Calculus 2 Lia Vas Integration by Parts Using integration by parts one transforms an integral of a product of two functions into a simpler integral. Divide the initial function into two parts called u and dv (keep dx in dv part). Then apply the following rule. Integration by parts: udv = uv vdu You were successful in choosing u and dv initially if the resulting integral vdu is simpler that the initial integral. If it is not, go back and rethink your choice of u and dv. Below are some hints on how to decompose the initial integral into udv: a) Integrals with the product of a polynomial and e ax. Try u = polynomial. b) Integrals with polynomial and sin ax (or cos ax). Try u = polynomial. c) e ax sin bx dx or e ax cos bx dx. You can start with u = e ax. You will need to do integration by parts twice and will end up with your initial integral after the second time. Solve for your initial integral then. d) Integrals with logarithmic functions. Try u = logarithmic function. e) Integrals with inverse trigonometric functions. Try u = inverse trigonometric function. The formula udv = uv vdu is really the product rule in disguise. To see why the integration by parts formula is true, start with the product rule (uv) = u v + v u that can be also written as d du (uv) = v + dv u. Integrating the product rule with respect to x, we have that dx dx dx d du dv dx (uv) dx = dx v dx + dx u dx uv = du v + dv u = v du + u dv. Solve for the term udv on the right side and obtain that uv vdu = udv. For definite integrals, the formula becomes b a udv = u v Practice Problems. Evaluate the following integrals. b a b a vdu.
. xe x dx 2. 3. x 2 e x dx Hypothesize on the number of integration by parts you would need to evaluate x n e x dx. 4. 2x sin 3x dx 5. (2x + 5) sin(2x + 5) dx 6. e x sin x dx 7. ln x dx 8. ln x x 2 dx 9. ln(2x ) (2x ) 2 dx.. tan x dx 3 sin 2x dx 2. 3. 4. In the following problems, sketch the given region and find its area in case the area is finite.
5. x, xe x y 6. Solutions: x, y ln x x 2. Following the hint for the first type of integrals, start by u = x and dv = e x dx. Then du = dx and v = e x dx = e x. Then use the formula for integration by parts and obtain xe x dx = u dv = uv v du = xe x e x dx. Note that this last integral is simpler that the initial integral indicating that you are on the right path. The last integral is equal to e x so your final answer is xe x e x + c. 2. Following the hint for the first type of integrals, start by u = x and dv = e 2x dx. Then du = dx and v = e 2x dx. To get v, you can use the substitution w = 2x dw = 2dx dw = dx 2 and so v = e 2x dx = 2 e w dw = 2 ew = 2 e 2x. Then = u dv = uv v du = 2 xe 2x + 2 e 2x dx. Note that this last integral is simpler that the initial integral indicating that you are on the right path. In fact, the last integral is the same as the one you evaluated when finding v so it is equal to 2 e 2x. Thus, the initial integral is equal to 2 xe 2x + 2 e 2x dx = 2 xe2x + e 2x + c = 2 xe 2x 4 e 2x + c 3. Start by u = x 2 and dv = e x dx. Then du = 2xdx and v = e x dx = e x. Then you have x 2 e x dx = u dv = uv v du = x 2 e x e x 2x dx = x 2 e x 2xe x dx. For this last integral, you need to use integration by parts again. Take u = 2x and dv = e x dx (alternatively, factor 2 out and take u = x). Then du = 2dx and v = e x dx = e x and so 2xe x dx = 2xe x 2e x dx = 2xe x 2e x. This gives you the final answer x 2 e x dx = x 2 e x (2xe x 2e x )+c = x 2 e x 2xe x + 2e x + c. Since xe x dx requires the integration by parts applied one time and x 2 e x dx requires the integration by parts applied two times, we can hypothesize that the integral x n e x dx requires the integration by parts applied n times. 4. Following the hint for the second type of integrals, start by u = 2x and dv = sin 3x dx. Then du = 2dx and v = sin 3x dx. To find v, use the substitution w = 3x dw = 3dx dw = dx 3 and so v = sin 3x dx = 3 sin wdw = cos w = cos 3x. Then you have 2x sin 3x dx = 3 3 2 2 x cos 3x 3 3 cos 3x dx. This last integral is similar to the one used to obtain v and it is equal to 2 sin 3x. Thus the initial integral is equal to x cos 3x+ 2 2 sin 3x = x cos 3x+ 2 sin 3x+c. 3 3 3 3 3 9 5. Following the hint for the second type of integrals, you can start with with u = 2x + 5 and dv = sin(2x + 5). In this case, du = 2dx and v = sin(2x + 5)dx. To find v, you can use substitution w = 2x + 5. In this case, dw = 2dx dw = dx. Obtain v = sin wdw = cos w = cos(2x+5). So, the integral becomes (2x+5) 2 cos(2x+5) cos(2x+5) 2dx. 2 This last integral simplifies to cos(2x + 5)dx (factor the negative out), you can evaluate it similarly as when finding v and obtain sin(2x + 5). So, the final answer is (2x + 5) cos(2x + 5) + sin(2x + 5) + c. 2
Method (2) Use substitution w = 2x + 5 first to simplify the integral and then use the integration by parts. In this case, dw = 2dx dw = dx. Obtain w sin w dw. Using the integration by parts with u = w and dv = sin wdw, we obtain w cos w + sin w + c = 2 (2x + 5) cos(2x + 5) + sin(2x + 5) + c. 6. This integral is of the third type. Let us denote the initial integral by I. You can start by u = e x an dv = sin xdx. Then dy = e x and v = sin xdx = cos x so that I = e x sin xdx = e x cos x+ e x cos xdx. Use the integration by parts again for this last integral. With u = e x and dv = cos xdx, you obtain du = e x dx and v = sin x so that e x cos xdx = e x sin x e x sin xdx. This last integral is our initial integral that we denoted by I. Thus I = e x cos x + e x cos xdx = e x cos x + e x sin x e x sin x = e x cos x + e x sin x I. Note that this gives you the equation I = e x cos x + e x sin x I. Solving for I gives you 2I = e x cos x + e x sin x I = 2 ( ex cos x + e x sin x) So, the final answer is I = 2 ex cos x + 2 ex sin x + c. 7. Following the hint for the fourth type of integrals, start by u = ln x and dv = dx. Then du = x dx and v = dx = x. Thus ln xdx = x ln x x xdx = x ln x dx = x ln x x + c. 8. Start by u = ln x and dv = dx = x 2 dx so that du = dx and v = x 2 dx = x = x 2 x Then ln xdx = ln x ln x dx = + x 2 dx = ln x + c. x 2 x x x x x x 9. Following the hint for the fourth type of integrals, you can start by u = ln(2x ) and dv = dx so that du = 2dx and v = (2x ) 2 dx. You can use substitution w = 2x (2x) x to find v. In this case, dw = 2dx dw = dx. Obtain v = w 2 dw = 2 w = =. 2w 2(2x) So, the integral becomes ln(2x ) 2 ln(2x) dx = + dx. This 2(2x) 2x 2(2x) 2(2x) (2x) 2 last integral is the same one you evaluated when finding v, and so it is. Thus, the final answer is ln(2x) 2(2x) 2(2x) + c. 2(2x) Method (2) Use substitution w = 2x first to simplify the integral. In this case, dw = 2dx dw = dx. Obtain ln w dw. Then use integration by parts with u = ln w and dv = dw. Then 2 w 2 du = dw and v = w 2 dw = w =. So, the integral becomes ln w dw = w w 2w 2 w w ln w + dw = ln w 2w 2 w w 2 w = ln w + c = ln(2x) + c. 2w 2w 2(2x) 2(2x). Following the hint for the fifth type of integrals, start by u = tan x and dv = dx. Thus du = dx and v = dx = x. So, the integral becomes x tan x x dx. Use the substitution +x 2 +x 2 w = + x 2 for this last integral and obtain that x dx = x dw = ln w = ln( + +x 2 w 2x x2 ). So, the initial integral is equal to x tan x ln( + 2 x2 ) + c.. Following the hint for the fifth type of integrals, you can start by u = sin (2x) and dv = 3dx so that du = (2x) 2dx = 2 2 4x 2 dx and v = 3dx = 3x. Thus the integral is 3x sin 2x 3x 2 4x 2 dx. Evaluate this integral using the substitution w = 4x 2 and obtain x.
6x w dw = 3 8x 4 w /2 dw = 3 4 2w/2 = 3 2 4x2. So, the final answer is 3x sin 2x + 3 2 4x2 + c. Method (2) Use the substitution w = 2x to simplify first and then use the integration by parts. With w = 2x dw = 2dx dx = dw 2, the integral reduces to 3 2 sin w dw. Then, using the integration by parts with u = sin w and dv = 3 2 dw (or u = 3 2 sin w and dv = dw), you obtain 3 2 w sin w + 3 2 w2 + c. Thus, the final answer is 3x sin 2x + 3 2 4x2 + c. 2. By problem 2, = 2 xe 2x 4 e 2x + c. Using this answer, = 2 xe 2x 4 e 2x 3. By problem 2, = 2 xe 2x 4 e 2x + c.. Thus = 2 xe 2x 4 e 2x = 2 e 2 4 e 2 + + 4 e = 4 3 4e 2.48. ( = lim x 2 xe 2x ) ( 4 e 2x 2 ()e 2() ) 4 e 2(). Let us consider first lim x 2 xe 2x write the function as a quotient x and note that the 2e 2x limit is of the form. Use L Hopital s rule to get lim x = =. 4e 2x Let us consider now lim x 4 e 2x. This limit evaluates as follows. lim x = =. 4e 2x Thus, the value of the antiderivative at the upper bound is. The value at the lower bound is 2 ()e 2() 4 e 2() = =. Hence, the integral is equal to =. The integral is 4 4 4 4 equal to a finite number and so it is convergent. 4. Since this integral is improper because of both bounds, you need to decompose it using any number a as a + a. Given the previous problem, you can use a = for example. Thus, xe 2x dx = xe 2x dx +. By the previous problem, the second integral is convergent. The first integral is xe 2x dx = ( 2 xe 2x 4 e 2x). To evaluate the antiderivative at the lower bound, write the antiderivative as a single fraction (2x+). Then 4e 2x plug the bounds as follows. (2x + ) 4e 2x = 4 = + 4 = + 4 = 4 =. Thus, the first integral is divergent and hence the sum of two integrals is divergent as well. 5. The condition x indicates the bounds of the integration < x. The second condition indicates the lower and upper y-curves: y = xe x is the lower and y = is the upper curve. Thus, the area A can be found as A = ( xe x )dx = Note that xe x dx = xe x e x + c by problem. Hence, xe x dx. A = ( xe x + e x ) = ( ()e + e ) lim x ( xex + e x ).
Use the graph of e x to determine that lim x e x =. For lim x xe x you need to use L Hopital s rule since this limit is of the form. Write your function as x to obtain e x form. Then use the L Hopital s rule to get lim x = =. e x Thus, A = e =. So, the area is finite and it is equal to. 6. The condition x indicates the bounds of the integration x <. The second condition indicates the lower and upper y-curves: y = ln x is the upper and y = is the lower curve. x 2 Thus, the area A can be found as A = By problem 8, ln xdx = ln x + c. Thus x 2 x x A = ln x x 2 dx = ( ln x x x) ( ) ln x x ln x dx = 2 x dx. 2 ( ln x = x lim x ) ( ln ). x Note that lim x = =. Since lim x x ln x is of the form, you need to use L Hopital s x /x rule. Get lim x = =. Hence, the value of the antiderivative at the upper bound produces. Thus, the area is A = ( ) =. So, the area is finite and it is equal to.