Substitution and Integration by-parts Support Workshop A 04-03-2014 In this session we look at two of the main tools for solving integrals. The substitution rule, which is almost analogous to the chain-rule and integration by-parts, which is almost analogous to the product rule in differentiation. Substitution If u(x) is differentiable and f(x) continuous then the indefinite integral Z du(x) Z f(u(x)) dx = f(u)du dx while the definite integral Z b du(x) Z u(b) f(u(x)) dx = f(u)du a dx u(a) Example: Z 2 Z du(x) ex 2xdx = eu(x) dx; dx where u(x) = x2: Thus Z 2 Z 2 ex 2xdx = eudu = eu = ex : Integration-by-parts: Given two differentiable functions u(x) and v(x), Z Z u(x)v0(x)dx = u(x)v(x) − v(x)u0(x)dx: (1) Example Let u(x) = x and v(x) = ex thus Z Z xexdx = xex − exdx = ex(x − 1): Sometimes we need to repeat, for instance with u(x) = x2 and v(x) = ex Z Z x2exdx = x2ex − 2 xexdx = x2ex + 2ex(1 − x): Repeated use of integration by parts kills-off polynomials. Repeated use of integration by parts is also used with cyclic functions such as cos(x) and sin(x). As an example where u(x) = cos(x) and v(x) = ex we have Z Z cos(x)exdx = cos(x)ex + sin(x)exdx : (2) | {z } I To solve I, apply by-parts Z Z sin(x)exdx = sin(x)ex − cos(x)exdx: 1 Using this in (2) we have Z Z cos(x)exdx = (cos(x) + sin(x))ex − cos(x)exdx: Rearranging Z 1 cos(x)exdx = (cos(x) + sin(x))ex + C: 2 1. Substitution. Solve the following indefinite integrals using substitution (a) R sin(10x)dx Z 1 (b) dx 1 − x Z x (c) dx 1 − x Z x3 (d) p dx x2 + 3 (e) R sin(x) cos(x)dx (f) R − cos2(x) sin(x)dx (g) R sin3(x)dx: First try and rearrange until a cos(x) and a sin(x) appear. Hint: cos2(x) + sin2(x) = 1: Z ln(x) (h) dx x (i) R esin(x) cos(x)dx Solution: Z Z du cos(u) cos(10x) (a) u(x) = 10x thus sin(10x)dx = sin(u) = − + C = − + C: 10 10 10 Z 1 Z 1 (b) u(x) = 1 − x thus dx = − du = − ln(juj) + C = − ln(j1 − xj) + C: 1 − x u Z x Z 1 − u Z 1 (c) u(x) = 1−x thus dx = − du = 1− du = u−ln(juj)+C = (1−x)−ln(j1− 1 − x u u xj) + C: (d) u(x) = x2 + 3 thus Z x2 Z u − 3 du p xdx = p x2 + 3 u 2 Z du = u1=2 − 3u−1=2 = 2=6u3=2 − 3u1=2 + C 2 = 2=6(x2 + 3)3=2 − 3(x2 + 3)1=2 + C (e) u(x) = sin(x) thus R sin(x) cos(x)dx = R udu = u2=2 + C = sin(x)2=2 + C: (f) u(x) = cos(x) thus R − cos2(x) sin(x)dx = R u2du = u3=3 + C = cos3(x)=3 + C: (g) First, note that sin3(x) = sin(x)(1−cos2(x)): Now substitute u = cos(x), we have R sin3(x)dx = R sin(x)(1 − cos2(x))dx = R −(1 − u2)du = −u + u3=3 + C = − cos(x) + cos3(x)=3 + C: Z ln(x) Z Z (h) u(x) = ln(x) thus dx = u(x)u0(x)dx = udu = u2=2 + C = ln(x)2=2 + C: x (i) u(x) = sin(x) thus R esin(x) cos(x)dx = R eu(x)u0(x)dx = R euu = eu + C = esin(x) + C: Page 2 2. Substitution. Solve the following definite integrals using substitution R 1 10x (a) 0 e dx Z 5 p (b) x2 x − 1dx 1 p R a 2 2 (c) 0 x a − x dx π Z 4 sin(x) (d) dx: Did you really have to calculate this? π cos(x) − 4 Z π jxj cos(e ) 311 (e) EXTRA, it's a trick! Try and \guess" the solution based on the last question 2 x dx −π tan(x) Solution: Z 1 Z 10 10x 1 u 0 u 10 10 (a) u(x) = 10x thus e dx = e u du = e j0 = e − 1: 0 10 0 Z 5 p Z 4 4 2 2 1=2 3=2 5=2 7=2 (b) u(x) = x − 1 thus x x − 1dx = (1 + u) u du = 2=3u + 4=5u + 2=7u 1 0 0 Wolfram alpha Failed to solve this!! p p 2 2 R a 2 2 R 0 −1 1 3=2 0 3 (c) u(x) = a − x thus 0 x a − x dx = a2 u 2 du = − 3 u ja2 = a =3: π p1 Z 4 sin(x) Z 2 1 p1 (d) u(x) = cos(x) thus dx = − du = − ln(juj)j 2 = 0! Think about this, is − p1 − π cos(x) − p1 u 2 4 2 tan(x) and odd or even function? jxj cos(e ) 311 (e) The function tan(x)2 is even, while x is odd, thus multiplied together form an odd functions. Integral of odd functions are odd. Use the fundamental theorem of calculus to show that the integral in zero (or ask a tutor). 3. Integration by parts. Solve the following integrals using integration by parts (a) R x sin(x)dx = (b) R e2xe5xdx = (c) R cos(x) sin(x)dx = (d) R x2 ln(x)dx = (e) Try u(x) = ln(x) and v(x) = 1 in R ln(x)dx = (f) Try u(x) = (ln(x))2 and v(x) = 1 in R (ln(x))2dx = R n (g) x log10(x)dx =, for any n 2 N? Solution: R 1 x (a) x sin(x)dx = 2 (cos(x) + sin(x))e + C: R 2x 5x R 7x 1 7x (b) e e dx = e dx = 7 e + C: If you used by-parts for this, I tricked you! R 2 R R 1 2 (c) cos(x) sin(x)dx = − cos(x) − cos(x) sin(x)dx; thus rearranging cos(x) sin(x)dx = − 2 cos(x) + C: Z 1 Z 1 (d) x2 ln(x)dx = x3 ln(x) − x3x−1dx = 19x3(3 ln(x) − 1) + C: 3 3 Page 3 R R x (e) Try u(x) = ln(x) and v(x) = 1 in ln(x)dx = ln(x)x − x dx = x(ln(x) − 1) + C: 2 R 2 2 R x 2 (f) Try u(x) = (ln(x)) and v(x) = 1 in (ln(x)) dx = (ln(x)) x − 2 ln(x) x dx = (ln(x)) x − 2x(ln(x) − 1) + C Z 1 Z 1 1 (g) xn log (x)dx = xn+1 log (x) − xn+1 = 10 n 10 n + 1 x ln(10) 1 1 xn+1 (n + 1) log (x) − (n + 1)2 10 ln(10) Page 4.