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REFRESHER 2: INTEGRATION TECHNIQUES MA26200, Gabriel Sosa June 23, 2014 - Monday 1. COMPOSITION WITH LINEAR FUNCTIONS Integration is the opposite of differentiation. R For example cos xdx = sin x + C, since (sin x + C)0 = cos x. Functions that are the derivative of a known function are easy to integrate. For example: Z Z 1 (sec x tan x)dx = sec x + C dx = ln |x| + C x Z Z 1 (sec2 x)dx = tan x + C dx = arctan x + C 1 + x2 We also possess the power rule for integrals Z xn+1 n x dx = + C whenever n 6= −1. n+1 Z Z 1 x−2 1 −3 So dx = x dx = + C = − + C. x3 −2 2x2 This also applies when we want to integrate: Z 2 Z Z 1 x5/2 x1/2 x +1 x2 3/2 −1/2 √ dx = + dx = x + x dx = + +C 5/2 1/2 x x1/2 x1/2 √ 2√ 5 = x +2 x+C 5 Sometimes you will have a known derivative composed with a linear function. In those cases integrate and divide by the coefficient of x in Z the linear function: cos (2x + 3) sin (2x + 3)dx = − +C 2 Z 1 ln |3−5x| dx = +C 3−5x −5 Refresher 2 MA26200 (Gabriel Sosa), Summer ’14 Page 1 of 12 2. THE SUBSTITUTION METHOD Question: When could substitution work? Answer: When the function to be integrated is one of the following: 1. A division. In this case try the denominator as the substitution. For example: Z Z cos x 1 u = sin x − 3 • dx = cos xdx. So du = cos xdx sin x − 3 sin x − 3 Z Z 1 cos x dx = du = ln |u| + C = ln | sin x − 3| + C sin x − 3 u Z Z 1 e2x u = 1 + ex 2x . Notice ex = u−1 dx = e dx. So • x x x du = e dx 1+e 1+e Z Z Z Z e2x 1 1 1 x x dx = e e dx = (u − 1)du = (1 − )du 1 + ex 1 + ex u u Z e2x dx = u − ln u + C = 1 + ex − ln (1 + ex ) + C x 1+e Z e2x dx = ex − ln (1 + ex ) + C. The constant absorbed the 1. 1 + ex In rare occasions, writing the division as a product is better and we apply: 2. A product. If we have the product of two functions where one is the derivative of the other. For example: Z Z ln x 1 1 u = ln x • dx = ln x dx. The derivative of ln x is , so du = x1 dx x x x Z Z Z ln x 1 u2 (ln x)2 dx = ln x dx = udu = +C = +C x x 2 2 R • sec2 x tan xdx. The derivative of tan x is sec2 x, so u = tan x 2 and Z du = sec xdx Z Z u2 2 2 sec x tan xdx = tan x sec xdx = udu = +C 2 Z tan2 x 2 So sec x tan xdx = +C 2 Refresher 2 MA26200 (Gabriel Sosa), Summer ’14 Page 2 of 12 3. A composition When there is one composition present. In this case, try the inner function being composed as the substitution. For example: u = ex • e sin e dx. Since sin e is a composition, we take du = ex dx R R R x e sin ex dx = (sin ex )ex dx = (sin u)du = − cos u + C R Hence ex sin ex dx = − cos ex + C R Z • x ecos x 2 x x cos x sin xdx. Since ecos 2 x is a composition, we consider du u = cos2 x and du = 2 cos x(− sin x)dx, or, = cos x sin xdx −2 Z Z −1 u 1 2 du 2 = e + C = − ecos x + C. ecos x cos x sin xdx = eu −2 2 2 Z u = sin x • sin3 x cos xdx. Since sin3 x is a composition, du = cos xdx Z Z 4 u4 sin x sin3 x cos xdx = u3 du = +C = +C 4 4 Z p √ u = x2 + 1 . • x3 x2 + 1dx. Since x2 + 1 is a composition, du = 2xdx du So we have u = x2 + 1, = xdx and x2 = u − 1. Z Z 2p Z p √ du 3 2 So x x2 + 1dx = x2 + 1x xdx = u(u − 1) 2 Z Z p 1 1 u5/2 u3/2 x3 x2 + 1dx = − u3/2 − u1/2 du = +C 2 2 5/2 3/2 Z p 1 2 1 1 2 3 5/2 3/2 x x2 + 1dx = u − u + C = u5/2 − u3/2 + C 2 5 3 5 3 Z p 3 1p 2 1p 2 x3 x2 + 1dx = (x + 1)5 − (x + 1) + C. 5 3 Refresher 2 MA26200 (Gabriel Sosa), Summer ’14 Page 3 of 12 3. INTEGRATION BY PARTS Question: When do we apply integration by parts? Answer: When we have a product of BASIC function whose derivatives are unrelated. (By BASIC I mean no compositions are present). Question: How do we apply it? Answer: Using the rule: You destroy vines, you vineman, less vines destroy you Z Z udv = uv − vdu along with the method to identify u-Log Alg Trig Exp Examples R (a) x3 ln xdx, by u-LATE, we get u = ln x and dv = x3 dx. Z 4 Z 4 4 x 1 x u = ln x v = x4 dx. , and x3 ln xdx = ln x − So 1 3 du = x dx dv = x dx 4 4x Z Z x4 ln x 1 x4 ln x 1 x4 x4 ln x x4 3 3 x ln xdx = x dx = − − +C = − +C 4 4 4 44 4 16 (b) R x sin xdx, by u-LATE, we get u = x and dv = sin xdx. R R u=x v = − cos x So , and x sin xdx = x(− cos x)− (− cos x)dx du = dx dv = sin xdcx R R x sin xdx = −x cos x + cos xdx = −x cos x + sin x + C. Refresher 2 MA26200 (Gabriel Sosa), Summer ’14 Page 4 of 12 Sometimes we need to apply the method more than once R x2 ex dx, by u-LATE we get u = x2 and dv = ex dx. R 2 x R x u = x2 v = ex 2 x So , and x e dx = x e − e 2xdx du = 2xdx dv = ex dx R 2 x R x e dx = x2 ex − 2 xex dx, so we have to do it again. R xex dx, by u-LATE, we get u = x and dv = ex dx R x R x u=x v = ex x So , and xe dx = xe − e dx = xex − ex du = dx dv = ex dx R R x2 ex dx = x2 ex − 2 xex dx = x2 ex − 2 [xex − ex ] + C R So x2 ex dx = x2 ex − 2xex + 2ex + C And, sometimes, we need to apply the method twice and be clever R ex cos xdx, by u-LATE we get u = cos x and dv = ex dx u = cos x v = ex , and So du = − sin xdx dv = ex dx R x R R e cos xdx = cos xex − ex (− sin x)dx = ex cos x + ex sin xdx, so we have to do it again: R ex sin xdx, by u-LATE we get u = sin x and dv = ex dx R x R x u = sin x v = ex x So , e sin xdx = sin xe − e cos xdx du = cos xdx dv = ex dx R R Hence ex cos xdx = ex cos x + ex sin x− ex cos xdx R R So ex cos xdx + ex cos xdx = ex cos x + ex sin x R And 2 ex cos xdx = ex cos x + ex sin x, which means Z ex (cos x + sin x) x e cos xdx = +C 2 Refresher 2 MA26200 (Gabriel Sosa), Summer ’14 Page 5 of 12 4. TRIGONOMETRIC SUBSTITUTION Question: When do we apply trigonometric substitution? Answer: When an expression involving an addition or a subtraction of squares appears AND substitution does not work). Z Z p x 3 For example, the integrals x 1 − x2 dx and dx should be x2 + 4 solved by substitution. [Check at the end of this section for details]. Examples: (a) Z 1 dx x x2 + 4 √ In this case we have x2 +22 ; since we have an addition, we will create a right triangle with legs x and 2 We need to relate the angle θ to x: x tan θ = 2 So, 2 tan θ = x and On the other hand, we also need to √ relate the angle θ to x2 + 4: √ x2 + 4 sec θ = 2 √ 2 sec2 θdθ = dx So, 2 sec θ = x2 + 4. Z Z Z 1 sec θ 1 2 √ dx = 2 sec θdθ = dθ Hence 2 tan θ2 sec θ 2 tan θ x x2 + 4 Z Z 1 Z Z 1 1 1 1 1 cos θ √ dx = dθ = dθ = csc θdθ sin θ 2 2 sin θ 2 x x2 + 4 cosθ Z 1 1 √ dx = ln | csc θ − cot θ| + C 2 x x2 + 4 √ x2 + 4 2 Finally from the triangle we know that csc θ = , while cot θ = . x √ x √ Z 1 1 x2 + 4 2 1 x2 + 4 − 2 √ − + C = ln So dx = ln +C 2 x x 2 x x x2 + 4 Refresher 2 MA26200 (Gabriel Sosa), Summer ’14 Page 6 of 12 (b) Z p 9 − x2 dx In this case we have 32 −x2 ; since we have a subtraction, we will create a right triangle with hypotenuse 3 and leg x We need to relate the On the other hand, we angle θ to x: also need to √ relate the angle θ to 9 − x2 : x sin θ = √ 3 9 − x2 cos θ = So, 3 sin θ = x and 3 √ 3 cos θdθ = dx So, 3 cos θ = 9 − x2 . Z Z p Z Z 2 Hence 9 − x2 dx = 3 cos θ3 cos θdθ = 9 cos θdθ = 9 cos2 θdθ The key to integrating cos2 θ lies on one of the identities in the Trigonometric Identitites Refresher : cos 2θ + 1 . cos 2θ = 2 cos2 θ − 1 or its equivalent cos2 θ = 2 Z p Z 9 9 sin 2θ So 9 − x2 dx = (cos 2θ + 1)dθ = +θ +C 2 2 2 Now remember sin 2θ = 2 sin θ cos θ, so: Z p 2 sin θ cos θ 9 9θ 9 + θ + C = sin θ cos θ + +C 9 − x2 dx = 2 2 2 2 √ x 9 − x2 Finally from the triangle we know that sin θ = , while cos θ = . 3 3 √ Z p x x 9 − x2 9 9 2 So 9 − x dx = + arcsin +C 2 3 3 2 3 √ Z p x x 9 − x2 9 2 And we conclude 9 − x dx = + arcsin +C 2 2 3 Refresher 2 MA26200 (Gabriel Sosa), Summer ’14 Page 7 of 12 (c) Z √ 1 x2 − 1 In this case we have x2 −12 ; since we have a subtraction, we will create a right triangle with hypotenuse x and leg 1 dx We need to relate the On the other hand, we also need to angle θ to x: √ relate the angle θ to x2 − 1: csc θ = x √ cot θ = x2 − 1 − csc θ cot θdθ = dx Z 1 Z 1 cot θ)dθ = − (− csc θ cot θ √ Hence dx = x2 − 1 Z 1 So √ dx = − ln |csc θ − cot θ| + C x2 − 1 Z csc θdθ Finally from the triangle we know that csc θ = x, while cot θ = Z p 1 2 √ dx = − ln x − x − 1 + C And we conclude x2 − 1 p x2 − 1. As for the integrals discussed at the beginning of the section R 3√ x 1 − x2 dx. This example shows that before trying trigonometric substitution one needs to check that substitution does not work. √ u = 1 − x2 Since 1 − x2 is a composition, we should try , so we du = −2xdx du have u = 1 − x2 , = xdx and x2 = 1 − u. −2 Z Z Z p p √ du x3 1 − x2 dx = x2 1 − x2 xdx = (1 − u) u −2 Z Z 3/2 5/2 p 1 1 u u x3 1 − x2 dx = − u1/2 − u3/2 du = − − +C 2 2 3/2 5/2 1 1 1 2 3/2 2 5/2 − u − u + C = − (1 − x2 )3/2 + (1 − x2 )5/2 + C 2 3 5 3 5 Refresher 2 MA26200 (Gabriel Sosa), Summer ’14 Page 8 of 12 Z x dx, this is a division, so we should first try the substitution x2 + 4 du u = x2 + 4, du = 2xdx, or equivalently = xdx. 2 Z Z Z Z 1 1 du 1 1 x 1 dx = xdx = = du = ln |u| + C x2 + 4 x2 + 4 u 2 2 u 2 Z p x 1 2 dx = ln |x + 4| + C = ln x2 + 4 + C x2 + 4 2 Notice that |x2 + 4| = x2 + 4, because x2 + 4 is always positive. Remark: You should have already memorized that Z x 1 1 dx = arctan +C x2 + a2 a a Refresher 2 MA26200 (Gabriel Sosa), Summer ’14 Page 9 of 12 5. PARTIAL FRACTIONS Question: When do we use partial fractions? Answer: When we have a division of polynomials, the degree of the numerator is LESS than the degree Z of the denominators AND substitux−2 tion does not work. For example dx should be integrated x2 − 4x − 5 using substitution, not partial fractions. Question: How do we find the partial fraction decomposition of a rational expression? Answer: Let’s look at the following two examples x−9 (a) 2 . The expression provided by: (x + 4)(x + 1) Irreducible quadratic factors x2 + 4 are of the form Ax + B . x2 + 4 C x+1 Ax + B C x−9 What we want then is 2 + = 2 x +4 x + 1 (x + 4)(x + 1) So (Ax + B)(x + 1) + C(x2 + 4) = x − 9. There are two ways of finding the values for A, B and C. Linear factors x + 1 are of the form Solution 1: If x = −1 If x = 0 If x = 1 then 5C = −10, then B + 4C = −9, (A + B)2 + 5C = −8, and C = −2 B−8 = −9, so B = −1 2(A − 1) − 10 = −8, 2A − 12 = −8, A = 2 Solution 2: (Ax + B)(x + 1) + C(x2 + 4) = x − 9 Ax2 + Ax + Bx + B + Cx2 + 4C = x − 9 (A + C)x2 + (A + B)x + (B + 4C) = x − 9 So A + C = 0, A + B = 1 and B + 4C = −9 Then C = −A, B = 1 − A and 1 − A − 4A = −9 So 5A = 10, and A = 2, B = −1 and C = −2 Refresher 2 MA26200 (Gabriel Sosa), Summer ’14 Page 10 of 12 2x − 1 2 x−9 = − , (x2 + 4)(x + 1) x2 + 4 x + 1 Z Z 2 x−9 2x − 1 and then dx = − dx 2 2 (x + 4)(x + 1) x +4 x+1 Z Z Z 2x 1 2 = dx − dx − dx x2 + 4 x2 + 4 x+1 Z x 1 x−9 2 dx = ln (x + 4) − arctan − 2 ln |x + 1| + C So (x2 + 4)(x + 1) 2 2 All these integrals should be known. Hence x2 − 2x + 3 (b) . The expression provided by: (x − 1)2 (x − 2) A Linear factors x − 2 are of the form x−2 Powers of linear factors (x − 1)2 are of the form B C + x − 1 (x − 1)2 A B C x2 − 2x + 3 So + + = and x − 2 x − 1 (x − 1)2 (x − 1)2 (x − 2) A(x − 1)2 + B(x − 2)(x − 1) + C(x − 2) = x2 − 2x + 3 Once away there are two ways to find the values of A, B and C: Solution 1: If x = 1, If x = 2, If x = 0, then then −C = 2, then A = 3 A + 2B − 2C = 3 so C = −2 so 3 + 2B + 4 = 3, and B = −2 Solution 2: A(x2 − 2x + 1) + B(x2 − 3x + 2) + Cx − 2C = x2 − 2x + 3 Ax2 − 2Ax + A + Bx2 − 3Bx + 2B + Cx − 2C = x2 − 2x + 3 (A + B)x2 + (C − 2A − 3B)x + (A + 2B − 2C) = x2 − 2x + 3 So A + B = 1, C − 2A − 3B = −2 and A + 2B − 2C = 3 Then B = 1−A, C −2A−3(1−A) = −2 or C −2A−3+3A = −2, which shows C = 1 − A. then B = C. We conclude A = 3 and B = C = −2 Refresher 2 MA26200 (Gabriel Sosa), Summer ’14 Page 11 of 12 x2 − 2x + 3 So the partial fractions representation for is (x − 1)2 (x − 2) 2 2 3 − − . x − 2 x − 1 (x − 1)2 Z Z x2 − 2x + 3 3 2 2 This implies that dx = − − dx (x − 1)2 (x − 2) x − 2 x − 1 (x − 1)2 Z Z Z 1 1 =3 dx − 2 dx − 2 (x − 1)−2 dx x−2 x−1 |x − 2|3 2 = 3 ln |x − 2| − 2 ln |x − 1| + 2(x − 1)−1 + C = ln + +C |x − 1|2 x − 1 One last persistent note: ALWAYS TRY SUBSTITUTION FIRST when integrating divisions and/or compositions. If that does not work you can try TRIGONOMETRIC SUBSTITUTION or PARTIAL FRACTIONS. INTEGRATION BY PARTS should only be applied to PRODUCTS of basic functions. To Z stress my point: x−2 dx should be solved by substituting u = x + 3. I invite you to (x + 3)2 try this byZ substitution and by partial fractions so you can see the difference. p While x 4 − x2 dx should be solved by substituting u = 4 − x2 . Try it by substitution and then by trigonometric substitution so you can tell the difference And remember, all the material in this refresher covers things you were supposed to have learned already. Hopefully this taught you a lot about how cumulative knowledge works. Refresher 2 MA26200 (Gabriel Sosa), Summer ’14 Page 12 of 12