9. Integration Standard integrals Basic rules. If u is a function of x, then the indenite integral
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1 9. Integrtion 9.. Stndrd integrls If u is function of, then the indenite integrl u d is function whose derivtive is u. The stndrd integrls re: ( If is constnt, then d = + c where c is constnt, clled the constnt of integrtion; this mens tht the derivtive of + c is. ( n d = n+ + c for n. n + For emple, d = / d = / / + c = + c. ( sin d = cos + c ; cos d = sin + c. (4 e d = e + c. (5 d = ln +c, or to llow for the possibility of being negtive, d = ln +c. (6 The inverse trigonometric functions lso give some integrls, like + d = tn + c. 9.. Bsic rules The rules for dierentition imply the following bsic rules for integrtion.. If is constnt then u d = u d ; for emple, cos d = cos d = sin + c. Note tht you don't need to write the constnt of integrtion until the nl nswer.. (u ± v d = u d ± v d For emple (e + d = e d + 44 d = e c.
2 Emple 9.. ( + d ( = ( + ( + ( d = ( d = c. 9.. Denite integrls The denite integrl of f( from to b is denoted by b f( d, nd it is equl to [F (] b = F (b F (, where F ( = f( d is the indenite integrl. You don't need to include the constnt of integrtion in F ( if you do it cncels out so there is no constnt of integrtion in the nl nswer. Emple 9.. Emple 9.. π/ cos d = [ d = [ ] π/ sin = sin π sin = =. ] = ( = 8 8 = 6. The Fundmentl Theorem of Clculus sys tht the indenite integrl b f( d mesures the re under the curve y = f( in the region b. Emple 9.4. Wht is the re of the tringulr region under y = m between = nd = b? It is b [ m ] b m d = = mb. Note tht this is just hlf the bse times the height, s epected Integrtion by substitution To integrte y d by substitution, do the following: ( Choose suitble substitution, sy u = g(. Use it to write in terms of u if necessry. (b Find d/du, either directly, or from the formul d/du = /(du/d. (c Rewrite the integrl using the formul y d = 45 y d du du
3 (this is vlid becuse of the chin rule. (d Try to epress the integrnd y(d/du s function of u lone, not involving, so tht the integrl becomes of the form h(udu. If you cn't do tht, the substitution is no use. (e Compute the integrl h(u du, then use the substitution u = g( to write it in terms of. Emple 9.5. Find cos d using the substitution u =. We hve = u, so d/du =. Therefore cos d = cos u du = sin u + c = sin + c sin Emple 9.6. Find d using the substitution u = cos +. cos + We hve du/d = sin, so d/du = / sin. sin sin cos + d = u sin du = du = ln u + c = ln cos + + c. u Emple 9.7. Find d. 9 Let = u, so tht u = /. We hve d/du =, so d = du 9 9 du = 9 9u u du = sin u + c = sin (/ + c. Emple 9.8. Using the substitution u = +, nd the re under the curve y = /( + between = nd =. It is + d. To nd the indenite integrl + d, use the substitution u = +. du/d =, so d/du = /(. Therefore = + d = +. du u du = ln u + c = ln + + c. 46
4 Therefore [ ] + d = ln + = ln ln = ln. Specil cse. If f( d = F ( + c, nd nd b re constnts, then f( + bd = F ( + b + c (this follows by putting u = + b. Emple d = ln( c Emple 9.. (4 + 5 d = ( c Integrtion by prts To integrte y d by prts, do the following: ( Write y s product of two terms, nd choose one of them tht you cn integrte. (b Let dv/d be the term you cn integrte, nd let v be the integrl. (c Let u be the other term, nd compute du/d. (d Rewrite the integrl using the formul y d = (this follows from the product formul. Emple 9.. Find cos d by prts. u dv d = uv d du d v d The integrnd cos is the product of nd cos. We cn integrte both terms, but if you choose, it won't help in fct, it mkes the integrl more complicted. Insted we choose to integrte cos. Let dv/d = cos, so tht v = sin. (You don't need to include the +c here Let u =. du/d =. cos d = du v d = sin d u dv d = uv d We know this lst integrl, so we get cos d = sin + cos + c. Emple 9.. Find the denite integrl ( + e d. First we wnt the indenite integrl ( + e d. 47 sin d.
5 Tke dv/d = e nd u = +. v = e nd du/d =. Therefore ( + e d = u dv du d = uv d d v d = ( + e + e d = ( + e e + c Emple 9.. Find sin d. = ( + e + c. ( + e d = [ ( + e ] = 7e. One needs to use integrtion by prts twice. Tke dv/d = sin, so v = cos, nd u =, so du/d =. sin d = u dv du d = uv d d v d = cos + cos d. Now we lredy found cos d = sin + cos + c using integrtion by prts, so we get sin d = cos + sin + cos + c. Emple 9.4. Find ln d. This doesn't look like product, but write it s ln = ln with dv/d = nd u = ln. v = nd du/d = /. Therefore du ln d = uv v d = ln d = ln + c. d 9.6. Integrtion of rtionl functions To compute the integrl of rtionl function, convert it into prtil frctions. use the following integrls: ( + b d = du d ln +b +c. Nmely, letting u = +b we hve =, so d du =. + b d = u du = ln u + c = ln + b + c. 48
6 ( ( ( + b d = + c. To see this, let u = + b; then we hve + b ( + b d = u ( du = + c = ( + c. u + b Emple 9.5. Wht is Write ( + ( + d? ( + ( + = ( ( + ( + d = ( + d + d + = ln + + ln + + c. + + d To integrte rtionl functions with the qudrtic + in the denomintor, use ( + d = tn + c ; (4 + d = ln( + + c. Emple d = 5 + d + + d. This ts the formule with =. Therefore + + d = 5 ln( + + tn ( + c. To integrte rtionl functions with more generl qudrtic in the denomintor, complete the squre, nd use substitution. Emple 9.7. Find We complete the squre: so d = ( = ( d = ( d.
7 Now use the substitution u = +. = u, so d/du =. Therefore d = u + u + 6 du = u u + 6 du + u + 6 du 9.7. Worked emples = ln(u tn (u/4 + c = ln( tn (( + /4 + c. Emple 9.8. Clculte the indenite integrls: (i tn d (ii ln d (iii sin ( d (iv e d Emple 9.9. Clculte the integrl ( + d. Emple 9.. Clculte the integrl π/4 cos sin cos + sin d. 5
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