Week #3 - Inegraion by Pars & Numerical Inegraion Secion 7. From Calculus, Single Variable by Hughes-Halle, Gleason, McCallum e. al. Copyrigh 5 by John Wiley & Sons, Inc. This maerial is used by permission of John Wiley & Sons, Inc. SUGGESTED PROBLEMS. sin d 3. Inegrae by pars, selecing: u = dv = sin d so du = d v = cos udv = uv vdu, sin d = cos ( cos ) d = cos + sin + C sin d Inegrae by pars, selecing: u = dv = sin d so du = d v = cos sin d = cos ( cos ) d = cos + cos d This new inegral is simpler (he power of wen from squared o linear), and if we inegrae by pars again, we should ge a solvable inegral: selecing: u = dv = cos d so du = d v = sin [ cos d = sin ] sin d = [ sin + cos ] + C Combining his wih he previous resul, he final inegral evaluaes o sin d = cos + sin + cos + C 9. x 3 ln x dx To inegrae by pars, we mus selec u = ln x, because we don know how o inegrae i:
selecing: u = ln x dv = x 3 dx so du = x dx x v = 7. (θ + )sin(θ + ) dθ x 3 ln x dx = x x ln x x x dx = ln x x 3 dx = x ln x 6 x + C Since everyhing seems o be in erms of (θ + ), i migh help o do a subsiuion firs: Le w = (θ +, so dw = dθ, and (θ + )sin(θ + ) dθ = w sin(w)dw This we can evaluae by inegraion by pars, selecing: u = w dv = sin w dw so du = dw v = cos w w sin(w)dw = w cos w ( cos w)dw = w cos w + sinw. + 7 5 d = (θ + )cos(θ + ) + sin(θ + ) + C Neiher subsiuion nor inegraion by pars looks o helpful a firs glance. When possible, i can help o separae sums in he numeraor o obain wo simpler inegrals: + 7 d = 5 d 5 }{{} I + 7 5 d } {{ } I The second inegral, I, can be done by subsiuion, or by guess and check: 7 d = 7 5 (5 ) / = (5 ) / + C The firs inegral, I, is more complicaed, bu wih wha we jus found, we should be able o use inegraion by pars: I = d = (5 ) / d 5
selecing: u = dv = (5 ) / d so du = d v = (5 ) / I = (5 ) / d = (5 ) / ( (5 ) / ) d = (5 ) / 3 (5 )3/ + C Combining I and I, and combining he consans ino C, we obain 5. xarcan x dx + 7 d = I + I = (5 ) / (5 )3/ (5 ) / +C 5 }{{ 3 }}{{} I I This looks like a classical subsiuion problem, given he x inside he arcan, and is derivaive, x ou fron. Le s sar wih ha subsiuion: Le w = x, so dw = x dx, so xarcan x dx = arcan w dw However, we only know he derivaive of arcan, no is inegral, so we need o use inegraion by pars, selecing: u = arcan w dv = dw so du = +w v = w arcan w dw = (w arcan w w + w dw) This new inegral can also be solved by subsiuion, since he derivaive of he denominaor will involve a w erm. Le z = + w, so dz = wdw: w + w dw = z dz = ln z = ln( + w ) 3
Combining all he pars, and hen subsiuing back ino he original x variable, xarcan x dx = (w arcan w w + w dw) = w arcan w ln( + w ) + C = x arcan x ln( + x ) + C 35. 5 ln( + ) d We only know how o differeniae ln, so rying inegraion by pars makes sense, selecing: u = ln( + ) dv = d so du = + d v = 5 ln( + ) d = ln( + ) + d The new inegral can be easily solved wih he subsiuion w = +, so = w and dw = d: w w + d = w dw = w dw w dw = w ln w + C = + ln( + ) + C so ln( + ) d = ln( + ) + d = ln( + ) ( + ln( + ) + C) Wih he limis, 5 = ( + )ln( + ) + C if we le C = C ln( + ) d = ( + )ln( + ) = (6ln 6 6) (ln() ) = 6ln 6 5 5.75 5 QUIZ PREPARATION PROBLEMS 38. For each of he following inegrals, indicae wheher inegraion by subsiuion or inegraion by pars is more appropriae. Do no evaluae he inegrals.
(a) (b) (c) (d) (e) (f) (g) xsin x dx x + x 3 dx xe x dx x cos(x 3 ) dx 3x + dx x sin x dx ln x dx (a) This inegral can be evaluaed using inegraion by pars wih u = x, dv = sin x dx. (b) We evaluae his inegral using he subsiuion w = + x 3. (c) We evaluae his inegral using he subsiuion w = x. (d) We evaluae his inegral using he subsiuion w = x 3. (e) We evaluae his inegral using he subsiuion w = 3x +. (f) This inegral can be evaluaed using inegraion by pars wih u = x,dv = sin x dx. (g) This inegral can be evaluaed using inegraion by pars wih u = ln x,dv = dx. 39. Find he exac value of he area under he firs arch of f(x) = xsin(x). You should have some sense of a skech of his graph: i will look like he sine graph, bu he ampliude will be growing wih linearly x because he sin(x) is muliplied by x. Because i is a produc of x and sin(x), he funcion will sill have y = values, crossing he x axis, a he same poins when sin(x) =. The firs arch of he curve is herefore beween x = and x = π, as hese are he firs wo poins when f(x) =. This means he area we re looking for will be represened by he inegral To evaluae his, we ll use inegraion by pars: selecing: u = x dv = sin x dx so du = dx v = cos x π π x sin(x) dx = x( cos x) π π cos(x) dx = x cos(x) + sin(x) π = [( π cos(π) + sin(π)] [( cos() + sin()] = π( ) = π xsin(x) dx. 5
7. x n cos ax dx = a xn sin(ax) n a x n sin(ax) dx We inegrae by pars. Le u = x n and dv = cos ax dx, so du = nx n dx and v = a sin ax. Then x n cos axdx = a xn sin ax (nx n )( sin ax) dx a = a xn sin ax n x n sin ax dx. a 5. Esimae f(x)g (x) dx if f(x) = x and g has he values in he following able. x 6 8 g(x).3 3.. 5.5 5.9 6. Since f (x) = x, inegraion by pars ells us ha f(x)g (x)dx = f(x)g(x) f (x)g(x)dx = f()g() f()g() xg(x)dx. We can use le and righ Riemann Sums wih x = o approximae xg(x)dx: Lef sum g() x + g() x + g() x + 6 g(6) x + 8 g(8) x = ((.3) + (3.) + (.) + 6(5.5) + 8(5.9)) = 5.6 Righ sum g() x + g() x + 6 g(6) x + 8 g(8) x + g() x = (3.) + (.) + 6(5.5) + 8(5.9) + (6.)) = 37.6. A good esimae for he inegral is he average of he lef and righ sums, so xg(x)dx Subsiuing values for f and g, we have 5.6 + 37.6 = 66.6. f(x)g (x)dx = f()g() f()g() xg(x)dx (6.) (.3) (66.6) = 76.8 77. 6