Question

大一不定积分

一定要选取适当代换计算下列不定积分

Answer 1

(1)

let
u=√(x-1)
2udu = dx
∫ dx/[x.√(x-1)]
=∫ (2udu)/[ u(u^2+1) ]
=2∫ du/(u^2+1)
=2arctanu + C
=2arctan√(x-1) + C
(2)
let
u=√x
2udu = dx
∫ dx/(√x +1)
=∫ (2udu)/(u +1)
=2∫ [1 -1/(u +1)] du
=2( u -ln|1+u| ) +C
=2( √x -ln|1+√x| ) +C
(3)
let

e^(x/2) = tanu
(1/2)e^(x/2) dx= (secu)^2 du
dx= [2(secu)^2/tanu] du
∫ dx/√(1+e^x)
=∫ [2(secu)^2/tanu] du/ (secu)
=2∫ [secu/tanu] du
=2∫ cscu du
=2ln|√(1+e^x)/ e^(x/2) - 1/e^(x/2) | +C
=2ln|√(1+e^x)-1| - x + C

Answer 2

∫(cos3xcos2x)dx
=(1/2)∫(cos3xcos2x+sin3xsin2x)+(cos3xcos2x-sin3xsin2x)dx
=(1/2)∫(cosx+cos5x)dx
=(sinx)/2+(sin5x)/10+C
类似∫(cosaxcosbx)dx、∫(sinaxcosbx)dx、∫(sinaxsinbx)dx 都可以这样做