MTH 252 Integral Calculus Chapter 8 Principles of Integral Evaluation Section 8 2 Integration by Parts Copyright 2005 by Ron Wallace all rights reserved Review The Product Rule d f x g x f x g x f x g x dx Example d d 2 2 2 d x tan x x tan x x tan x dx dx dx 2 x tan x x 2 sec 2 x Integration Equivalent d f x g x f x g x f x g x dx f x g x f x g x f x g x dx f x g x f x g x dx f x g x dx f x g x dx f x g x f x g x dx Integration Equivalent Let u f x and dv g x dx du f x dx and v g x dx g x u dv uv v du f x g x dx f x g x f x g x dx Integration by Parts f x g x dx u dv u dv du f x dx v g x dx u dv uv v du NOTE f x or g x can be equal to 1 but not both u dv uv v du Example xe 2x dx u x du dx 1 2x 1 2x xe e dx 2 2 dv e 2 x dx 1 2x v e 2 1 2x 1 2x xe e c 2 4 u dv uv v du Choosing u dv In general u should be such that du is simpler dv should must be easy to integrate Previous Example xe dx 2x 1 2 2x xe 2 u e 2 x du 2e 2 x dx xe 2 2x dx dv xdx 1 2 v x 2 This made the problem MORE difficult Repeated Integration by Parts u dv uv v du Sometimes the resulting integral is integrated by the same method Example x cos x dx 2 x sin x 2 x sin x dx 2 u x 2 du 2 x dx dv cos x dx v sin x u x dv sin x dx du dx v cos x x 2 sin x 2 x cos x cos x dx x 2 sin x 2 x cos x 2sin x c Circular Repeated Integration by Parts Example e cos x dx x e sin x e sin x dx x x u dv uv v du u e x du e x dx dv cos x dx v sin x u e x dv sin x dx du e x dx v cos x x x x x e cos x dx e sin x e cos x e cos x dx 2 e cos x dx e sin x e cos x x x x x x e sin x e cos x x e cos x dx c 2