Integration by parts two.pdf - Integration by Parts Dr...

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Integration by PartsDr. Philippe B. LavalKennesaw State UniversityAugust 21, 2008AbstractThis handout describes an integration method called Integration byParts.Integration by parts is to integrals what the product rule is toderivatives.1Integration by Parts (5.6)1.1General ConceptIffandgare two di/erentiable functions, then the product rule gives us:(f(x)g(x))0=f(x)g0(x) +f0(x)g(x)If we integrate both sides, we getZ(f(x)g(x))0dx=Zf(x)g0(x)dx+Zf0(x)g(x)dxFrom the fundamental theorem of calculus, we know that the integral of thederivative of a function is the function itself, therefore we have:f(x)g(x) =Zf(x)g0(x)dx+Zf0(x)g(x)dxOr, solving for the °rst integralZf(x)g0(x)dx=f(x)g(x)°Zf0(x)g(x)dxThis is the integration by parts formula. However, we usually do not rememberit this way. If we letu=f(x), thendu=f0(x)dx. Similarly, ifv=g(x), thendv=g0(x)dx. Doing the two substitutions gives us:Zudv=uv°Zvdu(1)1
The goal when using this formula is to pretend that the integral we are given isof the formRudv. We °nduanddvthat accomplish this. Once we haveuanddv, we °ndduandv. Then, we can rewrite the given integral asuv°Rvdu.This will work if the new integral we obtained can be evaluated or is easier toevaluate than the one we started with. Also, once we have selecteddv,vis anantiderivative ofdv. therefore, we must be able to °nd an antiderivative fordv.For de°nite integrals, the integration by parts formula becomesZbaf(x)g0(x)dx=f(x)g(x)jba°Zbaf0(x)g(x)dxWe illustrate this technique with several examples. For clarity, the quantitieswe select from the integral (uanddv) will be in bold characters. The quantitieswe deduce from our selection (duandv) will be in normal characters.Example 1FindRxsinxdxIf we selectu=xdu=dxv=°cosxdv= sinxdxThen, applying formula 1 gives us:Zxsinxdx=°xcosx°Z°cosxdx=°xcosx+Zcosxdx=°xcosx+ sinx+CRemark 2There is usually more than one way to selectuanddv. However,

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