Lesson 50 Integration by Parts

Transcription

1 5/3/07 Lesson 50 Integration by Parts Lesson Objectives Use the method of integration by parts to integrate simple power, eponential, and trigonometric functions both in a mathematical contet and in a real world problem contet 5/3/07 5/3/07 (A) Product Rule Recall that we can take the derivative of a product of functions using the product rule: d d f g f g g f So we are now going to integrate this equation and see what emerges (B) Product Rule in Integral Form We have the product rule as d d f g f g g f And now we will integrate both sides: d d f g d f g d g f d f g f g d g f d rearranging f g d f g g f d 5/3/07 3 5/3/07 4 (B) Product Rule in Integral Form (C) Integration by Parts Formula We will make some substitutions to simplify this equation: So we have the formula u dv uv f gd f g g f let u f and let v g then du f d and dv gd u dv uv d So what does it mean? It seems that if we are trying to solve one integral and we create a second integral!!!! Our HOPE is that the second integral is easier to solve than the original integral u dv 5/3/07 5 5/3/07 6

2 5/3/07 (D) Eamples (D) Eamples Integrate the following functions: ( a) ( b) ( c) ( d) e d e d ln( ) d ln( ) d The easiest way to master the method is by practicing, so determine e d We need to select a u and a dv So we have a choice.. Let u e and then dv d So du e d and v So we get : e d e e d Comment: Is the second integral any easier than the original??? 5/3/07 7 5/3/07 8 (D) Eamples So let s make the other choice as we determine Let u and then dv e d So du d and v e So we get : e d e e d Checkpoint: Is our second integral any easier than our first one??? e d e e d e e C Now verify by differentiating the answer 5/3/07 9 e d Eample b: e d e e d e u dv uv u du d e d u e e e d dv e d v e This is still a product, so we need to use integration by parts again. du d dv e d v e e e e C Eample c: ln d u dv uv (D) More Eamples u ln dv d Integrate the following functions: ln d ln C du d v ( a) ( b) ( c) ( d) cos( ) d e cos( ) d cos( ) d arcsin( ) d 5/3/07

3 5/3/07 Eample a: cos d u dv uv polynomial factor u sin sin d sin cos C du d dv cos d v sin Eample b: e cos d e sin sin e d e sin e cos cos e d uv u e du e sin e cos e cos d e d dv cos d u e dv sin d du e d vcos v sin This is the epression we started with! Eample b: CONTINUED) e cos d e cos d e cos d e sin e cos e sin sin e d u e du e sin e cos e cos d C e d e sin e cos cos e d dv cos d v sin u e dv sin d du e d vcos e sin e cos e cos d Eample b: e cos d e cos d e cos d e sin e cos e sin sin e d e sin e cos cos e d e sin e cos e cos d C This is called solving for the unknown integral. It works when both factors integrate and differentiate forever. e sin e cos e cos d (E) Further Eamples (E) Further Eamples For the following question, do it using the method requested. Reconcile your solution(s) d byparts d by Substitution Definite Integrals Evaluate: ( a) ( b) e 0 ln d tan ( ) d 5/3/07 7 5/3/07 8 3

4 5/3/07 Calculate Let Then, tan d 0 u tan dv d d du v tan d tan d tan 0 tan d d To evaluate this integral, we use the substitution t = + (since u has another meaning in this eample). Then, dt = d. So, d = ½ dt. When = 0, t =, and when =, t =. Hence, d 0 ln t dt t (ln ln) ln Therefore, tan d 4 ln d As tan - for 0, the integral in the eample can be interpreted as the area of the region shown here. 4

5 5/3/07 Further Eamples Further Eamples 5

6 5/3/07 Test Qs 5/3/07 3 Evaluate e sin d e does not become simpler when differentiated. Neither does sin become simpler. Nevertheless, we try choosing u = e and dv = sin Then, du = e d and v = cos. e sin d e cos e cos d So, integration by parts gives: The integral we have obtained, e cos d, is no simpler than the original one. At least, it s no more difficult. Having had success in the preceding eample integrating by parts twice, we do it again. 6

7 5/3/07 This time, we use u = e and dv = cos d Then, du = e d, v = sin, and e cos d e sin e sin d At first glance, it appears as if we have accomplished nothing. We have arrived at e sin d, which is where we started. However, if we put the epression for e cos d from Equation 5 into Equation 4, we get: e sin d e cos e sin e sin d This can be regarded as an equation to be solved for the unknown integral. Adding to both sides e sin d, we obtain: e sin d e cos e sin Dividing by and adding the constant of integration, we get: The figure illustrates the eample by showing the graphs of f() = e sin and F() = ½ e (sin cos ). e sin d e (sin cos ) C As a visual check on our work, notice that f() = 0 when F has a maimum or minimum. 7

8 5/3/07 (F) Internet Links Integration by Parts from Paul Dawkins, Lamar University Integration by Parts from Visual Calculus 5/3/