Transcription
Integration By U- SubstitutionAcademic Resource Center
Definition Integrals which are computed by change ofvariables is called U-substitution. In this we have to change the basic variable ofan integrand (like ‘x’) to another variable (like‘u’). This make the integral easy to determine.
Why U-Substitution It is one of the simplest integration technique. It can be used to make integration easier. It is used when an integral contains some function andits derivative, whenI f ( x) f 1 ( x)Let u f(x)du fʹ(x) dx
Why U-SubstitutionHence the integration becomesI uduu2 c2f 2 ( x) c2
When to use U-Substitution We have function and its derivative together. Where by use of simpler methods like POWERRULE , CONSTANT MULTIPLE RULE etc itsdifficult to solve integration.
When not to use U-Substitution If you fail to see such a pair of quantities,abandon this method.
Example 1 Lets Compute the following integralI e x2Put u x 2du 2xdx 2 x dx
Example 1(continued) the indefinite integral becomesI e duuu e c ex c2
Example 2 Consider the Following ExampleI 1dxx ln xPut lnx u1/x dx duHence the given Integral becomes
Example 2(Continued) Hence the given Integral becomes1I duu ln u ln(ln x ) c
Example 3 Consider the following IntegralI cos(2 x) sin(2 x)dxPutsin(2x) u2 cos(2x) dx ducos(2x) dx du/2
Example 3(Continued) Hence the given Integral becomes1I udu22 1 u c 2 2 u2 c4sin 2 2 x c4
Definite Integral Using U-Substitution When evaluating a definite integral using usubstitution, one has to deal with the limits ofintegration . So by substitution, the limits of integrationalso change, giving us new Integral in newVariable as well as new limits in the samevariable. The following example shows this.
Eample4 (Definite Integral) Consider the following Integral1I x ( x 1) dx2340Put x 1 u33 x dx du12x dx du32
Example 4(Continued) Hence the definite integral becomesWhen x 0, u 0When x 1,u 1Hence I 113udu 4051 u1 []035
Example4(continued)15 1I u ]015155 [1 0 ]151 15
Practice Problems1) I 3cos( x ) sin( x ) dx 22) I 2x4 x dx203) I x e dx x ( x 1) dx4) I 5) I289(x 3x 5)( 2 x 3) dx 3x48
Answer To Practice Problems4cox x1) I c4162) I 33) I ( x 2 3 x 5) 9 cx4e4) I c411105) I ( x 1) ( x 1) 9 c109
Important Tips for Practice Problem If you see a function and its derivative putfunction u e.g. in question 1 put sinx u andthen solve . Same is the case with question 2 and 3. For question 2 Put 4-x2 u and then solve. For question 3 Put x2 3x 5 u and then solve. For question 4 Put x4 u and then solve. For question 5 power rule fails because thereis additional x.
Important Tips for Practice Problem So we can reduce the integral in such a way sothat power rule works by using substitution. So in question 5 put x 1 u and then solve theintegral in variable u by using simple powerrule.
difficult to solve integration. When not to use U-Substitution If you fail to see such a pair of quantities, abandon this method. Example 1 Lets Compute the following integral Put u x 2 du 2xdx I ³ ex 2 2 x dx. Example 1(continued) the indefinite integral becomes e u c I ³ eu du ex2 c. Example 2 Consider the Following Example Put lnx u 1/x dx du Hence the given Integral .
