Integration Questions (1 Viewer)

appletini171 · Mar 23, 2009

Hey guys, I'm trying to do some work, and I'm struggling to get the answers out. Help would be massively appreciated.

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a) Use partial fractions to evaluate

b) Hence, and by using the substitution tan(<link rel="File-List" href="file:///C:%5CDOCUME%7E1%5CDefault%5CLOCALS%7E1%5CTemp%5Cmsohtmlclip1%5C01%5Cclip_filelist.xml"><link rel="themeData" href="file:///C:%5CDOCUME%7E1%5CDefault%5CLOCALS%7E1%5CTemp%5Cmsohtmlclip1%5C01%5Cclip_themedata.thmx"><link rel="colorSchemeMapping" href="file:///C:%5CDOCUME%7E1%5CDefault%5CLOCALS%7E1%5CTemp%5Cmsohtmlclip1%5C01%5Cclip_colorschememapping.xml"><!--[if gte mso 9]><xml> <w:WordDocument> <w:View>Normal</w:View> <w:Zoom>0</w:Zoom> <w:TrackMoves/> <w:TrackFormatting/> <w

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Q2
(a) Show that the reduction formula for:-

is

appletini171 · Mar 23, 2009

(b) Hence or otherwise, evaluate,

Sorry, it wouldn't let me post the last image in the first message

Drongoski · Mar 23, 2009

5647382910 · Mar 23, 2009

1.
a) using partial fractions:
integral from 1 to 0 (5/(2t + 1)(2 - t) dt
= integral from 1 to 0 [(2 / 2t + 1) + (1/2 - t)] dt
= [ln(2t + 1) - ln(2 - t)]
= ln6

b)let t = tan a/2
therefore da = (2/(1 + t^2)) dt
when a = pi/2 t = 1
when a = 0 t = 0
therefore:
= integral from one to zero ((1)/(-2t^2 + 3t + 2)) dt
= 1/5 integral from one to zero (5/(2t + 1)(2 - t) dt
which using part a) = 1/5 ln6

2. I = integral (1 x (ln x )^n)
integrating by parts
= x (ln x)^n - integral (n (ln x)^n-1)
= x(ln x)^n -nI_(n-1)

mys solution in 2 is differernt from wats required to prove
someone correct me if im wrong

Timothy.Siu · Mar 23, 2009

i know for 1. b) it says hence, but note that u can also do that using transformations
3sina+4cosa=5cos (a-pi/4)

0.2 S sec(a-pi/4) da = ln |sec(a-pi/4)+tan(a-pi/4)|

Integration Questions (1 Viewer)

appletini171

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appletini171

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Drongoski

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5647382910

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Timothy.Siu

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