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# Lesson 27: Integration by Substitution (worksheet solutions)

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Several problems using integration by substitution, with solutions

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### Lesson 27: Integration by Substitution (worksheet solutions)

1. 1. Solutions to Worksheet for Section 5.5 Integration by Substitution V63.0121, Calculus I April 27, 2009 Find the following integrals. In the case of an indeﬁnite integral, your answer should be the most general antiderivative. In the case of a deﬁnite integral, your answer should be a number. In these problems, a substitution is given. (3x − 5)17 dx, u = 3x − 5 1. 1 Solution. As suggested, we let u = 3x − 5. Then du = 3 dx so dx = du. Thus 3 1 1 1 18 1 (3x − 5)17 dx = u17 du = (3x − 5)18 + C · u +C = 3 3 18 54 4 x2 + 9 dx, u = x2 + 9 2. x 0 Solution. We have du = 2x dx, which takes care of the x and dx that appear in the integrand. The limits are changed to u(0) = 9 and u(4) = 25. Thus 25 4 1 25 √ 12 u du = · u3/2 x x2 + 9 dx = 29 23 0 9 1 98 = · (125 − 27) = . 3 3 √ ex √ √ dx, u = x. 3. x 1
2. 2. 1 Solution. We have du = √ dx, which means that 2x √ ex eu du = 2eu + C √ dx = 2 x cos 3x dx 4. , u = 5 + 2 sin 3x 5 + 2 sin 3x Solution. We have du = 6 cos 3x dx, so cos 3x dx 1 du 1 1 = ln |u| + C = ln |5 + 2 sin 3x| + C = 5 + 2 sin 3x 6 u 6 6 In these problems, you need to determine the substitution yourself. (4 − 3x)7 dx. 5. 1 Solution. Let u = 4 − 3x, so du = −3x dx and dx = − du. Thus 3 u8 1 1 (4 − 3x)7 dx = − u7 du = − + C = − (4 − 3x)8 + C 3 24 24 π/3 csc2 (5x) dx 6. π/4 1 Solution. Let u = 5x, so du = 5 dx and dx = du. So 5 π/3 5π/3 1 csc2 (5x) dx = csc2 u du 5 π/4 5π/4 5π/3 1 =− cot u 5 5π/4 1 5π 5π − cot = cot 5 4 3 √ 1 3 = 1+ 5 3 2
3. 3. 3 −1 x2 e3x 7. dx 1 Solution. Let u = 3x3 − 1, so du = 9x2 dx and dx = du. So 9 1 1 3 −1 x2 e3x eu du = eu + C dx = 9 9 1 3x3 −1 =e . 9 Sometimes there is more than one way to skin a cat: x 8. Find dx, both by long division and by substituting u = 1 + x. 1+x Solution. Long division yields x 1 =1− 1+x 1+x So x dx dx = x − 1+x 1+x To ﬁnd the leftover integral, let u = 1 + x. Then du = dx and so dx du = ln |u| + C = 1+x u Therefore x dx = x − ln |x + 1| + C 1+x Making the substitution immediately gives du = dx and x = u − 1. So u−1 x 1 1− dx = du = du 1+x u u = u − ln |u| + C = x + 1 − ln |x + 1| + C It may appear that the two solutions are “diﬀerent.” But the diﬀerence is a constant, and we know that antiderivatives are only unique up to addition of a constant. 2z dz , both by substituting u = z 2 + 1 and u = 3 √ z 2 + 1. 9. Find 3 z2 + 1 3
4. 4. Solution. In the ﬁrst substitution, du = 2z dz and the integral becomes du 3 2/3 3 u−1/3 du = u + C = (z 2 + 1)2/3 √= 2 2 u 3 In the second, u3 = z 2 + 1 and 3u2 du = 2z dz. The integral becomes 3u2 33 3 3u2 du = u + C = (z 2 + 1)2/3 + C. du = u 2 2 The second one is a dirtier substitution, but the integration is cleaner. Use the trigonometric identity cos 2α = cos2 α − sin2 α = 2 cos2 α − 1 = 1 − 2 sin2 α to ﬁnd sin2 x dx 10. Solution. Using cos 2α = 1 − 2 sin2 α, we get 1 − cos 2α sin2 α = 2 So 11 sin2 x dx = − cos 2x dx 22 x1 = − sin 2x + C. 24 cos2 x dx 11. Solution. Using cos 2α = 2 cos2 α − 1, we get 1 + cos 2α sin2 α = 2 So 11 sin2 x dx = + cos 2x dx 22 x1 = + sin 2x + C. 24 4
5. 5. 12. Find sec x dx by multiplying the numerator and denominator by sec x + tan x. Solution. We have sec x(sec x + tan x) sec x dx = dx sec x + tan x sec2 x + sec x tan x = dx tan x + sec x Now notice the numerator is the derivative of the denominator. So the substitution u = tan x+sec x gives sec x dx = ln |sec x + tan x| + C. 5