Lenarz Math 262 Exam #3 Review October 30, 2012 Name: Directions: Answer the following questions on a separate piece of paper. You may NOT use a calculator. If you do not show your work, you will receive no credit. 1. Evaluate the following integrals. Z 1 √ (a) dx x2 16 − x2 Solution: Use the trig substitution x = 4 sin θ so that dx = 4 cos θ dθ and √ 2 16 − x = 4 cos θ. Then we have Z Z 1 1 √ (4 cos θ) dθ dx = 2 2 2 (4 sin θ) (4 cos θ) x 16 − x Z 1 1 dθ = 2 16 Z sin θ 1 = csc2 θ dθ 16 1 = − cot θ + C 16 1 cos θ = − +C 16 sin θ ! √ 16−x2 1 4 +C = − x 16 4 √ 16 − x2 = − +C 16x Z (b) x3 1 dx +x Solution: Do a partial fraction decomposition: 1 1 A Bx + C = = + 2 x3 + x x(x2 + 1) x x +1 So we have 1 = A(x2 + 1) + (Bx + C)x = (A + B)x2 + Cx + A and hence A+B =0 C=0 A=1 Math 262 October 30, 2012 Page 2 Thus A = 1, B = −1 and C = 0 and we have Z Z 1 1 x dx = − dx x3 + x x x2 + 1 Z Z 1 x = dx − dx x x2 + 1 Z Z 1 1 1 = dx − du x 2 u 1 = ln |x| − ln |u| + C 2 1 = ln |x| − ln(x2 + 1) + C 2 where the substitution u = x2 + 1 (so that du = 2x dx) was used in the third line. Z (c) x2 x dx + 6x + 13 Solution: Complete the square in the denominator: Z Z x x dx = dx 2 x + 6x + 13 (x + 3)2 + 4 Z u−3 = du u2 + 4 Z Z u 1 = du − 3 du u2 + 4 u2 + 4 Z Z 1 1 1 = dw − 3 du 2 2 w u + 22 1 1 −1 u ln |w| − 3 tan = +C 2 2 2 1 3 2 −1 u = ln(u + 4) − tan +C 2 2 2 3 x+3 1 2 −1 +C = ln(x + 6x + 13) − tan 2 2 2 where the substitution u = x + 3 (so du = dx) was made in the second line, and the substitution w = u2 + 4 (so that dw = 2u du) was made in the fourth line. Math 262 Z (d) October 30, 2012 √ Page 3 x dx 4 − x2 Solution: Let u = 4 − x2 so that du = −2x dx. Then we have Z Z x 1 √ dx = − u−1/2 du 2 2 4−x 1 = − (2u1/2 ) + C 2 √ = − 4 − x2 + C Z (e) x2 7 dx − 6x + 18 Solution: Complete the square in the denominator. Z Z 7 7 dx = dx 2 x − 6x + 18 (x − 3)2 + 9 Z 1 = 7 du 2 u + 32 1 −1 u tan +C = 7 3 3 7 x−3 −1 = tan +C 3 3 where the substitution u = x − 3 (so that du = dx) was made in the second line. Z (f) √ x sin √ x dx √ Solution: Let w = x so that w2 = x and 2w dw = dx. Then we have Z Z √ √ x sin x dx = (w sin w)(2w) dw Z = 2w2 sin w dw Now use integration by parts with u = 2w2 dv = sin w dw so that du = 4w dw v = − cos w Math 262 October 30, 2012 and we have Z 2 Page 4 Z 2 2w sin w dw = (2w )(− cos w) − (− cos w)(4w) dw Z 2 = −2w cos w + 4w cos w dw Use integration by parts again with u = 4w dv = cos w dw so that du = 4 dw v = sin w and we have 2 −2w cos w + Z Z 2 4w cos w dw = −2w cos w + (4w)(sin w) − (sin w)(4) dw Z 2 = −2w cos w + 4w sin w − 4 sin w dw = −2w2 cos w + 4w sin w + 4 cos w + C √ √ √ √ √ = −2( x)2 cos x + 4 x sin x + 4 cos x + C √ √ √ = (−2x + 4) cos x + 4 x sin x + C Z (g) x3/2 ln x dx Solution: Use integration by parts with u = ln x so that du = 1 dx x dv = x3/2 dx 2 v = x5/2 5 Then we have Z Z 2 5/2 2 5/2 1 3/2 x ln x dx = (ln x) x − x dx 5 5 x Z 2 5/2 2 = x ln x − x3/2 dx 5 5 2 5/2 2 2 5/2 = x ln x − x +C 5 5 5 2 5/2 4 = x ln x − x5/2 + C 5 25 Math 262 Z (h) October 30, 2012 Page 5 x √ dx 3 x−1 √ √ Solution: Let u = 3 x − 1 so that u + 1 = 3 x and (u + 1)3 = x. This means that 3(u + 1)2 du = dx and we have Z Z x (u + 1)3 √ dx = (3(u + 1)2 ) du 3 u x−1 Z (u + 1)5 = du u Z 1 4 3 2 u + 5u + 10u + 10u + 5 + = 3 du u 1 5 5 4 10 3 2 = 3 u + u + u + 5u + 5u + ln |u| + C 5 4 3 √ √ 3 √ 15 √ = ( 3 x − 1)5 + ( 3 x − 1)4 + 10( 3 x − 1)3 + 15( 3 x − 1)2 5 4 √ √ +15( 3 x − 1) + 3 ln | 3 x − 1| + C Z (i) sin4 x cos3 x dx Solution: Let u = sin x so that du = cos x dx. Then we have Z Z 4 3 sin x cos x dx = sin4 x cos2 x cos x dx Z = sin4 x(1 − sin2 x) cos x dx Z = u4 (1 − u2 ) du Z = (u4 − u6 ) du 1 5 1 7 u − u +C 5 7 1 5 1 = sin x − sin7 x + C 5 7 =