So, if we cannot compute the integralh(x)dxdirectly, we often look for a new variableuand functionf(u) for whichh(x)dx=f(u(x))dudxdx=f(u)du,where the second integral is easier to evaluate than the first.NOTESIn deciding how to choose a newvariable, there are several thingsto look for:terms that are derivatives ofother terms (or pieces thereof )andterms that are particularlytroublesome. (You can oftensubstitute your troubles away.)EXAMPLE 6.2Using Substitution to Evaluate an IntegralEvaluate(x3+5)100(3x2)dx.SolutionYou probably cannot evaluate this as it stands. However, observe thatddx(x3+5)=3x2,which is part of the integrand. This leads us to make the substitutionu=x3+5, so thatdu=ddx(x3+5)dx=3x2dx. This gives us(x3+5)100u100(3x2)dxdu=u100du=u101101+c.We are not done quite yet. Since we invented the new variableu, we need to convert backto the original variablex, to obtain(x3+5)100(3x2)dx=u101101+c=(x3+5)101101+c.It’s always a good idea to perform a quick check on the antiderivative. (Remember thatintegration and differentiation are inverse processes!) Here, we computeddx(x3+5)101101=101(x3+5)100(3x2)101=(x3+5)100(3x2),which is the original integrand. This confirms that we have indeed found anantiderivative.
4-53SECTION 4.6..Integration by Substitution395INTEGRATION BY SUBSTITUTIONIntegration by substitution consists of the following general steps, as illustrated inexample 6.2.Choose a new variableu:a common choice is the innermost expression or“inside” term of a composition of functions. (In example 6.2, note thatx3+5 isthe inside term of (x3+5)100.)Computedu=dudxdx.Replace all termsin the original integrand with expressions involvinguanddu.Evaluatethe resulting (u) integral. If you still can’t evaluate the integral, youmay need to try a different choice ofu.Replace each occurrence ofuin the antiderivative with the correspondingexpression inx.Always keep in mind that finding antiderivatives is the reverse process of findingderivatives. In example 6.3, we are not so fortunate as to have the exact derivative we wantin the integrand.EXAMPLE 6.3Using Substitution: A Power Function Inside a CosineEvaluatexcosx2dx.SolutionNotice thatddxx2=2x.Unfortunately, we don’t quite have a factor of 2xin the integrand. This does not presenta problem, though, as you can always push constants back and forth past an integral sign.Notice that we can rewrite the integral asxcosx2dx=122xcosx2dx.We now substituteu=x2, so thatdu=2x dxand we havexcosx2dx=12cosx2cosu(2x)dxdu=12cosu du=12sinu+c=12sinx2+c.Again, as a check, observe thatddx12sinx2=12cosx2(2x)=xcosx2,which is the original integrand.EXAMPLE 6.4Using Substitution: A Trigonometric FunctionInside a PowerEvaluate(3 sinx+4)5cosx dx.