Integration by Change of Variables Use a change of variables to compute the following integrals. Change both the variable and the limits of substitution. � 4 √ a) 3x + 4 dx 0 � 3 b) x2 1 � x dx +1 π/2 sin5 x cos x dx c) 0 Solution � 4 √ a) 3x + 4 dx 0 � The integrand of 4 √ 3x + 4 dx consists of a (relatively complicated) square 0 root function with a simple expression inside it. Expressions used as input to complicated functions are good candidates for substitution; if we let u = 3x + 4 then du = 3x dx and dx = 13 du. When x = 0, u = 4 and when x = 4, u = 16. We now know enough to rewrite the integrand and limits of integration in terms of u, after which we can complete the integral. � x=4 � u=16 √ √ 1 3x + 4 dx = u du 3 x=0 u=4 � 16 1 = u1/2 du 3 4 �16 � 1 2 3/2 = u 3 3 4 � 2 � 3/2 16 − 43/2 = 9 2 (64 − 8) = 9 112 = ≈ 12 9 We can see from a rough graph of the function that the value of the integral must be between 8 and 16, so this is a reasonable answer. � b) 1 3 x dx x2 + 1 In this example, the most complicated operation is division by x2 + 1. We’ll try the substitution u = x2 + 1: u = x2 + 1, du = 2x dx, x dx = 1 du. 2 When x = 1, u = 2. When x = 3, u = 10. � 3 1 x dx x2 + 1 � x=3 = 1 · x dx x2 + 1 x=1 u=10 � = u=2 = = = 1 1 du u 2 1 u=10 ln(u)|u=2 2 1 (ln(10) − ln(2)) 2 � � √ 1 10 ln = ln( 5) ≈ 1 2 2 This answer is not easy to check, but by using an electronic graphing tool we can conﬁrm that it is approximately correct. � c) π/2 sin5 x cos x dx 0 Although the sine and cosine functions may be the most complicated part of this problem, performing the substitution u = x does not help us integrate this expression. We shift our focus to the next most complicated operation – exponentiation. Let u = sin(x). Then du = cos(x) dx and when x = 0, u = 0; when x = π/2, u = 1. � π/2 � 5 x=π/2 (sin x)5 · cos x dx sin x cos x dx = 0 x=0 � 1 u5 du = 0 �1 1 6 �� u � 6 � 0 � 1 6 1 6 ·1 − ·0 = 6 6 1 = 6 = Again the answer is hard to check without graphing. It seems reasonable given the fact that 0 ≤ sin x ≤ 1 on the interval in question.