ln7 1 5 2 ln7 11 2 ln11 7 2 ln7 2 438 17 1 e lnx 2 dx Put u lnx 2 Then dv dx du

# Ln7 1 5 2 ln7 11 2 ln11 7 2 ln7 2 438 17 1 e lnx 2 dx

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ln(7) 1 + 5 2 . ln|7| } = 11 2 . ln(11) – 7 2 . ln(7) – 2 4.38 . 17. 1 e ( ln(x) ) 2 dx . Put u = ( ln(x) ) 2 . Then dv = dx, du = 2 . ln(x) . 1 x dx , and v = x. = uv – v du = ( ln(x) ) 2 x x . 2 . ln(x) . 1 x dx = x . ( ln(x) ) 2 2 . ln(x) dx = x . ( ln(x) ) 2 – 2{ x . ln(x) – x } | e 1 = { e( ln(e) ) 2 – 2e . ln(e) + 2e } – { 1( ln(1) ) 2 – 2 . ln(1) + 2 } = e – 2 0.718 . 19. arcsin(x) dx . Put u = arcsin(x). Then dv = dx, du = 1 1 – x 2 dx , and v = x . = uv – v du = arcsin(x) . x – x . 1 1 – x 2 dx = x . arcsin(x) – x 1 – x 2 dx (use u–sub with u = 1 – x 2 ) = x . arcsin(x) – 1 u ( –1 2 ) du = x . arcsin(x) + u + C = x . arcsin(x) + 1 – x 2 + C .
Odd Answers: Chapter Eight Contemporary Calculus 6 6 21. x . arctan(3x) dx . Put u = arctan(3x). Then dv = x dx, du = 3 1 + 9x 2 dx, and v = 1 2 x 2 . = uv – v du = arctan(x) . 1 2 x 2 1 2 x 2 . 3 1 + 9x 2 dx = 1 2 x 2 . arctan(3x) – 1 2 1 3 1/3 1 + 9x 2 dx = 1 2 x 2 . arctan(3x) – 1 2 { x 3 1 9 . arctan(3x) } + C = 1 2 x 2 . arctan(3x) – 1 6 x + 1 18 arctan(3x) } + C . 23. 1 2 ln(x) x dx. Use u–substitution! Put u = ln(x). Then du = 1 x dx. = u du = 1 2 u 2 = 1 2 ( ln(x) ) 2 | 2 1 = 1 2 ( ln(2) ) 2 1 2 ( ln(1) ) 2 = 1 2 ( ln(2) ) 2 0.240 . 25. (a) sin 3 (x) dx = 1 3 { –S 2 . C + 2 S dx } = 1 3 { –S 2 . C – 2C } + K = 1 3 { –sin 2 (x) . cos(x) – 2 . cos(x) } + K . (b) sin 4 (x) dx = 1 4 { –S 3 . C + 3 S 2 dx } = 1 4 { –S 4 . C + 3[ 1 2 ( –SC + x ) ] } + K = 1 4 { –S 3 . C – 3 2 SC + 3 2 x } + K = 1 4 { –sin 3 (x) . cos(x) – 3 2 sin(x) . cos(x) + 3 2 x } + K (c) on your own. 27. (a) sec 3 (x) dx = 1 2 { sec(x)tan(x) + sec(x) dx } = 1 2 { sec(x) . tan(x) + ln| sec(x) + tan(x) | } + K (b) and (c) on your own. 29. cos 3 (2x + 3) dx . First do a substitution: u = 2x + 3. Then du = 2 dx and dx = 1 2 du . = 1 2 cos 3 ( u ) du = 1 2 { 1 3 ( C 2 . S + 2 C du ) } = 1 6 { C 2 . S + 2S } + K = 1 6 { cos 2 (u) . sin(u) + 2sin(u) } + K = 1 6 { cos 2 (2x + 3) . sin(2x + 3) + 2sin(2x + 3) } + K 31. x . (2x + 5) 19 dx . (a) By parts: put u = x. Then dv = (2x + 5) 19 dx , du = dx , and v = 1 40 (2x + 5) 20 . = uv – v du = x . 1 40 (2x + 5) 20 1 40 (2x + 5) 20 dx = x . 1 40 (2x + 5) 20 1 40 1 42 (2x + 5) 21 + C. (b) Substitution: put u = 2x + 5. Then du = 2 dx and dx = 1 2 du. Also, x = 1 2 ( u – 5 ) . = 1 2 ( u – 5 ) . u 19 1 2 du = 1 4 u 20 – 5u 19 du = 1 4 { 1 21 u 21 5 20 u 20 } + C = 1 84 (2x + 5) 21 5 80 (2x + 5) 20 + C. The answers (antiderivatives) in parts (a) and (b) look different, but you can check that the derivative of each answer is x . (2x + 5) 19 .
Odd Answers: Chapter Eight Contemporary Calculus 7 7 33. (a) Make an informed prediction. (b) 0 1 sin(x) dx = –cos(x) | 1 0 = (–cos(1)) – (–cos(0)) = cos(0) – cos(1) 1 – 0.54 = 0.46 . 0 1 x . sin(x) dx = –x . cos(x) + sin(x) | 1 0 = {–1 . cos(1) + sin(1)} – {–0 . cos(0) + sin(0)} = sin(1) – cos(1) 0.30 . 35. (a) Make an informed prediction. (b) See problem 17: V x–axis = x=1 e π ( ln(x) ) 2 dx = π (e – 2) 2.257 . V y–axis = y=0 1 2 π . y . e y dy = 2 π y=0 1 y . e y dy (use integration by parts with u = y, dv = e y dy : see Example 2) = 2 π ( y . e y – e y ) | 1 0 = 2 π (1 . e 1 – e 1 ) – 2 π (0 . e 0 – e 0 ) = 2 π (0) – 2 π (–1) = 2 π 6.283 .