Section6.1Larsonnotes - 1 Section 6.1 Integration by Parts Practice HW from Larson Textbook(not to hand in p 374 1-23 odd 31-35 odd Integration by Parts

# Section6.1Larsonnotes - 1 Section 6.1 Integration by Parts...

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Section 6.1: Integration by Parts Practice HW from Larson Textbook (not to hand in) p. 374 # 1-23 odd, 31-35 oddIntegration by PartsIntegration by parts undoes the product rule of differentiation.Suppose the have two functions uand v. Differentiating the product of these two functions by the product rule givesdxduvdxdvuuvdxd+=)(Integrating both sides with respect to xgives+=dxdxduvdxdxdvudxuvdxd)(or+=duvdvuuvSolving for dvugives the following integration by parts formula. Integration by Parts Formula - = du v uv dv u 1
Example 1: Integrate - dx xe x 3 Solution: 2
Example 2: Integrate dt t t ln 4 Solution: 3
Repeated Use of Integration by Parts Example 3: Integrate dx x x cos 2 Solution: 4
Example 4: Integrate dx x e x sin 2 Solution: 5
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Example 5: Integrate dx x tan 1 0 1 - 21-21-1+-=+-=-=-To compute dxxx12+, we must use u-dusubstitution. Here, we let dxxdudxxduxdxduxu212212===+=This gives CxCuduuududxxx++=+===+221ln21ln21121211. Going back to the integration parts problem, we haveCxxxdxxxxxdxx++-=+-=---212111ln21tan1tantan.We last evaluate the definite integral. This gives(continued on next page) 7
ln12ln2141)4tan(since4)1(tan:Note)(212ln2141ln212ln21)1(tan1ln212ln21)1(tan)(1ln21)(tan)()1(1ln21)1(tan)1(1ln21tantan1-112121211=-===--=--=---=+--+-=+-=------ππππxxxdxx 1 1 8
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