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

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

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

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