sept04pdf - Answers to Test 1 given on Sept 15, 2004 In 1-7...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Answers to Test 1 given on Sept 15, 2004 In 1-7 find each integral. 1. Z sin 2 3 xdx Use a half angle formula which is sin 2 = 1 2 (1- 2cos ) . So Z sin 2 3 xdx = Z 1 2 (1- cos6 x ) dx = 1 2 Z 1 dx- Z cos6 xdx Look at the second term. We want to use the formula Z cos udu = sin u + C. So u = 6 x du = 6 dx = 1 2 Z 1 dx- 1 6 Z cos6 x 6 dx = 1 2 x- 1 6 sin6 x + C = 1 2 x- 1 12 sin6 x + C 2. Z tan 4 x sec 6 xdx Z tan 4 x sec 6 xdx = Z tan 4 x sec 4 x sec 2 xdx = Z tan 4 x (sec 2 x ) 2 sec 2 xdx = Z tan 4 x (1 + tan 2 x ) 2 sec 2 xdx u = tan x du = sec 2 xdx Substituting 1 = u 4 (1 + u 2 ) 2 du = Z u 4 (1 + 2 u 2 + u 4 ) du = Z ( u 4 + 2 u 6 + u 8 ) du = Z u 4 du + 2 Z u 6 du + Z u 8 du Use formula Z u n du = u n +1 n + 1 = 1 5 u 5 + 2 7 u 7 + 1 9 u 9 + C = 1 5 tan 5 x + 2 7 tan 7 x + 1 9 tan 9 x + C 3. Z x 6 ln xdx . Use integration by parts. u = ln x dv = x 6 dx du = 1 x dx v = Z x 6 dx = x 7 7 uv- Z v du = (ln x ) x 7 7- Z x 7 7 1 x dx = x 7 7 ln x- 1 7 Z x 6 dx Use formula Z u n du...
View Full Document

This note was uploaded on 02/21/2010 for the course MAC 2312 taught by Professor Bonner during the Spring '08 term at University of Florida.

Page1 / 5

sept04pdf - Answers to Test 1 given on Sept 15, 2004 In 1-7...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online