# Math& 152 Section Integration by Parts

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1 Mth& 5 Section 7. - Integrtion by Prts Integrtion by prts is rule tht trnsforms the integrl of the product of two functions into other (idelly simpler) integrls. Recll from Clculus I tht given two differentible function f nd g, the product rule sttes d (f(x)g(x)) = f(x)g (x) + g(x)f (x) Integrting both sides of the eqution, we hve f(x)g(x) + constnt = (f(x)g (x) + g(x)f (x)) We cn drop constnt on the left side since other constnts of integrtion will be showing up lter nd they would just end up combining with this one. Thus, by properties of integrls we hve, f(x)g(x) = (f(x)g (x) + g(x)f (x)) = f(x)g (x) + g(x)f (x) Rerrnging the terms gives f(x)g (x) = f(x)g(x) g(x)f (x) Let u = f(x), du = f (x), v = g(x), nd dv = g (x). The bove integrl is commonly rewritten s: u dv = uv v du Remrk. Integrtion by Prts my be used to evlute definite integrls s follows: b u dv = uv b Exmple. Evlute the following integrls. () x cos x (b) 3 ln x (c) x cos x (d) e x cos x b v du

2 Exmple. Evlute the indefinite integrl x x + () Using Integrtion by Prts (b) Using Substitution Exmple 3. Prove the following reduction formul x n e x = x n e x n x n e x Use the bove formul to evlute x 3 e x.

3 Mth& 5 Section 7. - Trigonometric Integrls In this section we develop techniques for evluting integrls involving combintions of trigonometric functions. Cse : Evluting integrls of the form sin m x cos n x.. If m is odd nd positive (m = k + ), keep fctor of sin x, nd use the identity sin x = cos x nd write: sin m x cos n x = (sin x) k cos n x sin x = ( cos x) k cos n x sin x Integrte by substitution using u = cos x.. If n is odd nd positive (n = k + ), keep fctor of cos x, nd use the identity cos x = sin x nd write: sin m x cos n x = sin m x(cos x) k cos x = sin m x( sin x) k cos x Integrte by substitution using u = sin x. 3. If m nd n re both even, use the hlf-ngle formuls (repetedly, if necessry) nd write: sin x = ( cos x) nd cos x = ( + cos x) Cse : Evluting integrls of the form tn m x sec n x.. If m is odd nd positive (m = k + ), keep fctor of sec x tn x, nd use the identity tn x = sec x nd write: tn m x sec n x = (tn x) k sec n x sec x tn x = (sec x ) k sec n x sec x tn x Integrte by substitution using u = sec x. 3

4 . If n is even nd positive (n = k, k ), keep fctor of sec x, nd use the identity sec x = + tn x nd write: tn m x sec n x = tn m x(sec x) k sec x = tn m x( + tn x) k sec x Integrte by substitution using u = tn x. Cse 3: Evluting integrls of the form cot m x csc n x. This cse is similr to Cse. Cse 4: Evluting integrls of the form sin mx cos nx, sin mx sin nx, nd cos mx cos nx. Rewrite using the following identities: sin A cos B = [sin(a B) + sin(a + B)] sin A sin B = [cos(a B) cos(a + B)] cos A cos B = [cos(a B) + cos(a + B)] Exmple 4. Evlute the following integrls. () sin 3 x (b) sin 4 x cos 3 x (c) tn 3 x sec 7 x (d) π 4 tn3 x (e) π sin(4x) cos(3x) (f) tn 3 x (Rewrite tn 3 x = tn x tn x) (g) sec 3 x (Use IBP first with u = sec x) 4

5 Mth& 5 Section Trigonometric Substitution In this section we develop technique for evluting integrls of the form x, + x, nd x where is positive constnt. The ide is to convert integrls of these forms into n integrl involving trigonometric functions. Expression Substitution Domin Identity x x = sin θ π θ π sin θ = cos θ + x x = tn θ π < θ < π + tn θ = sec θ x x = sec θ θ < π or π θ < 3π sec θ = tn θ Remrk. In ech cse the restriction on θ is imposed to ensure tht the function tht defines the substitution is one-to-one. Exmple 5. Evlute the following integrls. () x (b) x (4 x ) 3/ (c) (d) x x 9 (x 6x+) (e) + x (f) +x x 5

6 Mth& 5 Section Integrtion by Prtil Frctions In this section we develop technique for integrting complicted rtionl functions by breking them up into sums nd differences of simpler rtionl functions tht cn be esily integrted. This process is clled prtil frction decomposition. Definition. A rtionl function f(x) is of the form where p nd q re polynomils. f(x) = p(x) q(x), q(x) Definition. Suppose f is the rtionl function, f(x) = p(x)/q(x), q(x). The function f is proper if deg(p) < deg(q). The function f is improper if deg(p) deg(q). If f is improper, we first simplify the expression using long division. Exmple 6. Evlute the integrl x x+4. If f is proper, we hve the following cses. Suppose f(x) = p(x)/q(x), where q(x) nd p nd q re polynomils. Cse : q(x) is the product of distinct liner fctors. We cn write f s: f(x) = where A, A,..., A k re constnts. A + A + + A k, x + b x + b k x + b k Cse : q(x) is the product of liner fctors, some of them repeted. Suppose q(x) hs the fctor ( x + b ) r, then the sum A A + x + b ( x + b ) + + A r ( x + b ) r occurs in the prtil frction decomposition of f(x). Exmple 7. We cn write the following prtil frction decomposition x 3 + x (x ) = A x + B x + 6 C x + D (x )

7 Cse 3: q(x) contins irreducible qudrtic fctors, no repets. We see tht f will hve term of the form for ech of those fctors. Ax + B x + bx + c Exmple 8. Set up but do not solve the prtil frction decomposition of 3x x(x 4)(x + )(x + 6x + ) Cse 4: q(x) contins repeted irreducible qudrtic fctor. Suppose q(x) hs the fctor (x + bx + c) r, then the sum A x + B x + bx + c + A x + B (x + bx + c) + + A r x + B r (x + bx + c) r occurs in the prtil frction decomposition of f(x). Exmple 9. Evlute the following integrls. () x 3 + x 4 (b) 3x 9 (x )(x+) (c) 8 (x+3)(x +9) (d) 8 (x+3)(x 9) (e) x+4 x (Let u = x + 4, see exmple 9 in the text) 7

8 Mth& 5 Section Approximte Integrtion The gol of this section is to pproximte the vlue of definite integrl when it cnnot be esily found using the Fundmentl Theorem of Clculus. We consider two cses:. Some functions do not hve n elementry ntiderivtive. For instnce, it s impossible to evlute the definite integrl ex exctly. Therefore, n pproximtion is necessry.. There my be cses when the function is determined from dt through scientific experiments. In this cse, n explicit form of the function my not be vilble. (See Exmple 5 in your text) We hve lredy discussed severl pproximtion methods bck in Section 5.. If we divide the closed intervl [, b] into n subintervls of equl length x = b, the left endpoint n nd right endpoint pproximtions, L n nd R n respectively, re given by: L n = R n = n f(x i ) x i= n f(x i ) x i= Another method for pproximting the definite integrl b f(x) is to choose our smple points to be the midpoint, denoted x i of the subintervl [x i, x i ]. Midpoint Rule: where x i = (x i + x i ). M n = x[f( x ) + f( x ) + + f( x n )], 8

9 Another type of pproximtion, clled the Trpezoidl Rule, T n, is obtined by verging L n nd R n : T n = (L n + R n ) [ n ] = n f(x i ) x + f(x i ) x i= i= [ n ] = x (f(x i ) + f(x i )) Trpezoidl Rule: i= = x [(f(x ) + f(x )) + (f(x ) + f(x )) + + (f(x n ) + f(x n ))] = x [f(x ) + f(x ) + f(x ) + + f(x n )] T n = x [f(x ) + f(x ) + f(x ) + + f(x n )] The error of ech pproximtion is given by: E T = b f(x) T n nd E M = b f(x) M n Exmple. Approximte the definite integrl ex using the midpoint nd trpezoidl rules (with n = 4). Find the error of ech estimte by compring to the ctul vlue. 9

10 Through numericl clcultions, we cn see tht the size of E M is bout hlf the size of E T (see Figure 5, pge 498). When doing pproximtions in pplied mthemtics, it s importnt to get n ide of how big our error estimtes might be. The gol, of course, it to mke our error s smll s possible. The following inequlities (proved in lter mth course clled Numericl Anlysis) depend on the size of the second derivtive of our function. According to Stewrt, this fct is not surprising becuse f (x) mesures how much the grph is curved. Error Bounds: Suppose f (x) K for x [, b]. We hve, E T K(b )3 n nd E M K(b )3 4n Exmple. Find the upper bound of E T for the pproximtion of ex (where n = 4). Exmple. How lrge should we tke the vlue of n to gurntee tht the midpoint pproximtion of ex is ccurte to within.? Our finl pproximtion method uses prbols to estimte res (see pp. 5 5 for more detiled explntion). Simpson s Rule: S n = x 3 [f(x ) + 4f(x ) + f(x ) + 4f(x 3 ) + + f(x n ) + 4f(x n ) + f(x n )] where n is even. Exmple 3. Use Simpson s Rule to pproximte ex (with n = 4). Find the error E S nd compre it to E T nd E M. Note: It cn be shown tht S n = 3 T n + 3 M n (see Exercise 48). We lso hve much better error bound, the proof of course will be omitted. Error Bound for Simpson s Rule: Suppose f (4) (x) K for x [, b]. We hve, E S K(b )5 8n 4

11 Mth& 5 Section Improper Integrls In this section we extend the concept of definite integrl to the following two types.. Infinite intervls (regions extended indefinitely in the horizontl direction).. Discontinuous integrnds (regions extended indefinitely in the verticl direction). The exmples below re clled improper integrls. An obvious grphicl interprettion exists in both cses. The question we wnt to nswer is whether the re described by the improper integrl is finite or infinite. If it is finite, we cll the integrl convergent. Otherwise, the integrl is divergent. Exmple 4. Evlute the following integrls. () e x (b) x Bsed on these exmples, we mke the following definitions. Type : Infinite intervls () If t f(x) exists for every number t, then f(x) = lim t t f(x) (b) If b f(x) exists for every number t b, then t b f(x) = lim t b t f(x) If the limits re finite, then the integrls re convergent. Otherwise, they re divergent. (c) If both f(x) nd f(x) re convergent, then we define f(x) = f(x) + f(x)

12 Type : Discontinuous integrnds () If f is continuous on [, b) (discontinuous t x = b), then b f(x) = lim t b t f(x) (b) If f is continuous on (, b] (discontinuous t x = ), then b f(x) = lim t + b t f(x) If the limits re finite, then the integrls re convergent. Otherwise, they re divergent. (c) If f is discontinuous t x = c, where < c < b, c f(x) <, nd b f(x) <, c then b f(x) = c f(x) + b c f(x) Exmple 5. Determine whether the following integrls re convergent or divergent. Find the exct vlue (if pplicble). () x (b) (c) (d) (e) (f) x x x ln x x +x

13 Here re two specil cses. Theorem. { x =, if p > (convergent) p p, if p (divergent) x p = {, if p < (convergent) p, if p (divergent) Proof. The proof of the first cse is Exmple 4 in your text. I will leve the second cse s n exercise. Sometimes it s impossible to find the exct vlue of n improper integrl. The following theorem llows us to determine convergence nd divergence. Agin, without going into the detils of the proof, it s possible to give grphicl explntion. Theorem (Comprison Theorem). Suppose f nd g re continuous functions with f(x) g(x), for x. () If (b) If f(x) is convergent, then g(x) is convergent. g(x) is divergent, then f(x) is divergent. Exmple 6. Determine whether the following integrls re convergent or divergent. () x 3 + (b) e x (c) x + 3

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