1.
Section 5.5 Integration by Substitution V63.0121.041, Calculus I New York University December 13, 2010Announcements ”Wednesday”, December 15: Review, Movie Monday, December 20, 12:00pm–1:50pm: Final Exam
2.
Announcements ”Wednesday”, December 15: Review, Movie Monday, December 20, 12:00pm–1:50pm: Final ExamV63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 2 / 37
3.
Resurrection Policy If your ﬁnal score beats your midterm score, we will add 10% to its weight, and subtract 10% from the midterm weight.Image credit: Scott Beale / Laughing Squid V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 3 / 37
4.
Objectives Given an integral and a substitution, transform the integral into an equivalent one using a substitution Evaluate indeﬁnite integrals using the method of substitution. Evaluate deﬁnite integrals using the method of substitution. V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 4 / 37
5.
OutlineLast Time: The Fundamental Theorem(s) of CalculusSubstitution for Indeﬁnite Integrals Theory ExamplesSubstitution for Deﬁnite Integrals Theory Examples V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 5 / 37
6.
Diﬀerentiation and Integration as reverse processesTheorem (The Fundamental Theorem of Calculus) 1. Let f be continuous on [a, b]. Then x d f (t) dt = f (x) dx a 2. Let f be continuous on [a, b] and f = F for some other function F . Then b f (x) dx = F (b) − F (a). a V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 6 / 37
7.
Techniques of antidiﬀerentiation?So far we know only a few rules for antidiﬀerentiation. Some are general,like [f (x) + g (x)] dx = f (x) dx + g (x) dx V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 7 / 37
8.
Techniques of antidiﬀerentiation?So far we know only a few rules for antidiﬀerentiation. Some are general,like [f (x) + g (x)] dx = f (x) dx + g (x) dxSome are pretty particular, like 1 √ dx = arcsec x + C . x x2 − 1 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 7 / 37
9.
Techniques of antidiﬀerentiation?So far we know only a few rules for antidiﬀerentiation. Some are general,like [f (x) + g (x)] dx = f (x) dx + g (x) dxSome are pretty particular, like 1 √ dx = arcsec x + C . x x2 − 1What are we supposed to do with that? V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 7 / 37
10.
No straightforward system of antidiﬀerentiationSo far we don’t have any way to ﬁnd 2x √ dx x2 + 1or tan x dx. V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 8 / 37
11.
No straightforward system of antidiﬀerentiationSo far we don’t have any way to ﬁnd 2x √ dx x2 + 1or tan x dx.Luckily, we can be smart and use the “anti” version of one of the mostimportant rules of diﬀerentiation: the chain rule. V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 8 / 37
12.
OutlineLast Time: The Fundamental Theorem(s) of CalculusSubstitution for Indeﬁnite Integrals Theory ExamplesSubstitution for Deﬁnite Integrals Theory Examples V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 9 / 37
13.
Substitution for Indeﬁnite IntegralsExampleFind x √ dx. x2 +1 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 10 / 37
14.
Substitution for Indeﬁnite IntegralsExampleFind x √ dx. x2 +1SolutionStare at this long enough and you notice the the integrand is thederivative of the expression 1 + x 2 . V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 10 / 37
15.
Say what?Solution (More slowly, now)Let g (x) = x 2 + 1. V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 11 / 37
16.
Say what?Solution (More slowly, now)Let g (x) = x 2 + 1. Then g (x) = 2x and so d 1 x g (x) = g (x) = √ dx 2 g (x) x2 +1 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 11 / 37
17.
Say what?Solution (More slowly, now)Let g (x) = x 2 + 1. Then g (x) = 2x and so d 1 x g (x) = g (x) = √ dx 2 g (x) x2 +1Thus x d √ dx = g (x) dx x2 + 1 dx = g (x) + C = 1 + x2 + C . V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 11 / 37
18.
Leibnizian notation FTWSolution (Same technique, new notation)Let u = x 2 + 1. V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 12 / 37
19.
Leibnizian notation FTWSolution (Same technique, new notation) √Let u = x 2 + 1. Then du = 2x dx and 1 + x2 = u. V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 12 / 37
20.
Leibnizian notation FTWSolution (Same technique, new notation) √Let u = x 2 + 1. Then du = 2x dx and 1 + x2 = u. So the integrandbecomes completely transformed into 1 √ x dx 2 du √ 1 √ du = = x2 + 1 u 2 u V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 12 / 37
21.
Leibnizian notation FTWSolution (Same technique, new notation) √Let u = x 2 + 1. Then du = 2x dx and 1 + x2 = u. So the integrandbecomes completely transformed into 1 √ x dx 2 du √ 1 √ du = = x2 + 1 u 2 u 1 −1/2 = 2u du V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 12 / 37
22.
Leibnizian notation FTWSolution (Same technique, new notation) √Let u = x 2 + 1. Then du = 2x dx and 1 + x2 = u. So the integrandbecomes completely transformed into 1 √ x dx 2 du √ 1 √ du = = x2 + 1 u 2 u 1 −1/2 = 2u du √ = u+C = 1 + x2 + C . V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 12 / 37
23.
Useful but unsavory variationSolution (Same technique, new notation, more idiot-proof) √Let u = x 2 + 1. Then du = 2x dx and 1 + x 2 = u. “Solve for dx:” du dx = 2x V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 13 / 37
24.
Useful but unsavory variationSolution (Same technique, new notation, more idiot-proof) √Let u = x 2 + 1. Then du = 2x dx and 1 + x 2 = u. “Solve for dx:” du dx = 2xSo the integrand becomes completely transformed into x x du √ dx = √ · x2 +1 u 2x V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 13 / 37
25.
Useful but unsavory variationSolution (Same technique, new notation, more idiot-proof) √Let u = x 2 + 1. Then du = 2x dx and 1 + x 2 = u. “Solve for dx:” du dx = 2xSo the integrand becomes completely transformed into x x du 1 √ dx = √ · = √ du x2 +1 u 2x 2 u V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 13 / 37
26.
Useful but unsavory variationSolution (Same technique, new notation, more idiot-proof) √Let u = x 2 + 1. Then du = 2x dx and 1 + x 2 = u. “Solve for dx:” du dx = 2xSo the integrand becomes completely transformed into x x du 1 √ dx = √ · = √ du x2 +1 u 2x 2 u 1 −1/2 = 2u du V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 13 / 37
27.
Useful but unsavory variationSolution (Same technique, new notation, more idiot-proof) √Let u = x 2 + 1. Then du = 2x dx and 1 + x 2 = u. “Solve for dx:” du dx = 2xSo the integrand becomes completely transformed into x x du 1 √ dx = √ · = √ du x2 +1 u 2x 2 u 1 −1/2 = 2u du √ = u+C = 1 + x2 + C . V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 13 / 37
28.
Useful but unsavory variationSolution (Same technique, new notation, more idiot-proof) √Let u = x 2 + 1. Then du = 2x dx and 1 + x 2 = u. “Solve for dx:” du dx = 2xSo the integrand becomes completely transformed into x x du 1 √ dx = √ · = √ du x2 +1 u 2x 2 u 1 −1/2 = 2u du √ = u+C = 1 + x2 + C .Mathematicians have serious issues with mixing the x and u like this.However, I can’t deny that it works. V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 13 / 37
29.
Theorem of the DayTheorem (The Substitution Rule)If u = g (x) is a diﬀerentiable function whose range is an interval I and fis continuous on I , then f (g (x))g (x) dx = f (u) duThat is, if F is an antiderivative for f , then f (g (x))g (x) dx = F (g (x))In Leibniz notation: du f (u) dx = f (u) du dx V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 14 / 37
30.
A polynomial exampleExampleUse the substitution u = x 2 + 3 to ﬁnd (x 2 + 3)3 4x dx. V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 15 / 37
31.
A polynomial exampleExampleUse the substitution u = x 2 + 3 to ﬁnd (x 2 + 3)3 4x dx.SolutionIf u = x 2 + 3, then du = 2x dx, and 4x dx = 2 du. So (x 2 + 3)3 4x dx V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 15 / 37
32.
A polynomial exampleExampleUse the substitution u = x 2 + 3 to ﬁnd (x 2 + 3)3 4x dx.SolutionIf u = x 2 + 3, then du = 2x dx, and 4x dx = 2 du. So (x 2 + 3)3 4x dx = u 3 2du = 2 u 3 du V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 15 / 37
33.
A polynomial exampleExampleUse the substitution u = x 2 + 3 to ﬁnd (x 2 + 3)3 4x dx.SolutionIf u = x 2 + 3, then du = 2x dx, and 4x dx = 2 du. So (x 2 + 3)3 4x dx = u 3 2du = 2 u 3 du 1 = u4 2 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 15 / 37
34.
A polynomial exampleExampleUse the substitution u = x 2 + 3 to ﬁnd (x 2 + 3)3 4x dx.SolutionIf u = x 2 + 3, then du = 2x dx, and 4x dx = 2 du. So (x 2 + 3)3 4x dx = u 3 2du = 2 u 3 du 1 1 = u 4 = (x 2 + 3)4 2 2 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 15 / 37
35.
A polynomial example, by brute forceCompare this to multiplying it out: (x 2 + 3)3 4x dx V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 16 / 37
36.
A polynomial example, by brute forceCompare this to multiplying it out: (x 2 + 3)3 4x dx = x 6 + 9x 4 + 27x 2 + 27 4x dx V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 16 / 37
37.
A polynomial example, by brute forceCompare this to multiplying it out: (x 2 + 3)3 4x dx = x 6 + 9x 4 + 27x 2 + 27 4x dx = 4x 7 + 36x 5 + 108x 3 + 108x dx V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 16 / 37
38.
A polynomial example, by brute forceCompare this to multiplying it out: (x 2 + 3)3 4x dx = x 6 + 9x 4 + 27x 2 + 27 4x dx = 4x 7 + 36x 5 + 108x 3 + 108x dx 1 = x 8 + 6x 6 + 27x 4 + 54x 2 2 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 16 / 37
39.
A polynomial example, by brute forceCompare this to multiplying it out: (x 2 + 3)3 4x dx = x 6 + 9x 4 + 27x 2 + 27 4x dx = 4x 7 + 36x 5 + 108x 3 + 108x dx 1 = x 8 + 6x 6 + 27x 4 + 54x 2 2Which would you rather do? V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 16 / 37
40.
A polynomial example, by brute forceCompare this to multiplying it out: (x 2 + 3)3 4x dx = x 6 + 9x 4 + 27x 2 + 27 4x dx = 4x 7 + 36x 5 + 108x 3 + 108x dx 1 = x 8 + 6x 6 + 27x 4 + 54x 2 2Which would you rather do? It’s a wash for low powers V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 16 / 37
41.
A polynomial example, by brute forceCompare this to multiplying it out: (x 2 + 3)3 4x dx = x 6 + 9x 4 + 27x 2 + 27 4x dx = 4x 7 + 36x 5 + 108x 3 + 108x dx 1 = x 8 + 6x 6 + 27x 4 + 54x 2 2Which would you rather do? It’s a wash for low powers But for higher powers, it’s much easier to do substitution. V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 16 / 37
42.
CompareWe have the substitution method, which, when multiplied out, gives 1 (x 2 + 3)3 4x dx = (x 2 + 3)4 2 1 8 = x + 12x 6 + 54x 4 + 108x 2 + 81 2 1 81 = x 8 + 6x 6 + 27x 4 + 54x 2 + 2 2and the brute force method 1 (x 2 + 3)3 4x dx = x 8 + 6x 6 + 27x 4 + 54x 2 2Is there a diﬀerence? Is this a problem? V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 17 / 37
43.
CompareWe have the substitution method, which, when multiplied out, gives 1 (x 2 + 3)3 4x dx = (x 2 + 3)4 + C 2 1 8 = x + 12x 6 + 54x 4 + 108x 2 + 81 + C 2 1 81 = x 8 + 6x 6 + 27x 4 + 54x 2 + +C 2 2and the brute force method 1 (x 2 + 3)3 4x dx = x 8 + 6x 6 + 27x 4 + 54x 2 + C 2Is there a diﬀerence? Is this a problem? No, that’s what +C means! V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 17 / 37
44.
A slick exampleExampleFind tan x dx. V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 18 / 37
45.
A slick exampleExample sin xFind tan x dx. (Hint: tan x = ) cos x V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 18 / 37
46.
A slick exampleExample sin xFind tan x dx. (Hint: tan x = ) cos xSolutionLet u = cos x . Then du = − sin x dx . So sin x tan x dx = dx cos x V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 18 / 37
47.
A slick exampleExample sin xFind tan x dx. (Hint: tan x = ) cos xSolutionLet u = cos x . Then du = − sin x dx . So sin x tan x dx = dx cos x V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 18 / 37
48.
A slick exampleExample sin xFind tan x dx. (Hint: tan x = ) cos xSolutionLet u = cos x . Then du = − sin x dx . So sin x tan x dx = dx cos x V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 18 / 37
49.
A slick exampleExample sin xFind tan x dx. (Hint: tan x = ) cos xSolutionLet u = cos x . Then du = − sin x dx . So sin x 1 tan x dx = dx = − du cos x u V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 18 / 37
50.
A slick exampleExample sin xFind tan x dx. (Hint: tan x = ) cos xSolutionLet u = cos x . Then du = − sin x dx . So sin x 1 tan x dx = dx = − du cos x u = − ln |u| + C V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 18 / 37
51.
A slick exampleExample sin xFind tan x dx. (Hint: tan x = ) cos xSolutionLet u = cos x . Then du = − sin x dx . So sin x 1 tan x dx = dx = − du cos x u = − ln |u| + C = − ln | cos x| + C = ln | sec x| + C V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 18 / 37
52.
Can you do it another way?Example sin xFind tan x dx. (Hint: tan x = ) cos x V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 19 / 37
53.
Can you do it another way?Example sin xFind tan x dx. (Hint: tan x = ) cos xSolution duLet u = sin x. Then du = cos x dx and so dx = . cos x V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 19 / 37
54.
Can you do it another way?Example sin xFind tan x dx. (Hint: tan x = ) cos xSolution duLet u = sin x. Then du = cos x dx and so dx = . cos x sin x u du tan x dx = dx = cos x cos x cos x u du u du u du = = = cos 2x 1 − sin2 x 1 − u2At this point, although it’s possible to proceed, we should probably backup and see if the other way works quicker (it does). V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 19 / 37
55.
For those who really must know allSolution (Continued, with algebra help)Let y = 1 − u 2 , so dy = −2u du. Then u du u dy tan x dx = 2 = 1−u y −2u 1 dy 1 1 =− = − ln |y | + C = − ln 1 − u 2 + C 2 y 2 2 1 1 = ln √ + C = ln +C 1 − u2 1 − sin2 x 1 = ln + C = ln |sec x| + C |cos x|There are other ways to do it, too. V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 20 / 37
56.
OutlineLast Time: The Fundamental Theorem(s) of CalculusSubstitution for Indeﬁnite Integrals Theory ExamplesSubstitution for Deﬁnite Integrals Theory Examples V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 21 / 37
57.
Substitution for Deﬁnite IntegralsTheorem (The Substitution Rule for Deﬁnite Integrals)If g is continuous and f is continuous on the range of u = g (x), then b g (b) f (g (x))g (x) dx = f (u) du. a g (a) V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 22 / 37
58.
Substitution for Deﬁnite IntegralsTheorem (The Substitution Rule for Deﬁnite Integrals)If g is continuous and f is continuous on the range of u = g (x), then b g (b) f (g (x))g (x) dx = f (u) du. a g (a)Why the change in the limits? The integral on the left happens in “x-land” The integral on the right happens in “u-land”, so the limits need to be u-values To get from x to u, apply g V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 22 / 37
59.
Example πCompute cos2 x sin x dx. 0V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 23 / 37
60.
Example πCompute cos2 x sin x dx. 0Solution (Slow Way)First compute the indeﬁnite integral cos2 x sin x dx and then evaluate.Let u = cos x . Then du = − sin x dx and cos2 x sin x dx V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 23 / 37
61.
Example πCompute cos2 x sin x dx. 0Solution (Slow Way)First compute the indeﬁnite integral cos2 x sin x dx and then evaluate.Let u = cos x . Then du = − sin x dx and cos2 x sin x dx V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 23 / 37
62.
Example πCompute cos2 x sin x dx. 0Solution (Slow Way)First compute the indeﬁnite integral cos2 x sin x dx and then evaluate.Let u = cos x . Then du = − sin x dx and cos2 x sin x dx V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 23 / 37
63.
Example πCompute cos2 x sin x dx. 0Solution (Slow Way)First compute the indeﬁnite integral cos2 x sin x dx and then evaluate.Let u = cos x . Then du = − sin x dx and cos2 x sin x dx = − u 2 du V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 23 / 37
64.
Example πCompute cos2 x sin x dx. 0Solution (Slow Way)First compute the indeﬁnite integral cos2 x sin x dx and then evaluate.Let u = cos x . Then du = − sin x dx and cos2 x sin x dx = − u 2 du = − 3 u 3 + C = − 1 cos3 x + C . 1 3 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 23 / 37
65.
Example πCompute cos2 x sin x dx. 0Solution (Slow Way)First compute the indeﬁnite integral cos2 x sin x dx and then evaluate.Let u = cos x . Then du = − sin x dx and cos2 x sin x dx = − u 2 du = − 3 u 3 + C = − 1 cos3 x + C . 1 3Therefore π π 1 cos2 x sin x dx = − cos3 x 0 3 0 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 23 / 37
66.
Example πCompute cos2 x sin x dx. 0Solution (Slow Way)First compute the indeﬁnite integral cos2 x sin x dx and then evaluate.Let u = cos x . Then du = − sin x dx and cos2 x sin x dx = − u 2 du = − 3 u 3 + C = − 1 cos3 x + C . 1 3Therefore π π 1 1 cos2 x sin x dx = − cos3 x =− (−1)3 − 13 0 3 0 3 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 23 / 37
67.
Example πCompute cos2 x sin x dx. 0Solution (Slow Way)First compute the indeﬁnite integral cos2 x sin x dx and then evaluate.Let u = cos x . Then du = − sin x dx and cos2 x sin x dx = − u 2 du = − 3 u 3 + C = − 1 cos3 x + C . 1 3Therefore π π 1 1 2 cos2 x sin x dx = − cos3 x =− (−1)3 − 13 = . 0 3 0 3 3 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 23 / 37
68.
Deﬁnite-ly QuickerSolution (Fast Way)Do both the substitution and the evaluation at the same time. Letu = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1 . So π cos2 x sin x dx 0 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 24 / 37
69.
Deﬁnite-ly QuickerSolution (Fast Way)Do both the substitution and the evaluation at the same time. Letu = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1 . So π cos2 x sin x dx 0 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 24 / 37
70.
Deﬁnite-ly QuickerSolution (Fast Way)Do both the substitution and the evaluation at the same time. Letu = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1 . So π cos2 x sin x dx 0 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 24 / 37
71.
Deﬁnite-ly QuickerSolution (Fast Way)Do both the substitution and the evaluation at the same time. Letu = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1 . So π −1 1 cos2 x sin x dx = −u 2 du = u 2 du 0 1 −1 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 24 / 37
72.
Deﬁnite-ly QuickerSolution (Fast Way)Do both the substitution and the evaluation at the same time. Letu = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1 . So π −1 1 cos2 x sin x dx = −u 2 du = u 2 du 0 1 −1 1 1 3 1 2 = u = 1 − (−1) = 3 −1 3 3 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 24 / 37
73.
Deﬁnite-ly QuickerSolution (Fast Way)Do both the substitution and the evaluation at the same time. Letu = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1 . So π −1 1 cos2 x sin x dx = −u 2 du = u 2 du 0 1 −1 1 1 3 1 2 = u = 1 − (−1) = 3 −1 3 3 The advantage to the “fast way” is that you completely transform the integral into something simpler and don’t have to go back to the original variable (x). V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 24 / 37
74.
Deﬁnite-ly QuickerSolution (Fast Way)Do both the substitution and the evaluation at the same time. Letu = cos x. Then du = − sin x dx, u(0) = 1 and u(π) = −1 . So π −1 1 cos2 x sin x dx = −u 2 du = u 2 du 0 1 −1 1 1 3 1 2 = u = 1 − (−1) = 3 −1 3 3 The advantage to the “fast way” is that you completely transform the integral into something simpler and don’t have to go back to the original variable (x). But the slow way is just as reliable. V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 24 / 37
75.
An exponential exampleExample √ ln 8Find √ e 2x e 2x + 1 dx ln 3 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 25 / 37
76.
An exponential exampleExample √ ln 8Find √ e 2x e 2x + 1 dx ln 3SolutionLet u = e 2x , so du = 2e 2x dx. We have √ ln 8 8√ 1 √ e 2x e 2x + 1 dx = u + 1 du ln 3 2 3 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 25 / 37
77.
About those limitsSince √ √ 2 e 2(ln 3) = e ln 3 = e ln 3 = 3we have √ ln 8 8√ 1 √ e 2x e 2x + 1 dx = u + 1 du ln 3 2 3 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 26 / 37
78.
An exponential exampleExample √ ln 8Find √ e 2x e 2x + 1 dx ln 3SolutionLet u = e 2x , so du = 2e 2x dx. We have √ ln 8 8√ 1 √ e 2x e 2x + 1 dx = u + 1 du ln 3 2 3Now let y = u + 1, dy = du. So 8√ 9 9 1 1 √ 1 u + 1 du = y dy = y 1/2 dy 2 3 2 4 2 4 9 1 2 1 19 = · y 3/2 = (27 − 8) = 2 3 4 3 3 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 27 / 37
79.
About those fractional powersWe have 93/2 = (91/2 )3 = 33 = 27 43/2 = (41/2 )3 = 23 = 8so 9 9 1 1 2 3/2 1 19 y 1/2 dy = · y = (27 − 8) = 2 4 2 3 4 3 3 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 28 / 37
80.
An exponential exampleExample √ ln 8Find √ e 2x e 2x + 1 dx ln 3SolutionLet u = e 2x , so du = 2e 2x dx. We have √ ln 8 8√ 1 √ e 2x e 2x + 1 dx = u + 1 du ln 3 2 3Now let y = u + 1, dy = du. So 8√ 9 9 1 1 √ 1 u + 1 du = y dy = y 1/2 dy 2 3 2 4 2 4 9 1 2 1 19 = · y 3/2 = (27 − 8) = 2 3 4 3 3 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 29 / 37
81.
Another way to skin that catExample √ ln 8Find √ e 2x e 2x + 1 dx ln 3SolutionLet u = e 2x + 1, V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 30 / 37
82.
Another way to skin that catExample √ ln 8Find √ e 2x e 2x + 1 dx ln 3SolutionLet u = e 2x + 1,so that du = 2e 2x dx. V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 30 / 37
83.
Another way to skin that catExample √ ln 8Find √ e 2x e 2x + 1 dx ln 3SolutionLet u = e 2x + 1,so that du = 2e 2x dx. Then √ ln 8 9√ 1 √ e 2x e 2x + 1 dx = u du ln 3 2 4 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 30 / 37
84.
Another way to skin that catExample √ ln 8Find √ e 2x e 2x + 1 dx ln 3SolutionLet u = e 2x + 1,so that du = 2e 2x dx. Then √ ln 8 9√ 1 √ e 2x e 2x + 1 dx = u du ln 3 2 4 9 1 = u 3/2 3 4 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 30 / 37
85.
Another way to skin that catExample √ ln 8Find √ e 2x e 2x + 1 dx ln 3SolutionLet u = e 2x + 1,so that du = 2e 2x dx. Then √ ln 8 9√ 1 √ e 2x e 2x + 1 dx = u du ln 3 2 4 1 3/2 9 = u 3 4 1 19 = (27 − 8) = 3 3 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 30 / 37
86.
A third skinned catExample √ ln 8Find √ e 2x e 2x + 1 dx ln 3SolutionLet u = e 2x + 1, so that u 2 = e 2x + 1 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 31 / 37
87.
A third skinned catExample √ ln 8Find √ e 2x e 2x + 1 dx ln 3SolutionLet u = e 2x + 1, so that u 2 = e 2x + 1 =⇒ 2u du = 2e 2x dx V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 31 / 37
88.
A third skinned catExample √ ln 8Find √ e 2x e 2x + 1 dx ln 3SolutionLet u = e 2x + 1, so that u 2 = e 2x + 1 =⇒ 2u du = 2e 2x dxThus √ ln 8 3 3 1 3 19 √ = u · u du = u = ln 3 2 3 2 3 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 31 / 37
89.
A Trigonometric ExampleExampleFind 3π/2 θ θ cot5 sec2 dθ. π 6 6 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 32 / 37
90.
A Trigonometric ExampleExampleFind 3π/2 θ θ cot5 sec2 dθ. π 6 6Before we dive in, think about: What “easy” substitutions might help? Which of the trig functions suggests a substitution? V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 32 / 37
92.
Solution θ 1Let ϕ = . Then dϕ = dθ. 6 6 3π/2 π/4 θ θ cot5 sec2 dθ = 6 cot5 ϕ sec2 ϕ dϕ π 6 6 π/6 π/4 sec2 ϕ dϕ =6 π/6 tan5 ϕNow let u = tan ϕ. So du = sec2 ϕ dϕ, and π/4 1 sec2 ϕ dϕ −5 6 =6 √ u du π/6 tan5 ϕ 1/ 3 1 1 3 =6 − u −4 √ = [9 − 1] = 12. 4 1/ 3 2 V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 33 / 37
93.
Graphs 3π/2 π/4 θ θ cot 5 sec 2 dθ 6 cot5 ϕ sec2 ϕ dϕ π 6 6 π/6 y y θ ϕ 3π π ππ 2 64The areas of these two regions are the same. V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 35 / 37
94.
Graphs π/4 1 −5 6 cot5 ϕ sec2 ϕ dϕ √ 6u du π/6 1/ 3 y y ϕ u ππ 1 1 64 √ 3 The areas of these two regions are the same. V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 36 / 37
95.
Summary If F is an antiderivative for f , then: f (g (x))g (x) dx = F (g (x)) If F is an antiderivative for f , which is continuous on the range of g , then: b g (b) f (g (x))g (x) dx = f (u) du = F (g (b)) − F (g (a)) a g (a) Antidiﬀerentiation in general and substitution in particular is a “nonlinear” problem that needs practice, intuition, and perserverance The whole antidiﬀerentiation story is in Chapter 6V63.0121.041, Calculus I (NYU) Section 5.5 Integration by Substitution December 13, 2010 37 / 37
Be the first to comment