Advanced Integration Techniques: Integration by Parts We may differentiate the product of two functions by using the product rule:

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1 Avance Inegraion Techniques: Inegraion by Pars We may iffereniae he prouc of wo funcions by using he prouc rule: x f(x)g(x) = f (x)g(x) + f(x)g (x). Unforunaely, fining an anierivaive of a prouc is no so sraighforwar. However, we will on occasion be able o use a echnique known as inegraion by pars when aemping o inegrae cerain ypes of proucs. Le s look a a paricular example: se f(x) = x an g(x) = sin x. Then x sin x = sin x + x cos x. x We coul use his o our avanage if we wishe o inegrae he funcion x cos x: since we know ha x cos x = x sin x sin x, x x cos xx = x sin x sin xx = x sin x + cos x. We can make he argumens above more general: since we know ha (f (x)g(x) + f(x)g (x))x = x f(x)g(x) = f (x)g(x) + f(x)g (x), f (x)g(x)x + f(x)g (x)x = f(x)g(x). So f (x)g(x)x + f(x)g (x)x = f(x)g(x) f(x)g (x)x = f(x)g(x) f (x)g(x)x. () In oher wors, he inegral of he prouc of funcions f(x) an g (x) can be rewrien as he sum of f(x)g(x) an he inegral of f (x)g(x). A firs glance his may no seem paricularly useful, bu i ofen occurs (as i i in he example above) ha f (x)g(x)x is much simpler o calculae han he original f(x)g (x)x. This observaion is ofen wrien in he more convenien form uv = uv vu. (2) The formula above correspons o he formula in () when u = f(x) an v = g (x)x; hen u = f (x)x (by iffereniaion) an v = g(x) (by inegraion). In paricular, since we will nee o evaluae v, we mus choose v so ha i is a funcion whose inegral we know.

2 Examples: Fin x cos xx. Unforunaely, we o no know an anierivaive of he funcion x cos x, nor woul rewriing he funcion be paricularly helpful. We migh be empe o ry inegraing by u-subsiuion, bu i shoul quickly become clear ha here are no goo choices for subsiuions; if we ry seing u = cos x, hen u = sin xx, which is singularly unhelpful; nor woul he subsiuion u = x cos x be useful. So inegraion by pars is our only available opion: recall ha he formula is uv = uv vu. We nee o make a choice for u an a choice for v. Le s ry seing Then we have u = x an v = cos xx. u = x u = x (by iffereniaion) v = cos xx v = sin x (by inegraion). Then formula (2) ells us ha x cos xx = x sin x sin xx. Forunaely for us, sin xx is easy o evaluae, sin xx = cos x + C. So x cos xx = x sin x sin xx = x sin x + cos x + C. To be cerain ha our answer is correc, we shoul iffereniae he resul: which confirms our calculaion. (x sin x + cos x + C) = sin x + x cos x sin x x = x cos x, Noice wha woul have happene if we ha swiche our choices for u an v; we woul have ha u = cos x u = sin xx (by iffereniaion) v = xx v = 2 x2 (by inegraion). 2

3 Then we woul have x cos xx = 2 x2 cos x 2 x 2 sin xx. Now 2 x 2 sin xx is more complicae han was he original inegral; so inegraion by pars i no simplify our problem bu acually mae i harer! In such a case, we shoul reurn o he original problem an ry anoher approach. In general, when using inegraion by pars, we shoul aemp o choose u an v so ha he resulan inegral vu is simpler han was he original inegral. Fin ln y y y. Le s concenrae on fining he inefinie inegral ln y y y before we evaluae he efinie inegral. We will nee o use inegraion by pars, an since we know how o inegrae y y, bu no ln y, le s se u = ln y v = y y u = y y (by iffereniaion) v = 2y 2 (by inegraion). Then ln y y = 2y y 2 2 ln y 2 y y y = 2y 2 ln y 2 y 2 y. Then since y 2 y = 2y 2, we have ln y y y = 2y 2 ln y y 2 + C. Reurning o he efinie inegral, we have ln y y = 2y 2 ln y y 2 y = 6 ln 2 ( ln 8) = 6 ln 2 ln 2 2 = 2 ln 8 ln 2. Fin ln xx.

4 Once again, we o no know an anierivaive of ln x, nor woul rewriing he funcion or using a u-subsiuion help, so inegraion by pars is our only oher choice. Since we know how o iffereniae ln x, ry u = ln x u = x (by iffereniaion) x v = x v = x (by inegraion). Then ln xx = x ln x x x = x ln x x Again, we may check our answer by iffereniaion: x = x ln x x + C. x (x ln x x + C) = ln x + x x = ln x. Exercise: Show ha x 2 e x = e x (x 2 2x + 2). Fin e 2θ sin(θ)θ. Again, i quickly becomes clear ha our only possible opion for inegraing he funcion is inegraion by pars. We woul probably like o break up he funcion as e 2θ an sin(θ), bu which of he wo shoul be u an which shoul be v? Separaely, boh of he componens are easy o inegrae, so here oes no seem o be an obvious way o make he choice. Le s ry u = e 2θ u = 2e 2θ θ (by iffereniaion) v = sin(θ)θ v = cos(θ) (by inegraion). Then e 2θ sin(θ)θ = e2θ cos(θ) ( 2 )e2θ cos(θ)θ = e2θ cos(θ) + 2 e 2θ cos(θ)θ. Now we nee o evaluae 2 e 2θ cos(θ)θ, which is no simpler han he inegral wih which we sare! However, i is no more complicae eiher (if i ha been more complicae, we woul probably wan o sar over an ry a ifferen approach). Le s ry evaluaing 2 e 2θ cos(θ)θ by pars: u = e 2θ u = 2e 2θ θ (by iffereniaion) v = cos(θ)θ v = sin(θ) (by inegraion).

5 Tha ells us ha 2 e 2θ cos(θ)θ = 2 ( e2θ sin(θ) 2 e2θ sin(θ)θ) = 2 e2θ sin(θ) e 2θ sin(θ)θ. We seem o be going in circles, since we have now reurne o e 2θ sin(θ)θ! To summarize, here is wha we have eermine: e 2θ sin(θ)θ = e2θ cos(θ) + 2 e 2θ cos(θ)θ = e2θ cos(θ) + 2 e2θ sin(θ) e 2θ sin(θ)θ. As we noe earlier, he unknown erm e 2θ sin(θ)θ shows up on boh sies. Wha happens if we a e 2θ sin(θ)θ o boh sies of he equaliy, in effec solving for he unknown? We en up wih e 2θ sin(θ)θ + e 2θ sin(θ)θ = e2θ cos(θ) + 2 e2θ sin(θ) e 2θ sin(θ)θ = e2θ cos(θ) + 2 e2θ sin(θ). e 2θ sin(θ)θ + e 2θ sin(θ)θ Recall ha we wane o eermine e 2θ sin(θ)θ; muliplying boh sies of he las equaion above by, we see ha e 2θ sin(θ)θ = e2θ cos(θ) + 2 e2θ sin(θ) + C. In general, a funcion ha is a prouc of wo of he funcions e kx, cos kx, an sin kx can be inegrae by using inegraion by pars wice, hen solving for he inegral. Some problems may require muliple echniques in orer o solve hem. Fin sec. Inegraion by pars seems naural here, an since we o no know wha sec is, le s se Then u = sec u = (by iffereniaion) 2 + v sec = 2 2 sec 2 v = = 2 2 (by inegraion). 2. Unforunaely, we on know immeiaely. However, we can use u-subsiuion o evaluae his inegral. Seing u = 2, we have 2u =. 2 So 5

6 2 2 = = u u u 2 u = 2 u 2 + C = C. Thus sec = 2 2 sec 2 2 = 2 2 sec C. Exercise: Evaluae e s+ s by saring wih a u-subsiuion. 6