Integration by Substitution

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1 Integrtion by Substitution Dr. Philippe B. Lvl Kennesw Stte University August, 8 Abstrct This hndout contins mteril on very importnt integrtion method clled integrtion by substitution. Substitution is to integrls wht the chin rule is to derivtives. Integrtion by Substitution (5.5) The Fundmentl Theorem of Clculus tells us tht in order to evlute n integrl, we need to nd n ntiderivtive of the function we re integrting (the integrnd). However, the list of ntiderivtives we hve is rther short, nd does not cover ll the possible functions we will hve to integrte. For exmple, R xe x dx is not in our list. Neither is R x p + x dx. Wht do we do then? One method, the one we will study in this hndout, involves chnging the integrl so tht it looks like one we cn do, by doing chnge of vrible, lso clled substitution. Substitution for integrls corresponds to the chin rule for derivtives.we look t some simple exmples to illustrte this. Before we see how to do this, we need to review nother concept, the di erentil.. The Di erentil You will recll from di erentil clculus tht the nottion dx ment smll chnge in the vrible x. It hs nme, it is clled the di erentil (of the vrible x). Now, if y f (x) nd f is di erentil function, we my lso be interested in nding the di erentil of y, denoted dy. De nition The di erentil dy is de ned by Exmple Find dy if y x By de nition dy f (x) dx dy x dx xdx

2 Exmple 3 Find dy if y sin x By de nition dy (sin x) dx cos xdx. The Substitution Rule for Inde nite Integrls We re now redy to lern how to integrte by substitution. Substitution pplies to integrls of the form R f (g (x)) g (x) dx. If we let u g (x), then du g (x) dx. Therefore, we hve f (g (x)) g (x) dx f (u) du () This is the substitution rule formul. Note tht the integrl on the left is expressed in terms of the vrible x. The integrl on the right is in terms of u. The key when doing substitution is, of course, to know which substitution to pply. At the beginning, it is hrd. With prctice, it becomes esier. Also, looking t eqution nd trying to understnd the pttern will mke things esier. In tht formul, it is ssumed tht we cn integrte the function f. Looking t the integrl on the left, one sees the function f. But the integrl lso hs extr expressions. Inside of f, there is n expression in term of x. Outside of f, is the derivtive of this expression. When this is the cse, the expression will be the substitution. For exmple, given R x sin x dx, one would use u x s the substitution. Given R cos x p sin xdx, one would use u sin x s the substitution. Let us look t some exmples. Exmple 4 Find R x sin x dx If u x, then du xdx, therefore x sin x dx sin x xdx sin udu The lst integrl is known formul sin udu cos u + C The originl problem ws given in terms of the vrible x, you must give your nswer in terms of x. Therefore, x sin x dx cos x + C

3 Exmple 5 Find R xe x dx If u x, then du xdx, therefore xe x dx e x xdx e u du e u du eu + C ex + C Exmple 6 Find R x 3 cos x 4 + dx If u x 4 +, then du 4x 3 dx, therefore x 3 cos x 4 + dx cos x 4 + x 3 dx cos u du 4 cos udu 4 4 sin u + C 4 sin x4 + + C Exmple 7 Find R tn xdx If we think of tn x s sin x nd let u cos x, then du cos x sin x tn xdx cos x dx u ( u du ) du ln juj + C ln jcos xj + C sin xdx, therefore This is not the formul usully remembered. Since cos x, nd, one of sec x 3

4 the properties of logrithmic functions sys tht ln b ln tn xdx ln ln jcos xj + C jsec xj + C ln ( ln jsec xj) + C ln jsec xj + C ln b, we hve Exmple 8 Find R x p x + dx If u x +, then du xdx, therefore x p x + dx pudu u du 3 u 3 + C 3 x C.3 The Substitution Rule for De nite Integrls With de nite integrls, we hve to nd n ntiderivtive, then plug in the limits of integrtion. We cn do this one of two wys:. Use substitution to nd n ntiderivtive, express the nswer in terms of the originl vrible then use the given limits of integrtion.. Chnge the limits of integrtion when doing the substitution. This wy, you won t hve to express the ntiderivtive in terms of the originl vrible. More precisely, b f (g (x)) g (x) dx g(b) g() We illustrte these two methods with exmples. f (u) du Exmple 9 Find R e ln x dx using the rst method. x First, we nd n ntiderivtive of the integrnd, nd express it in term of x. If 4

5 u ln x, then du dx. Therefore x ln x x dx It follows tht e ln x x u udu (ln x) dx (ln x) (ln e) e (ln ) Exmple Sme problem using the second method. The substitution will be the sme, but we won t hve to express the ntiderivtive in terms of x. Insted, we will nd wht the limits of integrtion re in terms of u. Since u ln x, when x, u ln. When x e, u ln e. Therefore, e ln x x dx u Remrk A specil cse of substitution is renming vrible in n integrl. You will recll tht R b R b f (u) du. In this cse, we just performed the trivil substitution u x, in other words, we simply renmed the vrible. This cn lwys be done, however, it does not ccomplish nything. Sometimes we do it for disply purposes, s we will see in the next theorem. udu.4 Integrting Even nd Odd Functions De nition A function f is even if f ( x) f (x). It is odd if f ( x) f (x) Exmple 3 f (x) x is even. In fct f (x) x n is even if n is even. Exmple 4 f (x) x 3 is odd. In fct f (x) x n is odd if n is odd. 5

6 Figure : Even Function Exmple 5 sin ( x) sin x, therefore sin x is odd. Exmple 6 cos ( x) cos x, therefore cos x is even. Exmple 7 From the previous two exmples, it follows tht tn x nd cot x re odd The grph of n even function is symmetric with respect to the y-xis. The grph of n odd function is symmetric with respect to the origin. Another wy of thinking bout it is the following. If f is even nd (; b) is on the grph of f, then ( ; b) is lso on the grph of f. If f is odd nd (; b) is on the grph of f, then ( ; b) is lso on the grph of f. This is illustrted on gure for even functions, nd on gure for odd functions. Knowing if function is even or odd cn mke integrting it esier. Theorem 8 Suppose tht f is continuous on [ ; ] then:. If f is even, then R R. If f is odd, then R 6

7 Figure : Odd Function Proof. Using the properties of integrls, we hve: + + In the rst integrl, we use the substitution u x so tht du dx, we obtin f ( u) du + For clrity, use the substitution u x to obtin f ( x) dx + We now consider the cses f is even nd odd seprtely. () cse : f is even. In this cse, f ( x) f (x). Therefore, eqution becomes + 7

8 cse : f is odd. In this cse, f ( x) f (x). Therefore, eqution becomes + Remrk 9 The rst prt of the theorem does not sve us lot of work. We still hve to be ble to nd n ntiderivtive in order to evlute the integrl. However, in the second prt, we only need to know the function is odd. If it is, then the integrl will be, there is no need to be ble to nd n ntiderivtive of the integrnd. Exmple Find R Let f (x) therefore tn x x 4 + x +..5 Things to Know tn x x 4 + x + dx The reder cn verify tht f is n odd function, tn x x 4 + x + dx Be ble to integrte using the substitution method. Know wht odd nd even functions re, be ble to recognize them nd know how to integrte them on n intervl of the form [ ; ]. Relted problems ssigned: #, 3, 7, 9,, 3, 5, 7, 9,, 3, 5, 9, 3, 37, 4, 49, 6,63, 64, 65 on pges 39, 393 Try this more chllenging problem: R xe x (x + ) dx 8

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