# X3 x2 x 2 ax b cx d 2 2 a d 0 b c 1

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Unformatted text preview: 2 −u ln 3 2 −√x e 3 +C =− +C ln 3 ln 3 sec u tan u du = sec u + C = sec(sin θ) + C 2 2 , du = − 2 dx, x x √ 22. e−u ln 3 du = − √ du = sin−1 u + C = sin−1 ex + C 1 − u2 2 sinh u du = −2 cosh u + C = −2 cosh(x−1/2 ) + C cos u du = 1 1 sin u + C = sin(x2 ) + C 2 2 2du = sin−1 u + C = sin−1 (ex /2) + C 4 − 4u2 , u = −x2 ln 4, du = −2x ln 4 dx = −x ln 16 dx, eu du = − 30. 2πx = eπx ln 2 , 1u 1 −x2 ln 4 1 −x2 e +C =− e 4 +C =− +C ln 16 ln 16 ln 16 2πx dx = 1 πx ln 2 1 e 2πx + C +C = π ln 2 π ln 2 31. (a) With u = x we get dx = sin−1 x + C 1 − x2 from (15), √ from (17), dx = tan−1 x + C 1 + x2 from (19), dx √ = sec−1 x + C (x &gt; 1) x x2 − 1 (|x| &lt; 1) Exercise Set 9.2 294 (b) With u = ax, du = adx, we get (for a &gt; 0) √ du = − u2 a2 a2 1 du =2 2 +u a du = u u2 − a2 √ adx √ = sin−1 x + C = sin−1 (u/a) + C a 1 − x2 1 1 adx = tan−1 x + C = tan−1 (u/a) + C 2 1+x a a 1 1 adx √ = sec−1 x + C = sec−1 (u/a) + C 2−1 a a x a2 x EXERCISE SET 9.2 1. u = x, dv = e−x dx, du = dx, v = −e−x ; 2. u = x, dv = e3x dx, du = dx, v = 1 3x e; 3 3. u = x2 , dv = ex dx, du = 2x dx, v = ex ; For xe−x dx = −xe−x + xe3x dx = 1 3x 1 xe − 3 3 x2 ex dx = x2 ex − 2 e3x dx = 1 3x 1 3x xe − e + C 3 9 xex dx. xex dx use u = x, dv = ex dx, du = dx, v = ex to get xex dx = xex − ex + C1 so x2 ex dx = x2 ex − 2xex + 2ex + C 1 4. u = x2 , dv = e−2x dx, du = 2x dx, v = − e−2x ; 2 For e−x dx = −xe−x − e−x + C 1 x2 e−2x dx = − x2 e−2x + 2 xe−2x dx xe−2x dx use u = x, dv = e−2x dx to get 1 1 xe−2x dx = − xe−2x + 2 2 1 1 e−2x dx = − xe−2x − e−2x + C 2 4 1 1 1 x2 e−2x dx = − x2 e−2x − xe−2x − e−2x + C 2 2 4 so 1 5. u = x, dv = sin 2x dx, du = dx, v = − cos 2x; 2 1 1 1 1 cos 2x dx = − x cos 2x + sin 2x + C x sin 2x dx = − x cos 2x + 2 2 2 4 1 sin 3x; 3 1 1 sin 3x dx = x sin 3x + cos 3x + C 3 9 6. u = x, dv = cos 3x dx, du = dx, v = x cos 3x dx = 1 1 x sin 3x − 3 3 7. u = x2 , dv = cos x dx, du = 2x dx, v = sin x; For x2 cos x dx = x2 sin x − 2 x sin x dx x sin x dx use u = x, dv = sin x dx to get x sin x dx = −x cos x + sin x + C1 so x2 cos x dx = x2 sin x + 2x cos x − 2 sin x + C 295 Chapter 9 8. u = x2 , dv = sin x dx, du = 2x dx, v = − cos x; x2 sin x dx = −x2 cos x + 2 x cos x dx; for x cos x dx = x sin x + cos x + C1 so x cos x dx use u = x, dv = cos x dx to get x2 sin x dx = −x2 cos x + 2x sin x + 2 cos x + C √ 2 1 x dx, du = dx, v = x3/2 ; x 3 √ 2 3/2 2 2 4 x1/2 dx = x3/2 ln x − x3/2 + C x ln x dx = x ln x − 3 3 3 9 9. u = ln x, dv = 10. u = ln x, dv = x dx, du = 1 1 dx, v = x2 ; x 2 11. u = (ln x)2 , dv = dx, du = 2 x ln x dx = ln x dx, v = x; x x dx = (ln x)2 dx = x(ln x)2 − 2 ln x dx = x ln x − Use u = ln x, dv = dx to get 12 1 x ln x − 2 2 12 1 x ln x − x2 + C 2 4 ln x dx. dx = x ln x − x + C1 so (ln x)2 dx = x(ln x)2 − 2x ln x + 2x + C √ 1 1 12. u = ln x, dv = √ dx, du = dx, v = 2 x; x x 13. u = ln(2x + 3), dv = dx, du = but 2x dx = 2x + 3 1− 2 dx, v = x; 2x + 3 3 2x + 3 dx = x − ln(2x + 3)dx = x ln(2x + 3) − x + 14. u = ln(x2 + 4), dv = dx, du = but x2 dx = +4 x2 1− √ ln x √ dx = 2 x ln x− 2 x ln(2x + 3)dx = x ln(2x + 3) − 3 ln(2x + 3) + C 2 ln(x2 + 4)dx = x ln(x2 + 4) − 2 dx = x − 2 tan−1 ln(x2 + 4)dx = x ln(x2 + 4) − 2x + 4 tan−1 x + C1 so 2 x +C 2 √ 15. u = sin−1 x, dv = dx, du = 1/ 1 − x2 dx, v = x; sin−1 x dx = x sin−1 x − x/ 1 − x2 dx = x sin−1 x + 16. u = cos−1 (2x), dv = dx, du = − √ cos−1 (2x)dx = x cos−1 (2x) + 1 − x2 + C 2 dx, v = x; 1 − 4x2 √ 2x dx 2x + 3 3 ln(2x + 3) + C1 so 2 2x dx, v = x; 2+4 x 4 x2 + 4 √ √ 1 √ dx = 2 x ln x− 4 x + C x 1 2x dx = x cos−1 (2x) − 2 2 1 − 4x 1 − 4x2 + C x2 dx x2 + 4 Exercise Set 9.2 296 2 dx, v = x; 1 + 4x2 2x 1 tan−1 (2x)dx = x tan−1 (2x) − dx = x tan−1 (2x) − ln(1 + 4x2 ) + C 2 1 + 4x 4 17. u = tan−1 (2x), dv = dx, du = 18. u = tan−1 x, dv = x dx, du = but x2 dx = 1 + x2 x tan−1 x dx = 1− 1 1 dx, v = x2 ; 1 + x2 2 1 1 + x2 ex cos x = ex sin x − ex sin x dx = ex (sin x − cos x) + C1 , 1 2x 2 e sin 3x − 3 3 1 2 e2x sin 3x dx = − e2x cos 3x + 3 3 e2x cos 3x dx = ex sin x dx = ex sin x dx so 1x e (sin x − cos x) + C 2 1 sin 3x; 3 e2x sin 3x dx. Use u = e2x , dv = sin 3x dx to get e2x cos 3x dx so 1 2x 2 4 e sin 3x + e2x cos 3x − 3 9 9 e2x cos 3x dx = ex cos x dx. ex sin x dx, 20. u = e2x , dv = cos 3x dx, du = 2e2x dx, v = 13 9 ex sin x dx = −ex cos x + ex cos x dx use u = ex , dv = cos x dx to get e2x cos 3x dx = e2x cos 3x dx, 1 2x e (3 sin 3x + 2 cos 3x) + C1 , 9 e2x cos 3x dx = 1 2x e (3 sin 3x + 2 cos 3x) + C 13 1 21. u = eax , dv = sin bx dx, du = aeax dx, v = − cos bx (b = 0); b 1 a eax sin bx dx = − eax cos bx + b b eax cos bx dx = 1 ax a e sin bx − b b eax cos bx dx. Use u = eax , dv = cos bx dx to get eax sin bx dx so 1 a a2 eax sin bx dx = − eax cos bx + 2 eax sin bx...
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