LESSON 0 Chpter 7.2 Trigonometric Integrls. Bsic trig integrls you should know. sin = cos + C cos = sin + C sec 2 = tn + C sec tn = sec + C csc 2 = cot + C csc cot = csc + C MA 6200 Em 2 Study Guide, Fll 204 tn = ln sec + C = ln cos + C sec = ln sec + tn + C cot = ln sin + C csc = ln csc cot + C 2. Products nd powers of trig functions. Usully the ide is to chnge the integrl to the form of u n ). u n du (or the integrl of liner comintions Integrnds contining powers of sine nd cosine functions. sin m cos n. Odd powers of sin fctor odd fctor of sin nd use the identity sin 2 = cos 2 to convert remining even powers of sin to cos. Then let u = cos. Odd powers of cos fctor odd fctor of cos nd use the identity cos 2 = sin 2 to convert remining even powers of cos to sin. Then let u = sin. No odd powers of sin or cos use cos 2 = 2 + 2 cos(2) or sin2 = 2 2 cos(2). After integrting the trig identity sin 2 = sin cos my e useful. Integrnds contining tngent nd secnt functions. tn m sec n. fctor sec 2 nd chnge remining fctors of sec 2 to tngent using sec 2 = tn 2 +. fctor tn sec nd chnge remining fctors of tn 2 to secnt using tn 2 = sec 2.
LESSONS nd 2 Chpter 7.3 Trig Susitution If integrnd contins Use this sustitution And use this identity 2 (u()) 2 u() = sin θ sin 2 θ = cos 2 θ 2 + (u()) 2 u() = tn θ + sin 2 θ = sec 2 θ (u())2 2 u() = secθ sec 2 θ = tn 2 θ * Sometimes the integrnd will contin powers of these roots. Completing the squre is sometimes necessry to get the forms 2 (u()) 2, 2 +(u()) 2, or (u()) 2 2. LESSONS 3 nd 4 Chpter 7.3 Prtil Frctions Let P() e rtionl function (P() nd Q() re oth polynomils) where P nd Q hve no Q() common fctors nd the degree of P is less thn the degree of Q. If P nd Q hve common fctors, cncel them. If degree of P not less thn degree of Q, then use long division to divide P y Q. Fctor the denomintor Q() into liner fctors (+) nd irreducile qudrtic fctors ( 2 ++c, where 2 4c < 0). m liner fctors ( + ) m give rise to m prtil frctions of the form A + + A 2 ( + ) 2 + A 3 ( + ) 3 + + A m ( + ) m + A m ( + ) m A,A 2,A 3,...,A n re constnts. When you set up your prtil frctions use A,B,C,D,... in the numertors insted. With the emples we use, you won t run out of letters! n irreducile qudrtic fctors ( 2 + + c) n give rise to n prtil frctions of the form A + B 2 + + c + A 2 + B 2 ( 2 + + c) 2 + A 3 + B 3 ( 2 + + c) 3 + + A n + B n ( 2 + + c) n + A n + B n ( 2 + + c) n A,A 2,A 3,...,A n nd B,B 2,B 3,...,B n re constnts. When you set up your prtil frctions use A + B, C + D, E + F,... in the numertors insted. With the emples we use, you won t run out of letters! Set the rtionl function P() equl to its sum of prtil frctions. Net multiply oth sides of this Q() eqution y the common denomintor Q(). Then solve for the constnts y either choosing vlues of tht will eliminte ll ut one constnt, nd/or equting the coefficients of like powers of to get system of liner equtions where the unknowns re the constnts. Solve tht system using your fvorite method.
When integrting the prtil frctions, you will usully e deling with the following kinds of integrls: = ln + C ( )n+ = ( ) n n + 2 + = tn + C + C, n LESSON 5 Chpter 7.6 Integtion using tles When using tle of integrls to evlute f(), rememer tht is not lwys equl to u. Sometimes you first must use sustitution u = u(). If you don t, your nswer will e incorrect y constnt fctor. Chpter 7.7 Approimte Integrtion ( ) To pproimte f(), sudivide the intervl [,] into n equl suintervls. Let i = +i. n Note tht 0 = nd n =. 0 2 3 n n Midpoint Rule: ( ) ( ( ) ( ) ( ) ( )) 0 + + 2 2 + 3 n + n f() M n = f + f + f + + f n 2 2 2 2 Trpeziodl Rule: ( ) f() T n = (f( 0 ) + 2f( ) + 2f( 2 ) + + 2f( n ) + f( n )). 2n Simpson s Rule: Note: n must e even f() ( ) S n = (f( 0 ) + 4f( ) + 2f( 2 ) + 4f( 3 ) + 2f( 4 ) + + 2f( n 2 ) + 4f( n ) + f( n )). 3n
LESSON 6 Chpter 7.8 Improper Integrls The integrl = = f() is improper if t lest one of the following is true: f hs n infinite discontinuity t some point c in the intervl [,] (the grph of f hs verticl symptote = c). We determine whether n improper integrl hs vlue y evluting the limit of proper integrls s follows: ( rememer to evlute the proper integrl on the right hnd side efore evluting the limit) Improper integrl of Type I t f() = lim f(). t f() = lim f(). t t Improper integrl of Type II If f hs discontinuity t, then If f hs discontinuity t, then f() = lim f(). t + t t f() = lim f(). t If the limit eists, then the limit vlue is the vlue of the improper integrl. If the limit does not eist, then the improper integrl is divergent. If f hs n infinite discontinuity t c where < c <, then f() = c f() + c f() provided BOTH improper integrls on the right hnd side eist. If either diverges, then diverges. An importnt integrl: is convergent if p > nd divergent if p. p f() Comprison Test for Improper Integrls When you cn t determine n nti-derivtive for n improper integrl, comprison with known convergent or divergent integrl is useful. Suppose tht f nd g re continuous functions with 0 g() f() for.
If If LESSON 7 Chpter 8. Arc length f() is convergent, then g() is divergent, then If y = f() nd, then L = If = g(y) nd c y d, then L = d Chpter 8.2 Are of surfce of revolution c g() is convergent. f() is divergent. + (f ()) 2 + (g (y)) 2 dy Rotte out the is y y = f() Rotte out the y is R y R y = f() R = rdius = f() R = rdius = Surfce Are = 2π(f()) + ( ) 2 dy Surfce Are = 2π() + ( ) 2 dy