π/2) -1 ]
= -(1/5) (e^(π/2) +1 )

(3)
∫x[sin(x/2)]^2 dx
=(1/2)∫x(1-cosx) dx
=(1/4)x^2 -(1/2)∫xcosx dx
=(1/4)x^2 -(1/2)∫xdsinx
=(1/4)x^2 -(1/2)x.sinx +(1/2)∫sinx dx
=(1/4)x^2 -(1/2)x.sinx -(1/2)cosx + C

(4)
∫(x^2+1).e^(-x) dx
=-e^(-x) +∫x^2.e^(-x) dx
=-e^(-x) -∫x^2.de^(-x)
=-e^(-x) -x^2.e^(-x) +2∫x.e^(-x)dx
=-e^(-x) -x^2.e^(-x) -2∫xde^(-x)
=-e^(-x) -x^2.e^(-x) -2x.e^(-x) +∫e^(-x)dx
=-e^(-x) -x^2.e^(-x) -2x.e^(-x) -e^(-x)dx + C
=-x^2.e^(-x) -2x.e^(-x) -2e^(-x)dx + C

(5)
∫e^(2x) .sinx dx
=-∫e^(2x) .dcosx
=-e^(2x) .cosx +2∫e^(2x) .cosx dx
=-e^(2x) .cosx +2∫e^(2x) dsinx
=-e^(2x) .cosx +2e^(2x).sinx -4∫e^(2x). sinx dx
5∫e^(2x) .sinx dx =-e^(2x) .cosx +2e^(2x).sinx
∫e^(2x) .sinx dx =(1/5) [-e^(2x) .cosx +2e^(2x).sinx] +C