Applied Calculus I. Lecture 29

Transcription

1 Applied Calculus I Lecture 29

2 Integrals of trigonometric functions We shall continue learning substitutions by considering integrals involving trigonometric functions.

3 Integrals of trigonometric functions We shall continue learning substitutions by considering integrals involving trigonometric functions. Find sin (x 2 3)xdx

4 Integrals of trigonometric functions We shall continue learning substitutions by considering integrals involving trigonometric functions. Find sin (x 2 3)xdx The expression that seems to complicate things here is x 2 3. So, let us make the substitution u = x 2 3. Then du = 2xdx, whic implies xdx = 1 2 du.

5 Integrals of trigonometric functions We shall continue learning substitutions by considering integrals involving trigonometric functions. Find sin (x 2 3)xdx The expression that seems to complicate things here is x 2 3. So, let us make the substitution u = x 2 3. Then du = 2xdx, whic implies xdx = 1 2 du. Then our integral becomes 1 2 sin udu = 1 2 cos u + C = 1 2 cos (x2 3) + C

6 Find tan xdx Examples

7 Examples Find tan xdx sin x Notice that tan xdx = dx. The problematic part is the cos x denominator. So, let u = cos x. Then du = sin xdx.

8 Examples Find tan xdx sin x Notice that tan xdx = dx. The problematic part is the cos x denominator. So, let u = cos x. Then du = sin xdx. Now we see that our integral becomes du = ln u + C = ln cos x + C u

9 Examples Find tan xdx sin x Notice that tan xdx = dx. The problematic part is the cos x denominator. So, let u = cos x. Then du = sin xdx. Now we see that our integral becomes du = ln u + C = ln cos x + C u Find cot xdx

10 Examples Find tan xdx sin x Notice that tan xdx = dx. The problematic part is the cos x denominator. So, let u = cos x. Then du = sin xdx. Now we see that our integral becomes du = ln u + C = ln cos x + C u Find cot xdx cos x Notice that cot xdx = dx. The problematic part is the sin x denominator. So, let u = sin x. Then du = cos xdx.

11 Examples Find tan xdx sin x Notice that tan xdx = dx. The problematic part is the cos x denominator. So, let u = cos x. Then du = sin xdx. Now we see that our integral becomes du = ln u + C = ln cos x + C u Find cot xdx cos x Notice that cot xdx = dx. The problematic part is the sin x denominator. So, let u = sin x. Then du = cos xdx. Now we see that our integral becomes du = ln u + C = ln sin x + C u

12 Find dx x ln x Examples

13 Find dx x ln x Examples Again, the part that complicates things is the denominator. But if we try the substitution u = x ln x we see that du = (ln x + 1)dx, and then we can t express everything in terms of u.

14 Find dx x ln x Examples Again, the part that complicates things is the denominator. But if we try the substitution u = x ln x we see that du = (ln x + 1)dx, and then we can t express everything in terms of u. The idea is to make a substitution for the part of your integrand whose derivative gives you the rest of the integrand (perhaps multiplied by a constant).

15 Find dx x ln x Examples Again, the part that complicates things is the denominator. But if we try the substitution u = x ln x we see that du = (ln x + 1)dx, and then we can t express everything in terms of u. The idea is to make a substitution for the part of your integrand whose derivative gives you the rest of the integrand (perhaps multiplied by a constant). In this case, we notice that (ln x) = 1. So, let us try u = ln x. Then x du = 1 dx, and our integral becomes x

16 Find dx x ln x Examples Again, the part that complicates things is the denominator. But if we try the substitution u = x ln x we see that du = (ln x + 1)dx, and then we can t express everything in terms of u. The idea is to make a substitution for the part of your integrand whose derivative gives you the rest of the integrand (perhaps multiplied by a constant). In this case, we notice that (ln x) = 1. So, let us try u = ln x. Then x du = 1 dx, and our integral becomes x du = ln u + C = ln ln x + C u

17 Find x8 3x2 +1 dx Examples

18 Find x8 3x2 +1 dx Examples Let u = 3x 2 + 1, then du = 6xdx. Hence, x8 3x2 +1 dx = 1 8 u du = u ln 8 + C = 83x2+1 6 ln 8 + C

19 Find x8 3x2 +1 dx Examples Let u = 3x 2 + 1, then du = 6xdx. Hence, x8 3x2 +1 dx = 1 8 u du = u ln 8 + C = 83x2+1 6 ln 8 + C Find sin 7 x cos xdx

20 Find x8 3x2 +1 dx Examples Let u = 3x 2 + 1, then du = 6xdx. Hence, x8 3x2 +1 dx = 1 8 u du = u ln 8 + C = 83x2+1 6 ln 8 + C Find sin 7 x cos xdx Let u = sin x, then du = cos xdx, yielding sin 7 x cos xdx = u 7 du = 1 8 u8 + C = 1 8 sin8 x + C

21 Find cos 3 x sin x + 1 dx Examples

22 Examples cos 3 x Find sin x + 1 dx Proceeding as usual, let u = sin x + 1, then du = cos xdx. So, cos xdx becomes du, the denominator becomes u, but we still have the factor cos 2 x that we need to express in terms of u.

23 Examples cos 3 x Find sin x + 1 dx Proceeding as usual, let u = sin x + 1, then du = cos xdx. So, cos xdx becomes du, the denominator becomes u, but we still have the factor cos 2 x that we need to express in terms of u. Recall that cos 2 x + sin 2 x = 1, so cos 2 x = 1 sin 2 x. Also, sin 2 x = (sin x + 1 1) 2 = (u 1) 2 = u 2 2u + 1. Therefore, cos 2 x = 2u u 2.

24 Examples cos 3 x Find sin x + 1 dx Proceeding as usual, let u = sin x + 1, then du = cos xdx. So, cos xdx becomes du, the denominator becomes u, but we still have the factor cos 2 x that we need to express in terms of u. Recall that cos 2 x + sin 2 x = 1, so cos 2 x = 1 sin 2 x. Also, sin 2 x = (sin x + 1 1) 2 = (u 1) 2 = u 2 2u + 1. Therefore, cos 2 x = 2u u 2. Thus, we obtain cos 3 x 2u u 2 sin x + 1 dx = du = (2 u)du = 2u 1 u 2 u2 + C = = 2(sin x + 1) 1 2 (sin x + 1)2 + C

25 Examples cos 3 x Find sin x + 1 dx Proceeding as usual, let u = sin x + 1, then du = cos xdx. So, cos xdx becomes du, the denominator becomes u, but we still have the factor cos 2 x that we need to express in terms of u. Recall that cos 2 x + sin 2 x = 1, so cos 2 x = 1 sin 2 x. Also, sin 2 x = (sin x + 1 1) 2 = (u 1) 2 = u 2 2u + 1. Therefore, cos 2 x = 2u u 2. Thus, we obtain cos 3 x 2u u 2 sin x + 1 dx = du = (2 u)du = 2u 1 u 2 u2 + C = = 2(sin x + 1) 1 2 (sin x + 1)2 + C In the hindsight, we could do cos 3 x (1 sin 2 sin x + 1 dx = x) cos x dx = (1 sin x) cos xdx 1 + sin x and then integrate.

26 Example An epidemic is growing in a region according to the rate N (t) = 100t t 2 + 2, where N(t) is the number of people infected after t days. Find a formula for the number of people infected after t dyas, given that 37 people were infected at t = 0.

27 Example An epidemic is growing in a region according to the rate N (t) = 100t t 2 + 2, where N(t) is the number of people infected after t days. Find a formula for the number of people infected after t dyas, given that 37 people were infected at t = 0. So, we need a formula for N(t) given that N(0) = 37. Clearly, N(t) is 100t some antiderivative of. So, let s find the whole family. t 2 + 2

28 Example An epidemic is growing in a region according to the rate N (t) = 100t t 2 + 2, where N(t) is the number of people infected after t days. Find a formula for the number of people infected after t dyas, given that 37 people were infected at t = 0. So, we need a formula for N(t) given that N(0) = 37. Clearly, N(t) is 100t some antiderivative of. So, let s find the whole family. t u=t2+2,du=2tdt 100t {}}{ 50 dt = t u du = 50 ln u + C = 50 ln t C

29 Example An epidemic is growing in a region according to the rate N (t) = 100t t 2 + 2, where N(t) is the number of people infected after t days. Find a formula for the number of people infected after t dyas, given that 37 people were infected at t = 0. So, we need a formula for N(t) given that N(0) = 37. Clearly, N(t) is 100t some antiderivative of. So, let s find the whole family. t u=t2+2,du=2tdt 100t {}}{ 50 dt = t u du = 50 ln u + C = 50 ln t C So, N(t) = 50 ln (t 2 + 2) + C for some specific value of C. To find this value, we need to use the fact that N(0) = 37.

30 Example An epidemic is growing in a region according to the rate N (t) = 100t t 2 + 2, where N(t) is the number of people infected after t days. Find a formula for the number of people infected after t dyas, given that 37 people were infected at t = 0. So, we need a formula for N(t) given that N(0) = 37. Clearly, N(t) is 100t some antiderivative of. So, let s find the whole family. t u=t2+2,du=2tdt 100t {}}{ 50 dt = t u du = 50 ln u + C = 50 ln t C So, N(t) = 50 ln (t 2 + 2) + C for some specific value of C. To find this value, we need to use the fact that N(0) = 37. Plugging t = 0 into the formula for N(t) we get 50 ln 2 + C = 37 that is, C = ln 2. Therefore, N(t) = 50 ln (t 2 + 2) ln 2. ( ) t This can be simplified to N(t) = 50 ln