Question

高数类试题，求解析。

高数类试题，求解析。只写打对勾的。

Answer 1

。

Answer 2

(1)∫(x/cos²x)dx
=∫xsec²xdx=∫xd(tanx)
=x·tanx-∫tanxdx
=x·tanx-∫(sinx/cosx)dx
=x·tanx+∫(1/cosx)d(cosx)
=x·tanx+ln|cosx|+C
(2)∫xe^(-x)dx=-∫xd(e^-x)
=-[x·(e^-x)-∫(e^-x)dx]
=-[x·(e^-x)+(e^-x)]+C
=-(x+1)·(e^-x)+C
(3)∫xsin²xdx=(1/2)∫x·(1-cos2x)dx
=(1/2)[∫xdx-∫xcos2xdx]
=(1/4)x²-(1/4)∫xd(sin2x)
=(1/4)x²-(1/4)[xsin2x-∫sin2xdx]
=(1/4)x²-(1/4)xsin2x+(1/8)cos2x+C
(4)∫x³e^(-x²)dx=(-1/2)∫x²d(e^-x²)
=(-1/2)[x²·e^(-x²)-∫(e^-x²)d(x²)]
=(-1/2)[x²·(e^-x²)+(e^-x²)]+C
=(-1/2)(x²+1)(e^-x²)+C
(6)∫ln³x/x²dx=-∫ln³xd(1/x)
=-[(ln³x)/x-∫(1/x)·3ln²x·(1/x)dx]
=-(ln³x)/x+3∫(ln²x)/x²dx
=-(ln³x)/x-3∫ln²xd(1/x)
=-(ln³x)/x-3[(ln²x)/x-∫(1/x)·2lnx·(1/x)dx]
=-(ln³x)/x-3(ln²x)/x+6∫(lnx)/x²dx
=-(ln³x)/x-3(ln²x)/x-6∫lnxd(1/x)
=-(ln³x)/x-3(ln²x)/x-6[(lnx)/x-∫(1/x²)dx]
=-(ln³x)/x-3(ln²x)/x-6(lnx)/x-(6/x)+C
=-(ln³x+3ln²x+6lnx+6)/x+C
(7)∫lnxdx=x·lnx-∫xd(lnx)
=x·lnx-∫x·(1/x)dx
=x·lnx-x+C
(8)∫xtan²xdx=∫x(sec²x-1)dx
=∫x²sec²xdx-∫xdx
=∫xd(tanx)-(1/2)x²
=xtanx-∫tanxdx-(1/2)x²
=xtanx-∫(sinx/cosx)dx-(1/2)x²
=xtanx+ln|cosx|-(1/2)x²
(10)∫(e^x)cosxdx=∫cosxd(e^x)
=e^x·cosx-∫e^xd(cosx)
=e^x·cosx+∫e^xsinxdx
=e^x·cosx+∫sinxd(e^x)
=e^x·cosx+[e^x·sinx-∫e^xd(sinx)]
=e^x·(sinx+cosx)-∫e^xcosxdx
所以，2∫e^xcosxdx=e^x·(sinx+cosx)+C
则，∫e^xcosxdx=(1/2)e^x·(sinx+cosx)+C
(15)∫x²sin²xdx=(1/2)∫x²(1-cos2x)dx
=(1/2)[∫x²dx-∫x²cos2xdx]
=(1/6)x³-(1/4)∫x²d(sin2x)
=(1/6)x³-(1/4)[x²·sin2x-∫sin2x·2xdx]
=(1/6)x³-(1/4)x²·sin2x-(1/4)∫xd(cos2x)
=(1/6)x³-(1/4)x²·sin2x-(1/4)[xcos2x-∫cos2xdx]
=(1/6)x³-(1/4)x²·sin2x-(1/4)xcos2x+(1/8)sin2x+C

Answer 3

10题算对勾吗？

追问

10题不算