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Chpter 6A Notes Pge of Introuction n Review Derivtives y = f(x) y x = f (x) Evlute erivtive t x = : y = x x= f f(+h) f() () = lim h h Geometric Interprettion: see figure slope of the line tngent to f t x =. Differentition formuls x xn = nx n x ln x = x x ex = e x sin(x) = cos(x) x cos(x) = sin(x) x x tn(x) = sec (x) sec(x) = sec(x) tn(x) x x sin (x) = x

Chpter 6A Notes Pge of x tn (x) = + x Review the sum, ifference, prouct n quotient rules! The chin rule Let F(x) = f(g(x)), then F (x) = f (g(x))g (x) Or let y = f(u), u = g(x), then y = y x u Exmple: u x x sin(x ) = cos(x )x Inefinite Integrls Inefinite integrls re lso known s ntierivtives. If F is the erivtive of F then we write Exmple: F (x)x = F(x) + C cos(x ) x x = sin(x ) + C Inefinite integrl formuls x n x = n + xn+ + C x x = ln x + C e x x = e x + C cos(x) x = sin(x) + C sin(x) x = cos(x) + C sec (x)x = tn(x) + C sec(x) tn(x) x = sec(x) + C

Chpter 6A Notes Pge 3 of Definite Integrls e Exmple: x x Geometric Interprettion: see figure Evlution theorem b F (x)x = F(x) b = F(b) F() Exmples: e x x = ln(x) e = ln(e) ln() = = cos(x ) x x = sin(x ) = sin() sin () Integrtion by substitution Suppose n integrl hs the form: I = f (g(x))g (x)x, let u = g(x), then u = g (x)x (this is ifferentil nottion) Substitute to obtin I = f (u)u = f(u) + C = f(g(x)) + C This is equivlent to tking the chin rule of ifferentition bckwrs! Exmple. I = cos(x ) x x Let u = x, u = x x I = cos(u) u = cos(u) u = sin(u) + C = sin(x ) + C. Exmple. J = cos(x ) x x 3

Chpter 6A Notes Pge 4 of Let u = x. If x = then u = 4. If x = 3 then u = 9. 9 J = cos(u) u 4 = sin (u) 4 9 6. Integrtion by prts Chin rule for ifferentition: x sin(x ) = (cos(x ))x Integrtion by substitution: let = x, u = x x (cos x )x x = cos(u) u = sin(u) + C = sin(x ) + C Prouct rule for ifferentition x [f(x)g(x)] = f (x)g(x) + f(x)g (x) Solve forf(x)g (x): f(x)g (x) = x [f(x)g(x)] f (x)g(x) Tke ntierivtives ( f(x)g (x)x = f(x)g(x) f (x)g(x)x (A) This is the rule for integrtion of inefinite integrls by prts. Shorthn formul: recll ifferentil nottion Box: u = f(x) v = g (x) x u = f (x)x v = g(x) Substitute into (A) bove to get u v = uv v u (B) An esy-to-remember form of the integrtion by prts formul!

Chpter 6A Notes Pge 5 of Exmple. Evlute I = x cos(x) x. Box: u = x v = cos(x) x u = x v = sin (x) I = u v = uv v u = x sin(x) sin (x)x = x sin(x) + cos(x) + C Check: x (x sin(x) + cos(x) + C) = sin(x) + x cos(x) sin(x) = x cos (x) In principle, pick u so tht u is simpler n pick v so tht it is possible to integrte. The following LIPET scheme for choosing u n v is often useful: (L)ogrithm (I)nverse Trig (P)olynomils or powers of x (E)xponentil functions (T)rig functions Whichever piece of the integrn is higher on the list is u n the other piece is v. Exmple. Integrtion by prts twice Evlute I = x sin(x) x. Box: u = x v = sin(x) x u = x x v = cos (x) I = uv v u = x cos(x) + cos(x) x x = x cos(x) + (x sin(x) + cos(x)) + C

Chpter 6A Notes Pge 6 of We use the result of exmple. Exmple 3. Recurrence of the originl integrl Evlute I = e x cos(x) x Box: u = e x u = e x x v = cos(x) x v = sin (x) I = uv v u = e x sin(x) + e x sin(x) x (C) Evlute J = e x sin(x) x Box: u = e x u = e x x v = sin(x) x v = cos (x) J = uv v u = e x cos(x) e x cos(x) x = e x cos(x) I (D) Combine (C) n (D) I = e x sin(x) + ( e x cos(x) I) I = e x (sin(x) cos(x)) + C I = e x (sin(x) cos(x)) + C We e n explicit constnt of integrtion when we put I on only one sie of the eqution. Since I represents n inefinite integrl, it hs n implicit constnt of integrtion. Exmple 4: combine prts n substitution Evlute I = sin (x) x x sin (x) = x Box: u = sin (x) u = x x v = x v = x I = u v = uv v u = x sin (x) x x x

Chpter 6A Notes Pge 7 of Evlute J = x x x Let t = x, t = x x, t = x x Then J = t t = ( t ) + C = t + C Combine to get I = x sin (x) + ( x ) + C Definite Integrls Evlution Theorem b F (x)x = F(x) b = F(b) F() Prouct Rule x f(x)g(x) = f (x)g(x) + f(x)g (x) f(x)g (x) = x [f(x)g(x)] f (x)g(x) integrte from x = to x = b: b f(x)g (x)x = b [f(x)g(x)]x = x b f (x)g(x)x b f(x)g (x)x = f(x)g(x) b b f (x)g(x)x (E) Shorthn using ifferentil nottion Box: u = f(x) v = g (x) x u = f (x)x v = g(x) Write s for n inefinite integrl: u v = u v v u Replce u, u, v n v with pproprite expressions in x n limits of integrtion to get (E)

Chpter 6A Notes Pge 8 of e Exmple : I = ln(x) x Box: u = ln (x) v = x u = x x v = x u v = u v v u e e ln(x) x = x ln(x) e x = (e ln(e) ln()) (e ) = (e ) (e ) = Exmple : Combine substitution with prts 4 I = e x x Let t = x = x t = x x x t = x t t = x If x = then t =. If x = 4 then t =. I = e t t t = e t t t Now use integrtion by prts Box: u = t v = e t t u = t v = e t u v = uv v u te t t = t e t e t t = (e e) e t = (e e) (e e)

Chpter 6A Notes Pge 9 of = e Then I = e is the nswer. 6.A Trigonometric Integrls Combine trig ientities n substitution to evlute some useful integrls. Recll the ientities: sin (x) + cos (x) = () x x sin(x) = cos (x) cos(x) = sin (x) We will consier two cses of integrls. The first is sin m (x) cos n (x)x, (A) where m n n re integers. If m or n re o, substitute u = sin (x) or u = cos (x). Exmple. Evlute I = sin 4 (x) cos 3 (x)x Cosine is rise to n o power. Fctor out one power of cosine. I = sin 4 (x) cos (x) cos(x) x Use () to express the rest of the integrn s powers of sine. I = sin 4 (x)( sin (x)) cos (x) x Substitute u = sin (x), u = cos(x) x I = u 4 ( u )u = u 4 u 6 u = 5 u5 7 u7 + C I = 5 sin5 (x) 7 sin7 (x) + C

Chpter 6A Notes Pge of Exmple. Evlute I = tn(x) x Recognize I = sin (x) cos (x) x Sine is rise to n o power. Let u = cos(x), u = sin(x) x I = u u = ln u + C = ln cos (x) + C If both m n n re even in cse (A), use the following hlf ngle ientities cos (x) = ( + cos(x)) () sin (x) = ( cos(x)) (3) sin(x) cos(x) = sin (x) (4) π Exmple. Evlute I = sin (x) cos (x)x By (4): sin (x) cos (x) = 4 sin (x) By (3): sin (x) = ( cos(4x)) Thus sin (x) cos (x) = ( cos(4x)) 8 I = 8 π ( cos(4x)) x Let u = 4x, u = 4 x If x = then u =, If x = π then u = π. I = 8 (π ) 8 π cos(u) u 4 = π 8 x π where cos(u)u = sin(u) π = 8 π cos(4x) x I = π 6

Chpter 6A Notes Pge of Recll the ientities tn (x) + = sec (x) (5) x tn(x) = sec (x) x sec(x) = tn(x) sec(x) The secon cse of integrls we consier is tn m (x) sec n (x) x (B) If n is even use the substitution u = tn (x) Exmple. Evlute I = tn 4 (x) sec 4 (x) x By (5): I = tn 4 (x)(tn (x) + ) sec (x) x Let u = tn (x), u = sec (x) x I = u 4 (u + )u = u 6 + u 4 u = 7 u7 + 5 u5 + C = 7 tn7 (x) + 5 tn5 (x) + C If m is o use the substitution u = sec (x) Exmple. Evlute I = tn 3 (x) sec 3 (x) x I = tn (x) sec (x) tn(x) sec(x) x = (sec (x) ) sec (x) tn (x) sec(x) x Let u = sec(x), u = tn(x) sec(x) x I = (u )u u = u 4 u u = 5 u5 3 u3 + C = 5 sec5 (x) 3 sec3 (x) + C If both m is even n n is o, express the integrn entirely in terms of sec (x). This is the hrest cse. Exmple. Evlute I = tn (x) sec(x) x From (5): tn (x) = sec (x) I = sec 3 (x)x sec(x) x

Chpter 6A Notes Pge of sec(x) x = sec(x) sec(x)+tn (x) sec(x)+tn (x) x = sec (x)+sec(x)tn (x) sec(x)+tn (x) Let u = sec(x) + tn(x) x then u x = tn(x) sec(x) + sec (x) n u = (tn(x) sec(x) + sec (x)) x Then sec(x) x = u u = ln u + C = ln sec(x) + tn (x) + C (6) The other integrl require to evlute I is trete by exmple 8 in section 6. of our text. The result is: sec 3 (x) x = (sec(x) tn(x) + ln sec(x) + tn(x) ) + D (7) Results (6) n (7) cn be combine to obtin n expression for I.

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