INTEGRATION TECHNIQUES Introdction We re looking for wys to compte the integrl of fnction f(x), f(x)dx. To pt it simply, wht we need to do is find fnction F (x) sch tht F (x) = f(x). Then if the integrl is definite, tht is, we he n interl oer which to integrte, the soltion is b f(x) dx = b F (x) dx = F (b) F (). For n indefinite integrl (no interl) the soltion is f(x) dx = F (x) dx = F (x) + c, where c is n rbitrry constnt. The integrls tht rise in differentil eqtion pplictions re mostly of the ltter type nd there the constnt c plys crcil prt nd shold not be forgotten. Selected specil cses Powers of x (possibly inclding constnt) re strightforwrd to integrte. The following forml works for both integer nd noninteger les of n. Ex. Ex. 3 Ex. 4 Ex. Ex. 6 (x + ) n dx = n+ (x + ) + c, n, ( = 0 when n = 0). n + dx = x + c x 3.2 dx = x4.2 4.2 + c (x 4 + 3x 3 2x 2 + x ) dx = x + 3x4 4 2x3 3 + x2 2 x + c (x + 3) 4 (x + 3) dx = + c (2x + 3) 4 dx = (2x + 3) + c (Note the extr coefficient.) 2 dx = (x ) 2 dx = 2(x ) 2 + c x
The deritie of the logrithm hs specil ppernce, which often mkes it esy to spot in n integrl. The forml looks like this: f (x) dx = ln f(x) + c. f(x) The bsolte le in the logrithm is there for reson. Both ln f(x) nd ln(f(x)) re soltions, bt only the one with positie le gies well defined nswer. Tking the bsolte le chooses the right soltion. sin x Ex. tn x dx = dx = ln cos x + c cos x x 3 x dx = 2x 2 2 3 x dx = 2 2 ln 3 x2 + c Ex. 3 dx = ln x + + c x + The fnctions e x2 nd e x2 re bit specil. It is necessry to know wht cn nd cnnot be done, nd lso to oid confsing it with similr fnctions. Ex. 2xe x2 dx = e x2 + c e x2 dx = π Ex. 3 e x2 dx No closed form soltion exists. Ex. 4 (e x ) 2 dx = e 2x dx = e2x 2 + c Integrtion by prts Integrtion by prts is especilly sefl when the integrnd hs two prts tht re not too complicted, bt together form something tht is hrd to integrte. The bsis for the method is the prodct rle for derities: d dx ((x)(x)) = (x)(x) + (x) (x). Integrting both sides with respect to x nd moing the terms rond yields dx = dx. 2
In the first exmple, we se this for cosine or sine fnction mltiplied by n exponentil fnction. e αx e αx cos βx dx = α eαx cos βx sin βx dx = α eαx sin βx α eαx (β sin βx) dx () α eαx (β cos βx) dx (2) Using (2) to replce the integrl in the right hnd side of () gies e αx cos βx dx = α eαx cos βx + β ( ) α α eαx sin βx α eαx β cos βx dx. By rerrnging the eqtion nd dding the constnt c we get the finl reslt e αx cos βx dx = eαx (α cos βx + β sin βx) + c. α 2 + β2 The integrl for the sine fnction cn be fond in ery similr wy. Here is n exmple tht is less tricky, bt reqires seerl steps of integrtion by prts ( nd re different fnctions in ech step). x 3 e x dx = x 3 e x = x 3 ex 3x2 = x 3 ex = x 3 ex ex 3x 2 dx e x 2 3x2 ex 2 3x2 ex 2 + + 6x + 6xex 3 ex 2 6x e x 3 6ex 4 6 + c ex 3 3
Prtil frctions Integrtion by prts does not lwys help. The integrl of rtionl fnction, pn (x) p n (x) dx, where p n (x) nd p n (x) re polynomils of degree n nd n, cn be compted sing prtil frctions. It is slly firly esy to pt n expression on common denomintor, for exmple, 3x + x 2 + 4 + x 2 x + 2 = (3x + )(x 2)(x + 2) + (x2 + 4)(x + 2) (x 2 + 4)(x 2) (x 2 + 4)(x 2)(x + 2) = 3x3 + x 2 2x + 2 = q(x). (3) x 4 6 Prtil frctions is the rt of doing the sme thing bckwrds. The procedre consists of three steps, here illstrted for the frction q(x) in (3).. Find the fctors in the denomintor (lmost the sme s finding the roots): (x 4 6) = (x 2 + 4)(x 2 4) = (x 2 + 4)(x + 2)(x 2) For polynomil with rel coefficients, ll fctors will be first or second order polynomils. 2. Write down the form of the prtil frctions. Ech second order denomintor gets first order nmertor nd ech first order denomintor gets constnt nmertor. Ax + B x 2 + 4 + C x + 2 + where A, B, C, nd D re nknown constnts. D x 2, 3. Determine the constnts so tht the sm of prtil frctions eqls the originl frction q(x). This cn be done in the following wy. Pt the frctions with constnts bck on common denomintor nd then set the old nd new nmertors to be eql: (Ax+B)(x 2 4)+C(x 2 +4)(x2)+D(x 2 +4)(x+2) = 3x 3 +x 2 2x+2. Choose nmber of good x-les to get conditions for the constnts. x = 2 32C = 32 C = x = 2 32D = 32 D = x = 0 4B 8C + 8D = 2 B = x = 3A 3B C + D = 8 A = 3 4
Now, when the originl expression is decomposed into prtil frctions it is possible to perform the integrtion. In or cse, we get 3x + q(x)dx = x 2 + 4 + x 2 x + 2 dx = 3 2x 2 x 2 + 4 dx + /2 2 (x/2) 2 + dx + x 2 dx x + 2 dx = 3 2 ln(x2 + 4) + 2 rctn x 2 + ln x 2 ln x + 2 + c Note : If the nmertor in the rtionl fnction is of higher degree thn the denomintor, polynomil diision shold be crried ot first. For exmple: x + 2x 4 + 3x 3 x 2 + x + x 3 = x 2 + 2x + 3 + 3x + 4 x 3 Note 2: A fctor tht is repeted more thn once in the denomintor gies specil terms: A fctor (x 3) n A (x 3) + A 2 (x 3) 2 + + A n (x 3) n. A fctor (x 2 + 4) n A x + B (x 2 + 4) + A 2x + B 2 (x 2 + 4) 2 + + A nx + B n (x 2 + 4) n.