Calculus: Integration by Parts

Integrating Products with Integration by Parts Let's consider the statement of the product rule for derivatives. We learn that: (u v)' = u' v + v' u Let's apply an integral sign to each term: °(uv)/= ° u/v + ° v /u The first term is the integral of a derivative and so is just uv. Now, let's rename things: u ' = du, v ' = dv So now we have u v = ° u/v + ° v/u Now, transpose the first integral on the right to get the integration by parts formula: ° udv = uv -° v du which is just the integration equivalent of the product rule for derivatives. This means that we will label one part of the integrand product u, and another part we will call dv. We will differentiate u to get du and we will antidifferentiate dv to get v. Then we'll just substitute in the parts formula. Rule of thumb: try not to set dv equal to something that require the power rule to get v since we wish to bring powers down, rather than raise them. For example if we have to integrate the product xe x , we will not set dv equal to x --since v would then equal (½ x²) and we will have raised powers from first to 2nd degree. An exponential antiderivative however is just an exponential function so there's no change in the exponent. In some cases, we have no choice since part of the integrand product cannot be labeled dv. For instance, if we have to integrate x ln x,we cannot set dv = ln x dx since we cannot antidifferentiate ln x. It isn't anybody's derivative. Let's do these 2 as examples. . integration by parts examples practice solutions . Integration by Parts PDF created with pdfFactory Pro trial version www.pdffactory.comExamples Example 1: Use integration by parts to find ° x ex dx ° xex dx = x ex -ex +C du = dx v = e x ° x ex dx = x ex -° ex dx u = x dv = e x dx ° u dv = u v -° v du Example 2: Use integration by parts to find ° x ln x dx . = ½ x 2 ln x – ¼ x 2 + C du = dx/x v = ½ x 2 ° x lnxdx = ½ x2 lnx -½° xdx u = ln x dv = x dx ° u dv = u v -° v du Note 1: When we set up our table with the u and the dv, notice that when there's a "d_" on the left of the equal sign, there's one on the right like in dv = x dx, or du = dx. Note 2: When we go from u to du, we're differentiating. When we go from dv to v, we're antidifferentiating or integrating. Example 3: Sometimes we apply the parts formula more than once. Use integration by parts to find  x 2 e 2x dx.° x2 e2x dx = 12x2 e2x -12xe2x -14e2x +C ° x2 e2x dx = 12x2 e2x -12xe2x -12du1 = dx v1 = ½ e 2 x ° e2x dx u1 = x dv1 = e 2 x dx ° x2 e2x dx = 12du = 2x dx v = ½ e 2 x x2 e2x -° x e2x dx u = x 2 dv = e 2 x dx ° u dv = u v -° v du Note 3: when repeating the parts formula in a question it's best to use "subscripts" --the little 1's below the line --as in u1 --when setting up the second or third equivalences. Note 4: the antiderivative of e 2 x = ½ e 2 x because the derivative of e 2 x = 2 e 2x so we multiply by ½ to eliminate the 2. Integration by Parts PDF created with pdfFactory Pro trial version www.pdffactory.comExample 4: Use integration by parts to find ° sec3 x dx Since this integrand is neither an even power of sec x nor an odd power of tan x, it doesn't fit any of the categories listed in standard techniques for doing trig integrals. Therefore, we'll have to improvise. First, we break sec 3 x up into sec x and sec 2 x. Since sec 2 x is the derivative of tan x, we'll set that = dv and we'll let u = sec x. Then we'll apply integration by parts. ° sec3 xdx = ° secx sec2 x dx ° sec3 xdx = 12integrate ° sec x dx , divide by 2 ( sec x tan x + ln | sec x + tan x | ) + C transpose -° sec3 x dx to left side 2 ° sec3 x dx = sec x tan x + ° sec x dx ° sec3 xdx = secx tanx -° sec3 xdx + ° secx dx ° sec3 xdx = secx tanx -° sec x (sec2 x -1) dx du = sec x tan x dx v = tan x ° sec3 x dx = sec x tan x -° sec x tan2 x dx u = sec x dv = sec 2 x dx ° u dv = u v -° v du Note 5: This is known as a BOOMERANG integral because it comes back at you. We started with ° sec3 x dx on the left side of the equation and then ran into it again on the right side. That's how we got 2 times it on the left side in the 2nd to last step. The same happens with integrands that include exponential functions mixed with other trig functions, especially sin x and cos x since they are both cyclical. A common approach is to name the initial integral I (in this case I = ° sec3 x dx ). Then, instead of rewriting the entire expression in each step as above, we simply write "I = " ** Watch for integrals like ° x e x 2 , because though it looks like a product that requires dx integration by parts, a substitution of u = x 2 will turn the integral into ½ ° eu du . . integration by parts examples practice solutions . Integration by Parts PDF created with pdfFactory Pro trial version www.pdffactory.comPractice 1/Use integration by parts on these integrals d) ° x ln x dx e) ° e-x sin x dx (boomerang) f) °(ln x)2 dx a) ° x e-x dx b) ° x sin x dx c) ° x sec x tan x dx 2/a) Find the area of the region bounded by y = ln x, the line x = e and the x-axis. Make a diagram. b) Find the volume of the solid generated when the region in part (a) is revolved about the x-axis. . integration by parts examples practice solutions .Solutions 1/Use integration by parts on these integrals° xe-x dx = -x e-x -e-x +C du = dx v = – e – x ° x e-x dx = -x e-x + ° e-x dx u = x dv = e – x dx  ° u dv = u v -° v du a) ° xe-x dx – x cos x + sin x + C du = dx v = -cos x ° x sinx dx = -x cos x + ° cosxdx u = x dv = sin x dx  ° u dv = u v -° v du b) ° x sinx dx . Integration by Parts PDF created with pdfFactory Pro trial version www.pdffactory.comx sec x – ln use integrals table for ° sec x dx . | sec x + tan x | + C du = dx v = sec x ° x sec x tanx dx = x sec x -° sec xdx u = x dv = sec x tan x dx ° u dv = u v -° v du c) ° x secx tanx dx I = 2/3 x 3 /2 ln x – 4/9 x 3 /2 + C I = 23x3/2 lnx -23du = dx /x v = 2/3 x 3 /2 ° x1/2 dx u = ln x dv = x 1 /2 dx ° u dv = u v -° v du d) ° x ln x dx transpose I , factor, divide by 2 I = – ½ e – x [cos x + sin x ] + C du1 = – e – x dx v1 = sin x I = – e– x cos x – e-xsin x -° e-x sin x dx (I) u1 = e – x dv1 = cos x dx du = – e – x dx v = – cos x I =– e–x cos x – ° e-x cos x dx u = e – x dv = sin x dx ° u dv = u v -° v du e) ° e-x sinx dx .. Integration by Parts PDF created with pdfFactory Pro trial version www.pdffactory.comI = x (du1 = dx /x v1 = x ln x) 2 – 2 x ln x + 2x + C u1 = ln x dv1 = dx I = x (ln x)2 -2 x ln x -° dx du = (2 ln x) dx /x v = x I = x (ln x)2 -2 ° ln x dx u = ( ln x )2 dv = dx ° u dv = u v -° v du f) °(ln x)2 dx 2/a) Find the area of the region bounded by y = ln x, the line x = e and the x-axis. Make a diagram. (1, 0) (e, 0) Area = = (e ln e – e) – (ln 1 – 1) = 1 sq. unit. e1 °lnx dx = x lnx -° dx = xlnx -x e1 | (see part 2 of question (1f) above for °(ln x)2 dx ) b) Find the volume of the solid generated when the region in part (a) is revolved about the x-axis. Volume = o e1 °(ln x)2dx = o [ x (lnx )2 -2x lnx + 2x] e1| = o(e -2) (see question (1f) above for °(ln x)2 dx ) integration by parts examples practice solutions Calculus II MathRoom Index Integration by Parts PDF created with pdfFactory Pro trial version www.pdffactory.com