exam2_fa00_answer

exam2_fa00_answer - Math 20B (Bender) Second Exam November...

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

View Full Document Right Arrow Icon
Math 20B (Bender) Second Exam November 2000 1. (80 pts.) Evaluate the following integrals. Remember to show your work! (a) [7.5, #6] Z (sin x ) (cos(cos x )) dx Ans. Use u = cos x to get rid of the messy cosine stuff. Then du = - sin xdx and so we have R - cos udu = - sin u + C , which gives - sin(cos x )+ C . (b) [7.5, #10] Z t 2 1 - t 2 dt Ans. Set t = sin θ . Then dt = cos θd θ and the integral becomes R sin 2 θd θ = (1 / 2) R (1 - cos 2 θ ) , which is ( θ/ 2) - (sin 2 θ ) / 4+ C . You now must substi- tute back in for θ : sin - 1 t 2 + sin(2 sin - 1 t ) 4 + C. You can leave your answer in this form; however, you could simplify it using sin(2 θ ) = 2 sin θ cos θ =2 t 1 - t 2 . (c) [7.5, #26] Z sin( t 1 / 2 ) dt Ans. Set t 1 / 2 = x to obtain dt =2 xdx and R 2 x sin xdx . This integral can be done by parts with x = u and sin xdx = dv . We obtain - 2 x cos x +2 R cos xdx
Background image of page 1

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

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 04/29/2010 for the course MATH MATH 20B taught by Professor Takeda during the Spring '07 term at UCSD.

Page1 / 2

exam2_fa00_answer - Math 20B (Bender) Second Exam November...

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

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