{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

sol3-0708

# sol3-0708 - MAT 137Y 2007–2008 Winter Session Solutions...

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, 2007–2008 Winter Session, Solutions to Term Test 3 1. Evaluate the following integrals, and simplify your answer. (8%) (i) Z ( x + 1 ) 3 x 2 dx . = Z x 3 + 3 x 2 + 3 x + 1 x 2 dx = Z x + 3 + 3 x + 1 x 2 dx = 1 2 x 2 + 3 x + 3ln | x |- 1 x + C . (8%) (ii) Z e 2 x 1 + e 4 x dx . Let u = e 2 x . Then du = 2 e 2 x dx and we have 1 2 Z du 1 + u 2 = 1 2 arctan u = 1 2 arctan ( e 2 x )+ C . (10%) (iii) Z x 2 ( x 2 + 64 ) 3 / 2 dx . By trigonometric substitution, let x = 8tan θ , so dx = 8sec 2 θ d θ . This gives us Z 64tan 2 θ 8 3 sec 3 θ · 8sec 2 θ d θ = Z tan 2 θ sec θ d θ = Z sec 2 θ- 1 sec θ d θ = Z sec θ- cos θ d θ = ln | sec θ + tan θ |- sin θ + C = ln √ x 2 + 64 8 + x 8- x √ x 2 + 64 + C = ln | p x 2 + 64 + x |- x √ x 2 + 64 + C . (10%) (iv) Z ln ( x 2- 9 ) dx . Let I be the integral. By parts, we let u = ln ( x 2- 9 ) , dv = dx , so du = 2 x x 2- 9 dx and v = x . Thus, I = x ln ( x 2- 9 )- Z 2 x 2 x 2- 9 dx = x ln ( x 2- 9 )- 2 Z 1 + 9 x 2- 9 dx = x ln ( x 2- 9 )- 2 x- 18 Z dx ( x + 3 )( x- 3 ) = x ln ( x 2- 9 )- 2 x- 18 Z A x + 3 + B x- 3 dx . Using partial fractions, we get A =- 1 6 and B = 1 6 . Therefore, I = x ln ( x 2- 9 )- 2 x + Z 3 x + 3- 3 x- 3 dx = x ln ( x 2- 9 )- 2 x + 3ln...
View Full Document

{[ snackBarMessage ]}

### Page1 / 4

sol3-0708 - MAT 137Y 2007–2008 Winter Session Solutions...

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

View Full Document
Ask a homework question - tutors are online