Supplementary Problem Set 2(Solution)

Supplementary Problem Set 2(Solution) - MAT 137Y 2009-10...

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Unformatted text preview: MAT 137Y 2009-10 Winter Session, Solutions to selected problems in Problem Set Supplement 2 5 (i) Let u = e 2 x dx , then du = 2 e 2 x dx . Then Z e 2 x 1 +( e 2 x ) 2 dx = 1 2 Z du 1 + u 2 = 1 2 arctan e 2 x + C . (ii) The integral involves power of trig functions, so Z π / 4 tan 5 θ sec 3 θ d θ = Z π / 4 ( tan 2 θ ) 2 sec 2 θ · sec θ tan θ d θ . Substituting u = sec θ = ⇒ du = sec θ tan θ d θ , we have Z √ 2 1 ( u 2- 1 ) 2 u 2 du = Z √ 2 1 ( u 6- 2 u 4 + u 2 ) du = 1 7 u 7- 2 5 u 5 + 1 3 u 3 √ 2 1 = 8 7 √ 2- 8 5 √ 2 + 2 3 √ 2- 1 7- 2 5 + 1 3 = 22 105 √ 2- 8 105 = 2 105 ( 11 √ 2- 4 ) . (iii) Let u = √ 1 + ln x , then u 2 = 1 + ln x and so 2 udu = 1 x dx . Hence, Z √ 1 + ln x x ln x dx = Z u u 2- 1 · 2 udu = 2 Z u 2 u 2- 1 du = 2 Z 1 + 1 u 2- 1 du and now we solve using partial fractions: 2 Z 1 + 1 u 2- 1 du = 2 u + Z 1 u- 1- 1 u + 1 du = 2 u + ln | u- 1 |- ln | u + 1 | + C = 2 √ 1 + ln x + ln | √ 1 + ln x- 1 |- ln | √ 1 + ln x + 1 | + C ....
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This note was uploaded on 10/13/2011 for the course MATHEMATIC MAT 137 taught by Professor Brainpigott during the Fall '10 term at University of Toronto.

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Supplementary Problem Set 2(Solution) - MAT 137Y 2009-10...

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