295
Exercise Set 5.6 23.
2
x ex dx ∫ Let
25.
u=x2 and dv = exdx. Thendu =2 xdx and v= . ex u dv u v du ∫v x2exdx = x2ex− ex· 2xdx ∫ Integration by Parts 2xexdx = ∫x2ex− ∫ Weevaluate 2 xexdx usingtheIntegrationbyPartsformula. ∫
2xex dx
x sin2xdx ∫ 2 Let u = x2 and dv = sin 2x dx. Then du = 2x dx and v = 1 − 2 cos2x. ∫ x2 sin2xdx Weevaluate Let
∫
=
x 2 · − cos2x + 1 ∫ cos2x·2xdx 2
2xcos2xdxusingtheIntegrationbyParts
u=xanddv=2cos2xdx. Then
Let u =2 and x dv = exdx.
du = dx and v = sin 2x.
Then du=2 dx and v= . ex u dv u v ∫vdu 2xexdx = 2 x·ex− 2exdx ∫ = 2 xex − 2e x + K (2 Thus, x2exdx = x2ex− xex− ex2K + ) ∫ = x2ex −xex 2 + 2 ex + C ( C = − K ) Sincewehaveanintegral ()() wheref(x),or fxgxdx ∫ x2 , can be differentiated repeatedly to a derivative that is eventually0and (ex ),or ,canbeintegratedrepeatedly gx easily, we can use tabular integration. f(x)and
∫ 2xcos2xdx =
x2+
repeatedintegrals
T ∫hus,x2 1 1 1 sin2xdx=−x2cos xsin2x− cos2x− K ∫ 2 x −2 2 4 f(x)g(x)dxwheref(x),or Sincewehaveanintegral eventually 0 and g(x), or ex, can be integrated repeatedly easily, we can use tabular integration. f(x)and
x2
x
2x − e ex 2 + ex 0
We add the products along the arrows, making the alter∫natesignchanges. ∫x2exdx=x2ex−2xex+2ex+C 24. (lnx)2dx Let u = (ln x)2 and dv = dx. Then
2lnx dxandv=x. d ∫u= x (lnx)2dx=x(lnx)2−
∫ ∫2lnxdx lnxdx =x(lnx)2−2(xlnx−x+C) =x(lnx)2−2 See Example 1. = x(ln x)2 −2xlnx+2x+K where K = − 2 C
g(x)and
repeatedderivatives
ex
∫ sin2xdx
1 = xsin2x+ cos2x+K 2
g(x)and
repeatedderivatives
xsin2x−
2x 2 0
repeatedintegrals
+ sin2x − 1 − 1 − 4sin2x + 1 8 cos 2x
We add the products along the arrows, making the alter∫natesignchanges. 2 ∫x2sin2xdx=−1x2cos x−5lnxdx 26.
1 1 2+2xsin2x+4cos2x+C x
Let u = ln x and dv = x−5 dx. Then du= ∫
1 x
x−4 − 4−4 . ln −4 1 x = x− ∫ − x dx − 4 4 · x =− x−4 lnx+ 1∫ x−5dx +C = x−4 x−4 4 lnx− 4 16 − 4
dxandv=
x−5lnxdx
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