Math191_Week11_section

Math191_Week11_section - a R xdx √ 4 x 2(1 1 2 R d(4 x 2...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Math 191 Problem Set for Week 10 Section 1. Evaluate the following integrals using integration by parts: a. R cos - 1 ( x 2 ) dx b. R e - 2 x sin 3 xdx 2. Evaluate the integrals. It may be necessary to use a substitution first. a. R dx x (1+ 3 x ) u = 3 x du = dx 3 x 2 / 3 dx = 3 u 2 du R 3 u 2 du u 3 (1+ u ) = 3 R du u (1+ u ) = 3 ln ± ± ± u u +1 ± ± ± + C = 3 ln ± ± ± 3 x 1+ 3 x ± ± ± + C b. R ds e s +1 u = e s + 1 du = e s ds 2 e s +1 ds = 2 udu u 2 - 1 R 2 udu u ( u 2 - 1) = 2 R du ( u +1)( u - 1) = R du u - 1 - R du u +1 = ln ± ± ± u - 1 u +1 ± ± ± + C = ln ± ± ± e s +1 - 1 e s +1+1 ± ± ± + C 3. Evaluate the integrals (1) without using a trigonometric substitution, (2) using a trigonometric substi- tution.
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: a. R xdx √ 4+ x 2 (1) 1 2 R d (4+ x 2 ) √ 4+ x 2 = √ 4 + x 2 + C (2) [ x = 2 tan y ] → R 2 tan y · 2 sec 2 ydy 2 sec y = 2 R sec y tan ydy = 2 sec y + C = √ 4 + x 2 + C b. R tdt √ 4 t 2-1 (1) 1 8 R d (4 t 2-1) 4 t 2-1 = 1 4 √ 4 t 2-1 + C (2) [ t = 1 2 sec θ ] → R 1 2 sec θ tan θ · 1 2 sec θdθ tan θ = 1 4 R sec 2 θdθ = tan θ 4 + C = √ 4 t 2-1 4 + C 4. R e t √ tan 2 e t + 1 dt 1...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online