Exam 2 Review Answers.pdf - Exam 2 Review Answers with Selected Solutions Evaluate the following integrals using integration by parts Z \u03c0\/2 \u03c0 1 x cos

Exam 2 Review Answers.pdf - Exam 2 Review Answers with...

This preview shows page 1 - 5 out of 15 pages.

Exam 2 Review Answers with Selected Solutions Evaluate the following integrals using integration by parts: 1. Z π/ 2 0 x cos x dx = π 2 - 1 2. Z ln x dx = x ln x - x + c 3. Z e 1 x 2 ln x dx = 2 e 3 9 + 1 9 4. Z 1 0 tan - 1 x dx Solution: Let u = tan - 1 x , dv = dx . Then du = 1 x 2 +1 dx and v = x , then Z 1 0 tan - 1 x dx = x tan - 1 x 1 0 - Z 1 0 x x 2 + 1 dx Letting w = x 2 + 1, then dw = 2 x dx . then we have x tan - 1 x 1 0 - Z 1 0 x x 2 + 1 dx = x tan - 1 x 1 0 - 1 2 Z 2 1 1 w dw = x tan - 1 x 1 0 - 1 2 ln | w | 2 1 = π 4 - 1 2 ln(2) 5. Z x 4 sin(3 x ) dx Solution: Using the tabular method: u dv + x 4 sin(3 x ) - 4 x 3 - cos(3 x ) 3 + 12 x 2 - sin(3 x ) 9 - 24 x cos(3 x ) 27 + 24 sin(3 x ) 81 - 0 - cos(3 x ) 243 So we have that Z x 4 sin(3 x ) dx = - x 4 cos(3 x ) 3 + 4 x 3 sin(3 x ) 9 + 12 x 2 cos(3 x ) 27 - 24 x sin(3 x ) 81 - 24 cos(3 x ) 243 + Z 0 · - cos(3 x ) 243 dx = - x 4 cos(3 x ) 3 + 4 x 3 sin(3 x ) 9 + 12 x 2 cos(3 x ) 27 - 24 x sin(3 x ) 81 - 24 cos(3 x ) 243 + c 6. Z 2 x 3 e ( x 2 ) dx Solution: Let u = x 2 , then du = 2 x dx , so Z 2 x 3 e ( x 2 ) dx = Z ue u du . Let v = u , dw = e u du , then dv = du and w = e u . This gives us that Z 2 x 3 e ( x 2 ) dx = Z ue u du = ue u - Z e u du = ue u - e u + c = x 2 e ( x 2 ) - e ( x 2 ) + c 1
Evaluate the following trigonometric integrals: 7. Z cos 3 x dx = sin x - sin 3 x 3 + c 8. Z π/ 2 π/ 8 sin 2 x dx = 3 π 16 - 1 4 2 9. Z sin 8 x cos 3 x dx = sin 9 x 9 - sin 11 x 11 + c 10. Z π/ 4 π/ 6 tan 2 x dx Solution: Z π/ 4 π/ 6 tan 2 x dx = Z π/ 4 π/ 6 (sec 2 x - 1) dx = tan x - x π/ 4 π/ 6 = tan π 4 - π 4 - h tan π 6 - π 6 i = 1 - 1 3 - π 12 11. Z sec 3 x dx = sec x tan x + ln | sec x + tan x | 2 + c 12. Z π/ 4 0 tan 7 / 3 x sec 4 x dx Solution: Z π/ 4 0 tan 7 / 3 x sec 4 x dx = Z π/ 4 0 tan 7 / 3 x sec 2 x sec 2 x dx = Z π/ 4 0 tan 7 / 3 x (tan 2 x + 1) sec 2 x dx and letting u = tan x , gives du = sec 2 x dx so we get that Z π/ 4 0 tan 7 / 3 x (tan 2 x + 1) sec 2 x dx = Z 1 0 u 7 / 3 ( u 2 + 1) du = Z 1 0 u 13 / 3 + u 7 / 3 du = 3 u 16 / 3 16 + 3 u 10 / 3 10 1 0 = 3 16 + 3 10 = 39 80 13. Z sin 2 x cos 2 x dx 2
Solution: Z sin 2 x cos 2 x dx = 1 4 Z (1 - cos(2 x ))(1 + cos(2 x )) dx = 1 4 Z 1 - cos 2 (2 x ) dx = 1 4 Z dx - 1 4 Z cos 2 (2 x ) dx = x 4 - 1 8 Z (1 + cos(4 x )) dx = x 4 - x 8 - sin(4 x ) 32 + c = x 8 - sin(4 x ) 32 + c Evaluate the following integrals using trigonometric substitution: 14. Z 8 4 2 p 64 - x 2 dx Solution: We use the following triangle to substitute for θ : 8 x 64 - x 2 θ x 8 = sin θ x = 8 sin θ dx = 8 cos θ dθ 64 - x 2 8 = cos θ 64 - x 2 = 8 cos θ So we have Z 8 4 2 p 64 - x 2 dx = Z x =8 x =4 2 (8 cos θ )(8 cos θ dθ ) = 64 Z x =8 x =4 2 cos 2 θ dθ = 64 2 Z x =8 x =4 2 1 + cos(2 θ ) = 32 θ + sin(2 θ ) 2 x =8 x =4 2 = 32 θ + 2 sin( θ ) cos( θ ) 2 x =8 x =4 2 = 32 " sin - 1 x 8 + x 8 · 64 - x 2 8 # 8 4 2 = 32 sin - 1 (1) + 8 8 · 64 - 64 8 - 32 sin - 1 1 2 + 1 2 · 64 - 32 8 = 32 " π 2 + 0 - π 4 - 4 2 8 2 # = 8 π - 16 15. Z 9 - x 2 x dx 3
Solution: We use the following triangle to substitute for θ : 3 x 9 - x 2 θ x 3 = sin θ x = 3 sin θ dx = 3 cos θ dθ 9 - x 2 3 = cos θ 9 - x 2 = 3 cos θ So we have Z 9 - x 2 x dx = Z 3 cos θ 3 sin θ (3 cos θ dθ ) = 3 Z cos 2 θ sin θ = 3 Z 1 - sin 2 θ sin θ = 3 Z csc θ dθ - 3 Z sin θ dθ = - 3 ln | csc θ + cot θ | + 3 cos θ + c And from the triangle above, we see that csc θ = 3 x and cot θ = 9 - x 2 x . So this gives us - 3 ln | csc θ + cot θ | + 3 cos θ + c = - 3 ln 3 x + 9 - x 2 x + 3 9 - x 2 3 + c = - 3 ln 3 x + 9 - x 2 x + p 9 - x 2 + c 16.

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture