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Integration by substitution then partial fractions, Exams of Mathematics
Integration by substitution then partial fractions
Typology: Exams
2016/2017
1 / 1
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1+cosx
tanx i dy
[~~ = | eco let y = cosx > dx = —-—_
1+ cosx cosx(1+cosx) Sinx
dy
y(1+ y)
There fore,— |
“ -1=AQy+1)+ By
Hence,when y = 0,A = —1,when y= -—1,B=1
tanx
As @ consequence, | dx = —In(cosx) + in(1+cesx)+ec
1+ ¢osx
MR.OSAMA A.A. ALSATTARI
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