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1.4.2 : Integration by parts Managing this process In this section we discuss the technique of integration by parts, which is essentially a reversal of the product rule of differentiation. Example 22 � Find x cos x dx. (uv )� = u � v + uv � . Suppose we integrate both sides here with respect to x. We obtain � � � � � (uv )� dx = u � v dx + uv � dx =⇒ uv = u � v dx + uv � dx. Solution How could x cos x arise as a derivative? Well, cos x is the derivative of sin x. So, if you were differentiating x sin x, you would get x cos x but according to the product rule you would also get � another term, namely sin x. Conclusion: This can be rearranged to give the Integration by Parts Formula : � � � uv dx = uv − u � v dx. x cos x dx = x sin x + cos x + C . Dr Rachel Quinlan What happened in this example was basically that the product rule was reversed. This process can be managed in general as follows. Recall from differential calculus that if u and v are expressions involving x, then MA180/MA186/MA190 Calculus Integration by Parts 61 / 222 Dr Rachel Quinlan MA180/MA186/MA190 Calculus Integration by Parts 62 / 222 An antiderivative for ln x Here is the first example again, handled according to this scheme. Example 23 Use the integration by parts technique to determine Solution: Write � Example 24 � ln x dx. Determine x cos x dx. Solution: Let u = ln x, v � = 1. Then u � = x1 , v = x. � � � ln x dx = uv � dx = uv − u � v dx � 1 x dx = x ln x − x = x ln x − x + C . u = x v � = cos x u � = 1 v = sin x Then � x cos x dx � � uv dx = uv − u � v dx � = x sin x − 1 sin x dx = � Note: This example shows that sometimes problems which are not obvious candidates for integration by parts can be attacked using this technique. = x sin x + cos x + C . Dr Rachel Quinlan MA180/MA186/MA190 Calculus Integration by Parts 63 / 222 Dr Rachel Quinlan MA180/MA186/MA190 Calculus Integration by Parts 64 / 222 Two Rounds of Integration by Parts Two rounds (continued) Sometimes two applications of the integration by parts formula are needed, as in the following example. Let I = Example 25 � x 2 e x dx. Evaluate x 2, � xe x dx To evaluate I apply the integration by parts formula a second time. v� u� ex . u = x v � = ex u� = 1 v = e x . � � x x Then I = xe dx = xe − e x dx = xe x − e x + C . Finally ex . Solution: Let u = = Then = 2x, v = � � � uv � dx = uv − u � v dx x 2 e x dx = � = x 2 e x − 2xe x dx � 2 x = x e − 2 xe x dx. Let I = � � x 2 e x dx = x 2 e x − 2xe x + 2e x + C . xe x dx. Dr Rachel Quinlan MA180/MA186/MA190 Calculus Integration by Parts 65 / 222 Dr Rachel Quinlan � An Example of Another Type Solution Let Then Dr Rachel Quinlan u= u� = e x u = ex u� = e x = cos x v = sin x. e x cos x dx = e x sin x − =⇒ 2 e x sin x dx. MA180/MA186/MA190 Calculus Integration by Parts =⇒ 67 / 222 � � v� � 66 / 222 e x sin x dx: Let and ex � � Then Example 26 � Determine e x cos x dx. Integration by Parts e x cos x dx (continued) For The next example shows another mechanism by which a second application of the integration by parts formula can succeed where the first is not enough. MA180/MA186/MA190 Calculus � � x v � = sin x v = − cos x. x e sin x dx = −e cos x + x x � � e x cos x dx, x e cos x dx = e sin x − −e cos x + e x cos x dx = e x sin x + e x cos x + C e x cos x dx = Dr Rachel Quinlan � x e cos x dx � 1 x (e sin x + e x cos x) + C 2 MA180/MA186/MA190 Calculus Integration by Parts 68 / 222 A Definite Integral Example 27 � 1 Evaluate (x + 3)e 2x dx. 0 Solution: Write u = x + 3, v � = e 2x ; � 1 (x + 3)e 2x dx = 0 = = = Dr Rachel Quinlan � u � = 1, v = 12 e 2x uv � dx = (uv )|10 − � 1 u � v dx 0 � � x + 3 2x ��1 1 1 2x e dx e � − 2 2 0 0 � � x + 3 2x ��1 1 1 2x ��1 e � − × e � 2 2 2 0 0 4 2 3 0 1 2 1 0 7 2 5 e − e − e + e = e − . 2 2 4 4 4 4 MA180/MA186/MA190 Calculus Integration by Parts 69 / 222