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Unformatted text preview: DANANG UNIVERSITY OF TECHNOLOGY CALCULUS WITH ANALYTIC GEOMETRY II Chapter 3 TECHNIQUES OF INTEGRATION I. INTEGRATION BY PARTS The Product Rule states that if f and g are differentiable functions, then In the notation for indefinite integrals this equation becomes [ ] ( ) ( ) ( ) '( ) ( ) '( ) d f x g x f x g x g x f x dx = + DANANG UNIVERSITY OF TECHNOLOGY CALCULUS WITH ANALYTIC GEOMETRY II or We can rearrange this equation as [ ] ( ) '( ) ( ) '( ) ( ) ( ) f x g x g x f x dx f x g x + = ( ) '( ) ( ) ( ) ( ) '( ) f x g x dx f x g x g x f x dx = ( ) '( ) ( ) '( ) ( ) ( ) f x g x dx g x f x dx f x g x + = DANANG UNIVERSITY OF TECHNOLOGY CALCULUS WITH ANALYTIC GEOMETRY II This formula is called the formula for integration by parts . It is perhaps easier to remember in the following notation. Let u=f(x) and v=g(x), then the formula for integration by parts becomes udv uv vdu = DANANG UNIVERSITY OF TECHNOLOGY CALCULUS WITH ANALYTIC GEOMETRY II Example sin x x dx g (i) Find Solution: Let Then and so sin u x dv x dx = = cos du dx v x = =  DANANG UNIVERSITY OF TECHNOLOGY CALCULUS WITH ANALYTIC GEOMETRY II } } } sin sin ( cos ) ( cos ) cos cos cos sin dv v v u u du x x dx x xdx x x x dx x x x dx x x x C = = =  + =  + + 6 64748 647 48 647 48 DANANG UNIVERSITY OF TECHNOLOGY CALCULUS WITH ANALYTIC GEOMETRY II Note Our aim in using integration by parts is to obtain a simpler integral than the one we started with. We usually try to choose u=f(x) to be a function that becomes simpler when differentiated as long as dv=g(x)dx can be readily integrated to give v. DANANG UNIVERSITY OF TECHNOLOGY CALCULUS WITH ANALYTIC GEOMETRY II (ii) Evaluate Solution: We choose Then Integration by parts gives 2 t t e dt g 2 , t u t dv e dt = = 2 , t du tdt v e = = 2 2 2 t t t t e dt t e te dt = DANANG UNIVERSITY OF TECHNOLOGY CALCULUS WITH ANALYTIC GEOMETRY II The integral that we obtained, , is simpler than the original integral but is still not obvious. Therefore, we use integration by parts a second time , this time with Then and t te dt g , t u t dv e dt = = , t du dt v e = = . t t t t t te dt te e dt te e C = = + DANANG UNIVERSITY OF TECHNOLOGY CALCULUS WITH ANALYTIC GEOMETRY II Putting together, we get ( 29 2 2 2 2 2 2 2 2 , where 2 . t t t t t t t t t t e dt t e te dt t e te e C t e te e C C C = = + = + + =  DANANG UNIVERSITY OF TECHNOLOGY CALCULUS WITH ANALYTIC GEOMETRY II (iii) Evaluate Solution: Neither nor sinx becomes simpler when differentiated, but we try choosing and dv=sinxdx anyway. Then Integration by parts gives sin x e x dx g x e x u e = , cos , x du e dx v x = =  sin cos cos x x x e x dx e x e x dx =  + DANANG UNIVERSITY OF TECHNOLOGY...
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 Spring '09
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