HW7 Solution

# HW7 Solution - Kim Jin – Homework 7 – Due 3:00 am –...

This preview shows pages 1–4. Sign up to view the full content.

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

View Full Document

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: Kim, Jin – Homework 7 – Due: Oct 17 2007, 3:00 am – Inst: Diane Radin 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Evaluate the definite integral I = Z π/ 2 sin 2 x cos 3 xdx. 1. I = 4 15 2. I = 1 15 3. I = 2 15 correct 4. I = 8 15 5. I = 2 5 Explanation: Since sin 2 x cos 3 x = sin 2 x cos 2 x cos x = sin 2 x (1- sin 2 x )cos x, we see that I can be written as the sum I = Z π/ 2 sin 2 x (1- sin 2 x )cos xdx = Z π/ 2 sin 2 x cos xdx- Z π/ 2 sin 4 x cos xdx, of two integrals, both of which can be eval- uated using the substitution u = sin x . For then du = cos xdx, while x = 0 = ⇒ u = 0 x = π 2 = ⇒ u = 1 . Thus I = Z 1 u 2 du- Z 1 u 4 du = h 1 3 u 3- 1 5 u 5 i 1 . Consequently, I = 2 15 . keywords: Stewart5e, indefinite integral, powers of sin, powers of cos, trig substitu- tion, 002 (part 1 of 1) 10 points Find the value of the definite integral I = Z π 6 cos 4 xdx. 1. I = 9 π 32- 1 4 2. I = 3 π 16 3. I = π 8 4. I = π 4 5. I = π 16 + 9 √ 3 64 correct 6. I = 3 π 32 + 1 64 Explanation: Since cos 2 x = 1 2 (1 + cos2 x ) , we see that cos 4 x = 1 4 (1 + cos2 x ) 2 = 1 4 ( 1 + 2cos2 x + cos 2 2 x ) . Thus by trig identities yet again, cos 4 x = 1 8 (3 + 4cos2 x + cos4 x ) . Kim, Jin – Homework 7 – Due: Oct 17 2007, 3:00 am – Inst: Diane Radin 2 Consequently, I = Z π 6 1 8 (3 + 4cos2 x + cos4 x ) dx = • 3 8 x + 1 4 sin2 x + 1 32 sin4 x ‚ π 6 But sin π 3 = √ 3 2 , sin 2 π 3 = √ 3 2 Consequently, I = π 16 + 9 √ 3 64 . keywords: trig identity, integral 003 (part 1 of 1) 10 points Determine the indefinite integral I = Z x (3 cos 2 x + sin 2 x ) dx. 1. I = x 2- 1 2 x sin2 x + 1 2 cos2 x + C 2. I = x 2 + 1 2 x sin2 x + 1 4 cos2 x + C correct 3. I = 1 2 x 2- x cos2 x- 1 2 sin2 x + C 4. I = x 2- 1 2 x cos2 x- 1 4 sin2 x + C 5. I = 1 2 x 2 + x sin2 x + 1 2 cos2 x + C Explanation: Since cos 2 x = 1 2 (1 + cos2 x ) and sin 2 x = 1 2 (1- cos2 x ) , we see that I = 1 2 Z x { 3(1 + cos2 x ) +1- cos2 x } dx = 2 Z xdx + Z x cos2 xdx = x 2 + Z x cos2 xdx. But after integration by parts, Z x cos2 xdx = 1 2 x sin2 x- 1 2 Z sin2 xdx = 1 2 x sin2 x + 1 4 cos2 x + C . Consequently, I = x 2 + 1 2 x sin2 x + 1 4 cos2 x + C . keywords: trigonometric identities, integra- tion by parts 004 (part 1 of 1) 10 points Evaluate the indefinite integral I = Z 1- sin x cos x dx. 1. I =- ln(1- cos x ) + C 2. I = ln(1 + sin x ) + C correct 3. I =- ln(1 + cos x ) + C 4. I = ln(1 + cos x ) + C 5. I = ln(1- sin x ) + C Explanation: Kim, Jin – Homework 7 – Due: Oct 17 2007, 3:00 am – Inst: Diane Radin 3 Z 1- sin x cos x dx = Z (sec x- tan x ) dx = ln | sec x + tan x | - ln | sec x | + C = ln | (sec x + tan x )cos x | + C = ln(1 + sin x ) + C keywords: trig integral 005 (part 1 of 1) 10 points Find the volume obtained by rotating the region bounded by the curves y = sin x , x = π 2 , x...
View Full Document

## This note was uploaded on 04/12/2010 for the course PHY 58195 taught by Professor Turner during the Spring '09 term at University of Texas.

### Page1 / 11

HW7 Solution - Kim Jin – Homework 7 – Due 3:00 am –...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online