7.1 Integration by Parts (...or, undoing the product rule.)

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1 Integration by Parts (...or, undoing the product rule.) Integration by Parts Recall the differential form of the chain rule. If u and v are differentiable functions. Then (1) d(uv) = du v +u dv Solving the above equation for u dv and integrating yields () udv = uv vdu Equation () is known as Integration by Parts. This formula is also written in the following form. (3) f(x)g (x)dx = f(x)g(x) f (x)g(x)dx and the definite integral form is given by (4) b a f(x)g (x)dx = f(x)g(x) b a b a f (x)g(x)dx

2 7.1 Example 1. Compute the integrals. a) xe x dx Let u = x and dv = e x dx then du = dx and v = e x so that xe x dx = xe x e x dx Notice that the integral on the right-hand side is easy to compute. Now we check: = xe x e x +C d dx (xex e x ) = e x +xe x e x = xe x

3 7.1 3 b) x sinxdx This time we let u = x and dv = sinxdx then du = xdx and v = cosx Then (5) x sinxdx = x cosx+ xcosxdx Now we are still stuck with an integral that we can not compute directly but we seem to be making some progress. Let s try integration by parts on xcosxdx. Let u = x and then du = dx and dv = cosxdx v = sinx Then xcosxdx = xsinx sinxdx = xsinx+cosx+c It follows that (5) reduces to x sinxdx = x cosx+ xcosxdx = x cosx+xsinx+cosx+c Of course we should confirm that D x ( x cosx+xsinx+cosx ) =? x sinx

4 7.1 4 c) π π xsinnxdx Let u = x and dv = sinnxdx. Then π π xsinnxdx = xcosnx n = 1 n π π + 1 n π π cosnxdx (πcosnπ +πcos( nπ))+ sinnx n = πcosnπ n + 1 n(sinnπ sin( nπ)) π π and if n is a positive integer = πcosnπ n +0 = π( 1)n n This integral occurs frequently in Fourier Analysis.

5 7.1 5 Tabular Integration Repeated applications of Integration by Parts can become a bookkeeping nightmare. The next example uses tabular integration to help us keep things in order. Example. Tabular Integration Evaluate x 3 cosxdx. Let f(x) = x 3 and g(x) = cosx. Then f(x) and its derivatives g(x) and its integrals x 3 3x 6x cosx sinx cosx 6 sinx 0 cosx It follows that x 3 cosxdx = x 3 sinx+3x cosx 6xsinx 6cosx+C

6 7.1 6 Example 3. Tabular Integration - Part II Evaluate x e x dx. Let f(x) = x and g(x) = e x. Then f(x) and its derivatives g(x) and its integrals x e x x e x / e x /4 0 e x /8 It follows that x e x dx = x ex x ex + ex 4 +C = ex 4 (x x+1)+c

7 7.1 7 Solving for the Integral Consider the following equation. (6) a f(x)dx = g(x)+b f(x)dx Now if a b then (a b) f(x)dx = g(x) f(x)dx = g(x) a b Example 4. Suppose that a and b are nonzero real numbers. Evaluate e ax sinbxdx Let u = e ax du = ae ax dx dv = sinbxdx v = 1 b cosbx Thus I = def. e ax sinbxdx = 1 b eax cosbx+ a b e ax cosbxdx...and it looks like we are going nowhere...

8 7.1 8 What the heck, let s try again. So let u = e ax du = ae ax dx dv = cosbxdx v = 1 b sinbx Then Thus I = 1 b eax cosbx+ a b = 1 b eax cosbx+ a b e ax cosbxdx ( 1 b eax sinbx a b = 1 b eax cosbx+ a b eax sinbx a b I a +b I = 1 (asinbx bcosbx) b eax b ) e ax sinbxdx It follows that (7) e ax sinbxdx = eax a +b (asinbx bcosbx)

9 7.1 9 Example 5. Integral of the Natural Logarithm Evaluate lnxdx. There are several ways to evaluate this integral. We ll start with integration by parts. Let u = lnx du = dx x dv = dx v = x Then lnxdx = xlnx x dx x = xlnx x+c

10 Substitution and Integration by Parts Recall the first example. Notice that the integral is of the form lnudu. }{{} x }{{} e x dx, (here u = e x ) lnu du This observation leads us to another approach for lnxdx. Example 6. a) Evaluate Substitution and Integration by Parts lnxdx. We let x = e u then dx = e u du. Thus lnxdx = ue u du = ue u e u +C, (from Example 1) = xlnx x+c as we saw in the previous example. (Also, see Example 7.)

11 b) Evaluate z(lnz) dz. Let z = e u. Then dz = e u du and z(lnz) dz = u e u du So by Example 3 we have u e u du = eu 4 ( u u+1 ) +C = z 4 ( (lnz) lnz +1 ) +C We should check this one: ( d z ( (lnz) lnz +1 )) dz 4 = z ( (lnz) lnz +1 ) + z 4 ( 4lnz z z ) = z(lnz) zlnz + z +zlnz z = z(lnz)

12 7.1 1 Integrating Inverses Example 7. Evaluate cos 1 xdx. Let y = cos 1 x. Then cosy = x and dx = sinydy. Thus cos 1 xdx = ysinydy Now apply integration by parts with u = y and dv = siny to obtain = ycosy + cosydy = ycosy siny +C = xcos 1 x sin ( cos 1 x ) +C = xcos 1 x +C Let s verify this. ( D x xcos 1 x ) = cos 1 x x x = cos 1 x

13 Example 8. In the handout discussing integration techniques, we used geometry to conclude that (8) 1 0 dx = π/4 and we mentioned that we would eventually evaluate this integral using trigonometric substitution. Actually, there is another way. First, apply partial integration to the left-hand side of the (seemingly) unrelated integral below. Let u = x and dv = xdx/. Then v = and (9) x dx = x dx Now we return to the integral in (8). We start by rationalizing the numerator of the integrand. dx = = dx dx = sin 1 x x dx x dx = sin 1 x+x dx Here the last line follows from (9). Now solve for the integral and divide by to conclude (10) dx = 1 (sin 1 x+x ) Rather than check this one by differentiating, let s use this result to verify the following proposed solution to the gas tank problem.

14 Find the area A of the shaded region in the sketch below. In other words, find (11) A = a 0 dx Observe that this area is the sum of the areas of the sector OPQ and the triangle OPB. Q P By the Alternate Interior Angle theorem (from high school geometry), OPB = POQ. Now let θ = OPB and recall that the area of a sector is one-half the central angle times the square of the radius, i.e., A S = θ r /. Notice in triangle OPB that sinθ = a. O θ 1 a θ 1 a B Putting this all together we have A = A S +A T = 1 θ(1) + 1 a 1 a = 1 θ + 1 a 1 a (sin 1 a+a 1 a ) = 1 Now by (10) and (11), we have A = a 0 dx = 1 ( sin 1 x+x ) a 0 = 1 [(sin 1 a+a ) ] 1 a 0 Observe also that 1 (sin 1 a+a ) 1 a a=1 = 1 sin 1 1 = 1 π confirming our observation in (8).

15 Example 9. Evaluate e x dx. Since nothing jumps out, let s try multiplying by a convenient form of one followed by a change of variables. e x x dx = e x dx x = ue u du, (here u = x) = ue u e u +C, (again from Example 1) = xe x e x +C Let s check ( D x xe x e ) ( x e x ) = x + xe x 1 x e x 1 x = e x