Math 107H Topics for the first exam. csc 2 x dx = cot x + C csc x cotx dx = csc x + C tan x dx = ln secx + C cot x dx = ln sinx + C e x dx = e x + C


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1 Integrtion Mth 07H Topics for the first exm Bsic list: x n dx = xn+ + C (provided n ) n + sin(kx) dx = cos(kx) + C k sec x dx = tnx + C sec x tnx dx = sec x + C /x dx = ln x + C cos(kx) dx = sin(kx) + C k csc x dx = cot x + C csc x cotx dx = csc x + C tn x dx = ln secx + C cot x dx = ln sinx + C e x dx = e x + C dx x + = Arctn( x ) + c Bsic integrtion rules: for k=constnt, k = k sec x dx = ln sec x + tnx + C csc x dx = ln csc x + cotx + C dx = Arcsin( x x ) + c dx x x = Arcsec( x ) + c (f(x) ± g(x) dx = ± g(x) dx The Fundmentl Theorem of Clculus x f(t) dt = F(x) is function of x. F(x) = the re under grph of f, from to x. FTC : If f is cts, then F (x) = f(x) (F is n ntideriv of f!) Since ny two ntiderivtives differ by constnt, nd F(b) = b f(t) dt, we get FTC : If f is cts, nd F is n ntideriv of f, then b = F(b) F() = F(x) b Integrtion by substitution. The ide: reverse the chin rule! g(x) = u, then d dx f(g(x))= d dx f(u) = f (u) du dx, so f (u) du dx dx = f (u) du = f(u)+c f(g(x))g (x) dx ; set u = g(x), then du = g (x) dx, so f(g(x))g (x) dx = f(u) du, where u = g(x) Exmple: x(x 3) 4 dx ; set u = x 3, so du=x dx. Then x(x 3) 4 dx = (x 3) 4 x dx = u 4 du u=x 3 = u c u=x 3 = (x 3) 5 + c 0 The three most importnt points:. Mke sure tht you clculte (nd then set side) your du before doing step!. Mke sure everything gets chnged from x s to u s 3. Don t push x s through the integrl sign! They re not constnts! We cn use usubstitution directly with definite integrl, provided we remember tht b relly mens x=b x=, nd we remember to chnge ll x s to u s! x(+x ) 6 dx; set u = +x, du = x dx. when x =, u = ; when x =, u = 5; so x( + x ) 6 dx = 5 u 6 du =...
2 Bsic integrtion formuls (AKA dirty tricks) complete the squre x + bx + c = (x + rx) + c = (x + r/) + (c (r/) ) x + x + dx = (x + ) + dx use trig identities sin x + cos x =, tn x + = sec x, sin(x) = sinxcos x, tn x =sin x, etc. sec x sin x cos cos x dx = x dx cos x pull frctions prt; put frctions together! x + dx = x + x 3 dx x 3 do polynomil long division x 3 x dx = x + x x dx dd zero, multiply byone sec x(tnx + sec x) sec x dx = dx sec x + tnx Integrtion by prts Product rule: d(uv) = (du)v + u(dv) reverse: u dv = uv v du x cosx dx : set u=x, dv=cos x dx du=dx, v = sin x (or ny other ntiderivtive) So: x cos x = x sinx sin x dx specil cse: ; u = f(x), dv=dx = xf(x) xf (x) dx x Arcsin x dx = x Arcsin x x The bsic ide: integrte prt of the function ( prt tht you cn), differentite the rest. Gol: rech n integrl tht is nicer. x 3 lnx dx = (x 4 /4) lnx (x 4 /4)(/x) dx Trig substitution Ide: get rid of squre roots, by turning the stuff inside into perfect squre! x : set x = sin u. dx = cosu du, x = cos u x x dx = cos u sin u cosu du x=sin u + x : set x = tnu. dx = sec u du, + x = sec u sec u dx = (x + 4) 3/ ( sec u) du x= 3 tn u x : set x = sec u. dx = secutnu du, x = tnu
3 sec u tnu x x dx = sec u tnu du x=sec u Undoing the usubstitution : use right tringles! (Drw right tringle!) x = sinu, then ngle u hs opposite = x, hypotenuse =, so djcent = x. So cos u = ( x )/, tnu = x/ x, etc. Trig integrls: Wht trig substitution usully leds to! sin n x cos m x dx If n is odd, keep one sinx nd turn the others, in pirs, into cosx (using sin x = cos x), then do usubstitution u = cos x. If m is odd, reverse the roles of sin x nd cos x. If both re even, turn the sinx into cos x (in pirs) nd use the double ngle formul cos x = ( + cos(x)) This will convert cos m x into bunch of lower powers of cos(x); odd powers cn be delt with by substitution, even powers by nother ppliction of the ngle doubling formul! sec n x tn m sin m x x dx = cos n+m x dx If n is even, set two of them side nd convert the rest to tnx using sec x = tn x +, nd use u = tnx. If m is odd, set one ech of sec x, tnx side, convert the rest of the tnx to sec x using tn x = sec x, nd use u = sec x. If n is odd nd m is even, convert ll of the tnx to sec x (in pirs), leving bunch of powers of sec x. Then use the reduction formul: sec n x dx = n secn x tnx + n sec n x dx n At the end, rech sec x dx = tnx + C or sec x dx = ln sec x + tnx + C A little trick worth knowing: the substitution u = π x, since sin(π x) = cos x nd cos(π x) = sin x, will reverse the roles of sin x nd cos x, so will turn cot x into tnu nd csc x into sec u. So, for exmple, the integrl cos 4 x sin 7 x dx = csc 3 x cot 4 x dx, which our techniques don t cover, becomes sec 3 u tn 4 u du, which our techniques do cover. 3
4 Prtil frctions rtionl function = quotient of polynomils Ide: integrte by writing function s sum of simpler functions Procedure: f(x) = p(x)/q(x) (0): rrnge for degree(p)<degree(q); do long division if it isn t (): fctor q(x) into liner nd irreducible qudrtic fctors (): group common fctors together s powers (3): for ech group (x ) n dd together: (3b): for ech group (x + bx + c) n dd together: x + b x + + n (x ) n n x + b n (x + bx + c) n x + bx + c + + (4) set f(x) = sum of ll sums; solve for the undetermined coefficients put sum over common denomentor (=q(x)); set numertors equl. lwys works: multiply out, group common powers, set coeffs of the two polys equl x + 3 = (x ) + b(x ) = ( + b)x + ( b); = + b, 3 = b liner term (x ) n : set x =, will llow you to solve for coefficient if n, tke derivtives of both sides! set x=, gives nother coeff. Improper integrls x (x ) (x + ) = A x + Fund Thm of Clc: b B (x ) + Cx + D x + = A(x )(x + ) + B(x + ) + (Cx + D)(x ) (x ) (x + ) = F(b) F(), where F (x) = f(x) Problems: =, b = ; f blows up t or b or somewhere in between integrl is improper ; usul technique doesn t work. Solution to this: = lim b b b b (blow up t ) = lim r + (similrly for blowup t b (or both!)) b = lim s b s r = lim ɛ 0 + b b = lim = lim ɛ 0 + b ɛ r (blows up t c (b/w nd b)) = lim r c The integrl converges if (ll of the) limit(s) re finite Comprison: 0 f(x) g(x) for ll x; if if b g(x) dx converges, so does diverges, so does 4 +ɛ b b + lim s c + s g(x) dx
5 Applictions of integrtion Volume by slicing. To clculte volume, prroximte region by objects whose volume we cn clculte. Volume (volumes of cylinders ) = (re of bse)(height) = (re of crosssection) x i. So volume = right (re of cross section) dx Solids of revolution: disks nd wshers. Solid of revolution: tke region in the plne nd revolve it round n xis in the plne. tke crosssections perpendiculr to xis of revolution; crosssection = disk (re=πr ) or wsher (re=πr πr ) rotte round xxis: write r (nd R) s functions of x, integrte dx rotte round yxis: write r (nd R) s functions of y, integrte dy Otherwise, everything is s before: volume = right A(x) dx or volume = top region bottom rotte A(y) dy The sme is true if xis is prllel to x or y xis; r nd R just chnge (we dd constnt). Cylindricl shells. Different picture, sme volume! Solid of revolution; use cylinders centered on the xis of revolution. The intersection is cylinder, with re = (circumference)(height) = πrh volume = right (re of cylinder) dx or top bottom (re of cylinder) dy!) region revolve round verticl line: integrte dx revolve round horizontl line: integrte dy r h rotte 5
6 region in plne between y = 4x, y = x, revolved round yxis =0, right=4, r = x, h = (4x x ) volume = Arclength nd surfce re 4 0 πx(4x x ) dx rotte Arclength. Ide: pproximte curve by lots of short line segments; length of curve sum of lengths of line segments. Line segment between (c i, f(c i )) nd (c i+, f(c i+ )): + ( f(c i+) f(c i ) ) c i+ c (c i+ c i ) + (f (c i )) x i i right So length of curve = + (f (x)) dx The problem: integrting + (f (x))! Sometimes, + (f (x)) turns out to be perfect squre... Surfce re. Ide: find the re of surfce (of revolution) by pproximting the surfce by things whose re we cn figure out. Frustrum of cone! re of frustrum = π (f(c i+ ) + f(c i ) ) ( f(ci+ ) f(c i ) ) + (ci+ c i ) c i+ c i πf(c i + (f (c i )) x i. So re of surfce = right πf(x) + (f (x)) dx The problem: sme problem! But sometimes it s possible to do... for f(x) = r x, the thing to integrte simplifies to: πr! 6
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