Supplementary Problem Set 2(Solution)

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

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 ....
View Full Document

## 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.

### Page1 / 3

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online