MTH 252 Integral Calculus Chapter 6 Integration Section 6 3 Integration by Substitution Copyright 2005 by Ron Wallace all rights reserved A Differentiation Example d 2 sin x 3 dx d sin u dx where u x 2 3 du cos u dx cos x 2 3 2 x Therefore 2 2 2 x cos x 3 dx sin x 3 c Let u x 2 3 du 2 xdx cos x 2 3 2 x dx cos u du sin u c sin x 2 3 c This example demonstrates the process known as Integration by Substitution Substitution Reversing the Chain Rule Chain Rule d f g x f g x g x dx Integration Equivalent f g x g x dx f g x c Let u g x du g x dx f u du f u c The Substitution Method f x dx 1 Find u g x and du g x dx where f x dx h g x g x dx h u du 2 Evaluate h u du H u c 3 Subustute u g x giving f x dx H g x c Example 1 x 5 dx 3 Let u x 5 du dx 3 43 u du u c x 5 dx 4 3 13 3 43 x 5 c 4 Example 2 Off by a constant multiple 7 x 5 dx 3 Let u 7 x 5 du 7dx 1 3 1 13 u du 7 x 5 dx 7 7 x 5 7dx 7 3 1 3 43 u c 7 4 3 43 7 x 5 c 28 Example 2 Alternative approach 7 x 5 dx 3 Let u 7 x 5 du 7dx 1 du dx 7 1 1 13 u du u du 7 x 5 dx 7 7 3 13 1 3 43 u c 7 4 3 43 7 x 5 c 28 Example 3 e x 1 e Let u e x ex 1 e du e x dx 1 2x 2x dx 1 dx e dx du 2 x 2 1 u 1 e 1 sin u c x sin 1 e x c Example 4 dx dx 1 dx dx 2 2 2 9 x x2 9 x 99 1 x 1 3 9 x 1 Let u du dx 3du dx 3 3 1 3du 1 1 2 2 du 9 1 u 3 1 u 1 1 tan u c 3 1 1 x tan c 3 3