Applied Calculus I Lecture 29
Integrals of trigonometric functions We shall continue learning substitutions by considering integrals involving trigonometric functions.
Integrals of trigonometric functions We shall continue learning substitutions by considering integrals involving trigonometric functions. Find sin (x 2 3)xdx
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.
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
Find tan xdx Examples
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.
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
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
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.
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
Find dx x ln x Examples
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.
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).
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
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
Find x8 3x2 +1 dx Examples
Find x8 3x2 +1 dx Examples Let u = 3x 2 + 1, then du = 6xdx. Hence, x8 3x2 +1 dx = 1 8 u du = 1 6 6 8 u ln 8 + C = 83x2+1 6 ln 8 + C
Find x8 3x2 +1 dx Examples Let u = 3x 2 + 1, then du = 6xdx. Hence, x8 3x2 +1 dx = 1 8 u du = 1 6 6 8 u ln 8 + C = 83x2+1 6 ln 8 + C Find sin 7 x cos xdx
Find x8 3x2 +1 dx Examples Let u = 3x 2 + 1, then du = 6xdx. Hence, x8 3x2 +1 dx = 1 8 u du = 1 6 6 8 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
Find cos 3 x sin x + 1 dx Examples
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.
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.
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
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.
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.
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
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 u=t2+2,du=2tdt 100t {}}{ 50 dt = t 2 + 2 u du = 50 ln u + C = 50 ln t2 + 2 + C
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 u=t2+2,du=2tdt 100t {}}{ 50 dt = t 2 + 2 u du = 50 ln u + C = 50 ln t2 + 2 + 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.
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 u=t2+2,du=2tdt 100t {}}{ 50 dt = t 2 + 2 u du = 50 ln u + C = 50 ln t2 + 2 + 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 = 37 50 ln 2. Therefore, N(t) = 50 ln (t 2 + 2) + 37 50 ln 2. ( ) t This can be simplified to N(t) = 50 ln 22 + 1 + 37