Definite, indefinite integrals, and improper integrals.

May 19th, 2015
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how to use partial fractions, substitution and finding the arc length. Also examples and explanations on undetermined forms in Calculus are found in this document

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Mathematics 112 Spring 11: Solutions to homework #1Definite and indefinite integralsZi)2xdx. Using partial fraction,(x 1)(x2 + 1)2xABx + C=+ 2,2(x 1)(x + 1)x1x +1which leads to A = C = 1, B = 1. Hence,ZZZ2x1x + 1dx =dx +dx2(x 1)(x + 1)x1x2 + 1ZZZx11dx dx +dx=22x1x +1x +11= ln |x 1| ln |x2 + 1| + arctan x + C2Z2xii)dx. Again, using partial fraction,(x 1)(x + 1)2ABC2x=++,2(x 1)(x + 1)x 1 x + 1 (x + 1)2which leads to A = 12 , B = 21 , C = 1. Hence,ZZZZZ2x11111dx =dx dx +dx2(x 1)(x + 1)2x12x+1(x + 1)2111= ln |x 1| ln |x + 1| 22x+1/2iii)sin3 (x) cos3 (x)dx. We want to use substitution./6Z/23/2Z3sin3 (x) cos2 (x) cos(x)(x)dxsin (x) cos (x)dx =/6/6/2Zsin3 (x)(1 sin2 (x)) cos(x)(x)dx=/61Z321Z(u3 u5 )duu (1 u )du =u = sin(x), du = cos(x)dx =1212u4 u6=469=128 1 11111 =4 664 384212Z /2iv)x3 sin(x2 )dx. We need to use both substitution and integration by parts. First let0t = x2 , dt = 2xdx, so thatZ /20/2Z1x sin(x )dx =232t sin(t)dt0Then we integrate by parts:((u=tdu = dtdv = sin(t)dtv = cos(t)Z /2Z/2 1 /211t sin(t)dt = t cos(t) +cos(t)dt(remember that cos(/2) = 0)2 022 00/2 11= sin(t)

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