rev-ans - P Vojta Math 1B Solutions to Review Problems 9...

Info iconThis preview shows pages 1–4. 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

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: P. Vojta Math 1B Solutions to Review Problems 9 December 2009 1. (a). Substitute u = sin x : Z sin 4 x cos 3 x dx = Z u 4 (1- u 2 ) du = u 5 5- u 7 7 + C = sin 5 x 5- sin 7 x 7 + C. (b). Integrate by parts with u = arcsin x dv = dx du = dx √ 1- x 2 v = x The indefinite integral is Z arcsin x dx = x arcsin x- Z x dx √ 1- x 2 = x arcsin x + 1 2 Z du √ u = x arcsin x + √ u + C = x arcsin x + p 1- x 2 + C, so Z 1 arcsin x dx = x arcsin x + p 1- x 2 1 = π 2- 1 . (Note: had we computed the definite integral directly, this would have involved an improper integral after the integration by parts.) (c). Integrate by parts with u = sin2 x dv = e x dx du = 2cos2 x dx v = e x Then Z e x sin2 x dx = e x sin2 x- 2 Z e x cos2 x dx. Integrate by parts again with u = cos2 x dv = e x dx du =- 2sin2 x dx v = e x : Z e x sin2 x dx = e x sin2 x- 2 e x cos2 x + 2 Z e x sin2 x dx ! 5 Z e x sin2 x dx = e x sin2 x- 2 e x cos2 x + C Z e x sin2 x dx = e x 5 ( sin2 x- 2cos2 x ) + C. 1 2 (d). Let u = 4- x 2 ; then du =- 2 x dx and Z 2 x 3 p 4- x 2 dx =- 1 2 Z 4 (4- u ) √ u du =- 1 2 8 3 u 3 / 2- 2 5 u 5 / 2 4 =- 32 3 + 32 5 = 64 15 . (e). Integrate by parts with u = x dv = cos2 x dx du = dx v = 1 2 sin2 x to get Z x sin 2 x dx = 1 2 Z x (1- cos2 x ) dx = x 2 4- 1 2 Z x cos2 x dx = x 2 4- 1 2 x 2 sin2 x- 1 2 Z sin2 x dx = x 2 4- x 4 sin2 x- 1 8 cos2 x + C. (f). Substitute u = tan 2 θ : Z π/ 4 tan 2 θ sec 4 θ dθ = Z 1 u 2 ( u 2 + 1) du = u 5 5 + u 3 3 1 = 1 5 + 1 3 = 8 15 . (g). Let x = tan θ ; then dx = sec 2 θ : Z dx (1 + x 2 ) 2 = Z sec 2 θ sec 4 θ dθ = Z cos 2 θ dθ = 1 2 Z (1 + cos2 θ ) dθ = θ 2 + 1 4 sin2 θ + C = 1 2 arctan x + 1 4 sin(2arctan x ) + C = 1 2 ( arctan x + sin(arctan x )cos(arctan x ) ) + C = 1 2 arctan x + tan(arctan x ) sec 2 (arctan x ) + C = 1 2 arctan x + x x 2 + 1 + C. 3 It’s OK if you got as far as 1 2 arctan x + 1 4 sin(2arctan x ) + C . (h). The denominator factors as ( x +1)( x 2- 4 x +7). So we look for partial fractions: x 2 + 7 x- 6 ( x + 1)( x 2- 4 x + 7) = A x + 1 + Bx + C x 2- 4 x + 7 ; x 2 + 7 x- 6 = A ( x 2- 4 x + 7) + ( Bx + C )( x + 1) . Successively setting x equal to- 1, 0, and 1 gives the equations- 12 = 12 A,- 6 = 7 A + C, and 2 = 4 A + 2( B + C ) . Therefore we can read off: A =- 1, C = 1, and B = 2. Thus Z x 2 + 7 x- 6 ( x + 1)( x 2- 4 x + 7) dx =- Z dx x + 1 + Z 2 x + 1 x 2- 4 x + 7 dx =- ln | x + 1 | + Z 2( x- 2) + 5 ( x- 2) 2 + 3 dx =- ln | x + 1 | + ln | x 2- 4 x + 7 | + 5 √ 3 arctan x- 2 √ 3 + C. (i). Let u = 1 + √ x ; then x = ( u- 1) 2 ; dx = 2( u- 1) du and the integral is Z dx (1 + √ x ) 3 = Z 2( u- 1) du u 3 =- 2 u + 1 u 2 + C =- 2 1 + √ x + 1 (1 + √ x ) 2 + C. 2. Completing the square gives x 4- 4 x = ( x- 2) 2- 4, so substitute u = x- 2 and apply Formulas 39 and 40 from the table of integrals: Z x 2 p x 2- 4 x dx = Z ( u + 2) 2 p u 2- 4 du = Z 4 u p u 2- 4 du + u 8 (2 u 2- 4) p u 2- 4- 2ln u + p u 2- 4 + 4 u 2 p u 2- 4- 2ln u + p u 2- 4 = 4 3 ( u 2- 4) + u 4 ( u 2 + 6) p u 2- 4- 10ln...
View Full Document

This note was uploaded on 03/25/2011 for the course MATH 1B taught by Professor Reshetiken during the Fall '08 term at Berkeley.

Page1 / 20

rev-ans - P Vojta Math 1B Solutions to Review Problems 9...

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

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