# 7. Indefinite Integrals

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1 7. Indefinite Integrls These lecture notes present my interprettion of Ruth Lwrence s lecture notes (in Herew) 7. Prolem sttement By the fundmentl theorem of clculus, to clculte n integrl we need to find function F stisfying F = f. Then, f f = F() F(). Such function F is clled primitive function (%/&\$8 %*78&5), n nti-derivtive or n indefinite integrl (.*&2/!- -9#)*!) of f. This chpter presents techniques for clculting indefinite integrls. Given function f, we denote its indefinite integrl y f. To denote tht we evlute primitive function of f t we write f. A more stndrd nottion is f (t).

2 00 Indefinite Integrls The stndrd nottion is useful when rther thn putting the nme of function in the integrnd we write its functionl form. For emple, for f () = 2 we my either write f or t 2. Recll tht n indefinite integrl is only determined up to n dditive constnt. Thus, when we write n identity like f = g we men tht ny primitive of f is lso primitive of g, nd when we write n identity like f = h+ g we men tht ny primitive of g plus h is primitive of f. Comment 7. It is not generlly true tht if F nd G re primitive functions of f then they differ y constnt. Consider the function f () = 2 R{0}. Then F() = nd G() = + < 0 +5 > 0 re oth primitive functions of f. Generlly, if function is defined on domin consisting of disconnected components, its indefinite integrl is unique up to different dditive constnt in ech connected component. The most elementry method for finding indefinite integrls is to differentite functions nd crete tle of functions nd their derivtive. Then, we cn red this tle ckwrd. This is not very prcticl men, ut still, we hve t the outset few functions whose indefinite integrls we know. Emple 7. Since we know tht sin = cos, it follows tht cost= sin. It is lso true tht cost= 5+sin, ut it is certinly not true tht sin = 5+sin. Recll tht the equlity relly mens tht sin elongs to the set of primitive functions of cos. In fct, we should hve written sin cos, ut we won t e doing it.

3 7.2 Elementry functions 0 We present elow tle of elementry integrls", which re otined y differentition: f e sin cos tn f() e cos sin cos 2 ln sin 2 cos 2 tn (+ 2 ) ln(sin) cot ln(cos) tn sinh cosh cosh sinh tnh tnh 2 sinh 2 + cosh 2 tnh ( 2 ) This tle is n strting point for clculting indefinite integrls of mny more functions. 7.2 Elementry functions The functions for which we hve nmes powers, trigonometric functions, hyperolic functions, the eponentil nd the logrithm, roots, inverse trigonometric functions nd inverse hyperolic functions re clled commonly elementry functions. So is ny function tht cn e otined y dding, multiplying, dividing nd composing elementry functions (note the recursive definitions). Emple 7.2 The function is n elementry function. f () = sin 2 e +sin (tnh 3 ) 3 5 +ln(tn ()) By the lws of derivtion we know how to differentite every elementry function, getting gin n elementry function. Note, however, tht there re mny more functions tht re not elementry. We simply don t hve nmes for them (sometimes we do). It turns out tht mny elementry functions don t hve elementry

4 02 Indefinite Integrls indefinite integrls. A theorem y Liouville in 835 provided the first proof tht non-elementry nti-derivtives of elementry functions eist. Emple 7.3 The following elementry functions don t hve elementry ntiderivtives: f () = 4 f () = ln(ln) f () = sin nd f () = e 2. It should e cler, however, tht these functions hve nti-derivtives. We simply cn t epress them s elementry functions. In fct, the function e t2 is sufficiently importnt in sttistics tht it hs nme the error function. 7.3 Integrtion y prts (.*8-(" %*79#)*!) Recll the Leiniz rule, (uv) = u v+uv. Since uv is primitive function of (uv), we hve uv = u v+ uv, or u v = uv uv (7.) which in more stndrd nottion reds u (t)v(t) = u()v() u(t)v (t). This innocent-looking identity is useful strting point for evluting indefinite integrls. Emple 7.4 Consider the indefinite integrl I() = lnt. We evlute it s follows: set u() = nd v() = ln. Then, I() = lnt = ln u (t) v(t) u() v() t t u(t) v (t) = ln.

5 7.3 Integrtion y prts (.*8-(" %*79#)*!) 03 Note tht we could hve s well tken u() = + 5, in which cse we would hve gotten I() = lnt = ( +5) ln u() v() u (t) v(t) (t +5) u(t) t v (t) = ln, so s epected, it doesn t mtter which nti-derivtive of u we tke. Emple 7.5 For n N consider the indefinite integrls I n () = t n lnt. Here we ll tke u () = n nd v() = ln, hence I n () = t n lnt= n+ n+ ln t n+ n+ n+ n+ = ln t n+ (n+) 2. Emple 7.6 For n N consider the indefinite integrls I n () = t n e t. Tking e = u () nd n = v() we get I n () = n e n t n e t = n e ni n (). This is recurrence reltion (%#*2 2(*). For n = 0, Then, It is esy to see tht I = e e = ( )e I 0 () = e. I 2 = 2 e 2( )e = ( )e I 3 = 3 e 3( )e = ( )e = I n () = n! n n! n (n )! + n 2 (n 2)! +( )n e = n!e n ( ) n k k. k! k=0 Emple 7.7 Consider the sequence of indefinite integrls I n () = sin n t.

6 04 Indefinite Integrls We tret it s follows: I n () = sin n t sint = sin n cos +(n ) = sin n cos +(n ) sin n 2 t cos 2 td sin n 2 t( sin 2 t)d Hence, we get recurrence reltion This time we need oth I 0 nd I : = sin n cos +(n )I n 2 () (n )I n (). I n () = n n I n 2() n sinn cos. I 0 () = nd I () = cos. Then, I 2 () = 2 I 0() 2 sin cos I 3 () = 2 3 I () 3 sin2 cos I 4 () = 3 4 I 2() 4 sin3 cos, nd so on. 7.4 Sustitution method (%"7%% ;)*:) Recll the chin rule, from which follows tht (F g) = (F g)g F g = (F g)g For F = f (t), g() f = ( f g)g (7.2) nd it more stndrd nottion, f (g(t))g (t) = g() f (t). As we will see, this identity is useful for finding mny indefinite integrls.

7 7.4 Sustitution method (%"7%% ;)*:) Sustitution of liner functions Suppose tht we know primitive function of f nd we wnt to evlute I() = f (t +), where is constnt. Set g() = +, nd use the sustitution formul (7.2), I() = f (g(t))g (t) = g() f (t) = + f (t). Comment 7.2 The physicists" wy of doing the sme procedure goes s follows: they define, sy, u = +, nd then write Then, du = du d = d. d f ( +)d = f (u)du, nd it is understood tht the right hnd side is evluted t u = +. Emple 7.8 cos(t +) = sin( +). Emple 7.9 Find I() = 2 t 2. First note tht I() = (t) 2. Setting g() =, we use the sustitution formul (7.2) to get I() = g 2 (t) = g (t) g 2 (t) = g() t 2 = sin (g()) = sin.

8 06 Indefinite Integrls Emple 7.0 As similr emple, consider I() = 8+2t t 2. The trick here is to first complete the squre under the root, I() = = 9 (t ) 2 3. ((t )3) 2 Set then Then, I() = 3 g 2 (t) = g() = 3. g (t) g 2 (t) = g() = t 2 sin Other emples Emple 7. Consider gin the integrls I n () = sin n t, this time only for odd vlues of n = 2m +. We ve lredy solved these integrls using integrtion y prts. This time we will use the method of sustitution. Strt with I n () = sint(sin 2 t) m d = sint( cos 2 t) m d. Setting we get f () = ( 2 ) m nd g() = cos, I n () = f (g(t))g cos (t) = ( t 2 ) m. Tke for emple n = 3. Then, I 3 () = cos cos3 3 Emple 7.2 Consider integrls of the form I() = r(t) t 2,

9 7.5 Integrls of rtionl functions 07 where r() is rtionl function. Setting g() = sin or equivlently = sing(), we get I() = r(sin(g(t)))g (t) = sin r(sin(t)). Tke for emple r() = 2. Then, Thus, r(sin(t)) = sin 2 t= 2 ( cos2t) = 2 4 sin2 = ( sin cos). 2 t 2 t 2 = 2 sin Integrls of rtionl functions Consider integrls of the form p(t) q(t), where p nd q re polynomils ( function tht is the rtio of two polynomils is clled rtionl function). If deg p degq then we cn divide the polynomils to get p() s() = r()+ q() q(), where r nd s re polynomils, nd degs < degq. Since we know how to integrte polynomils, it remins to lern how to integrte rtionl functions with deg p < degq Liner denomintors When q() is liner function, we end with n integrl of the form = ln + t + Emple 7.3 Consider the following indefinite integrl I() = t 3 t +5.

10 08 Indefinite Integrls Since it follows tht 3 +5 = ( +5) = ( +5)2 5( +5) = ( +5)2 5( +5)+25( +5) = , I() = ln Qudrtic denomintors When q() is qudrtic function there re three possiilities:. q hs two distinct rel-vlued roots, in which cse we otin n integrl of the form t + (t c)(t d), 2. The two roots of q coincide, in which cse we otin n integrl of the form t + (t c) 2 3. q is irreducile, i.e., hs two comple-vlued root, then the integrl cn e rought to the form t + (t c) 2 +d 2 Cse : Two distinct rel-vlued roots In this cse we split the the integrnd s follows: t + (t c)(t d) = A t c + Mtching coefficients we otin B t d A(t d)+b(t c) =. (t c)(t d) A+B = nd Ad Bc =. This is liner system tht lwys hs solution provided c d: A = c+ c d nd B = d + d c

11 7.5 Integrls of rtionl functions 09 Then, (t +) (t c)(t d) = c+ d + ln c+ ln d c d d c Emple 7.4 Consider the cse = 0, =, c = nd d = 2, (t +)(t +2) = ln + ln +2 Cse 2: Two equl roots We split the integrnd s follows, t + (t c)+(+c) = (t c) 2 (t c) 2. Thus, nmely, (t +) (t c) 2 = t c +(c+) (t +) (t c) 2 = ln c c+ c (t c) 2, Emple 7.5 2t +5 t 2 +4t +4 = 2t +5 = 2 ln +2 (t +2) Cse 3: Two comple-vlued roots Remins the cse where q is irreducile (with rel numers). In this cse we cn ring q y squre completion to the form q(t) = (t c) 2 +d 2. Now, Hence, t + (t c)+(c+) (t c) 2 = +d2 (t c) 2 +d 2. (t +) (t c) 2 +d 2 = (t c) (t c) 2 +d 2 +(c+) ( c) 2 +d 2. The first integrl equls 2 ln(t c)2 +d 2,

12 0 Indefinite Integrls wheres the second integrl equls, using the method of sustitution To conclude, c+ d 2 [(t c)d] 2 + = c+ d tn c d. (t +) (t c) 2 +d 2 = 2 ln( c)2 +d 2 + c+ d tn c d Denomintors of higher degree Let s consider n emple. Other follow similr route: Emple 7.6 (t +5) t 3 t 2 t + To do something we must e le to decompose the denomintor. In the present cse t 3 t 2 t + = (t +)(t ) 2. Since we hve simple root, ( ), nd doule root,, we look for decomposition, Mtching the numertors from which we get tht Then, t +5 t 3 t 2 t + = A t + B (t ) 2 + C t +. t +5 = A(t 2 )+B(t +)+C(t 2 2t +), A = C B 2C = nd A+B+C = 5. A = B = 3 nd C =, from which we conclude tht (t +5) 3 t 3 t 2 = ln +ln +. t Definite integrls We now turn to evlute definite integrls. By the Newton-Leiniz theorem it suffices to find primitive function F of f, in which cse f = F. In this section we will see how to pply the integrtion methods for the clcultion of definite integrls.

13 7.6 Definite integrls 7.6. Integrtion y prts Since it follows tht uv = (uv) = u v+ uv, u ()v()d = u()v() u()v ()d Emple 7.7 Clculte 6 5 ln Method of sustitution Let F = f. Since If follows tht i.e., F g = Emple 7.8 Clculte Letting g(t) = t 2, we hve I = We proceed y writing (F g) = f (g(t))g (t), f (g(t))g (t) = F g() g(), f (g(t))g (t) = I = g (t) (+g(t))(2+g(t)) = g(2) 2 g(0) otining A = B =. Hence I = g() g() t (+t 2 )(2+t 2 ). f (t) (+t)(2+t) = (+t)(2+t) = A +t + B 2+t = A(2+t)+B(+t), (+t)(2+t) 0 4 +t 0 4 (+t)(2+t). 2+t = 2 (ln5 ln ln6+ln2) = 2 ln 5 3.

14 2 Indefinite Integrls 7.7 Improper integrls Improper integrls (.**;*/!!-.*-9#)*!) re definite integrls in which something infinite" tkes plce: either the domin of integrtion is infinite or the integrnd diverges within the domin of integrtion. The first instnce comprises of integrls of the form f = lim f f lim y y f = lim f = lim Emple 7.9 Compre the two cses t 2 nd Note tht the Riemnn sums tht correspond to Dt = re the respectively convergent nd divergent series, nd f. + = p2 6 t =. The second instnce comprises the emples which we interpret s 0 lim 0 + Emple 7.20 Consider the integrl Since sin is nti-symmetric, for every nd t t nd lim sint. sint= 0. 0 t, 0 + It my seem therefore tht the integrl of sin over the whole line is zero, ut this is not the cse s sint= cosy cos does not hve limit s y nd y t.

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