Ejercicios De Calculo
´ Usar la sustitucion indicada para evaluar las integrales dadas. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 2 x (x2 + 1)23 dx; u = x2 + 1cos3 x sen x dx; u = cos x e2t dt; u = 2t cos(3x) dx; u = 3x 15. sec2 (2x + 1) dx; u = 2x + 1 16. y y 2 + 4 dy; u = y + 4 17. sen(πθ) cos(πθ) dθ; u = sen(πθ) dx ; u = 3x + 5 3x + 5 √ x2 3 + x dx; u = 3+ x [sec(sen θ)]2 cos θ dθ; u = sen θ 3x √ dx; u = 4x2 + 5 4x2 + 5 18. 19. 20. 21.
2
12. 13. 14.
√ √ cos x √ dx; u = x x 5x4 dx; u = x5 + 1 (x5 + 1)2 dx ; u = ln x x ln x sen 3θ dθ; u = 1 +cos 3θ 1 + cos 3θ e2x dx; u = 4 + e2x 4 + e2x x2 dx; u = x3 9 + x6 dx ; u = ln x x 1 − (ln x)2 dx ; u = 3x 9x2 − 1 √ dx √ ; u= x x (1 + x) x √ ex dx; u = ex 4 + e2x
22. 23. 24. 25. 26.
0
dx ; u =ax + b ax + b x dx ; u = ax + b ax + b x dx ; u = ax + b (ax + b)2 x2 dx ; u = ax + b (ax + b)3
1
(x + 2)(x + 3)5 dx; u = x + 3
3
27.
1 1
(2x − 5)6 dx; u = 2x − 5 x3
0 √ 2 3
28.
x2+ 8 dx; u = x2 + 8 tan−1 √
x 2
29.
0
3π (4 + x2 ) √
dx; u = tan−1
x 2
√ ln(2/ 3)
30.
ln 2
e−x dx ; u = e−x 1 − e−2x
Aplicar el m´todo de cambio de variable para hallar laintegral dada. e 1. 2. 3.
1
(3x + 1)4 dx (2x2 − 3)5 x dx
3
14. 15. 16.
0
sec2 3θ dθ cos x ln(sen x) dx sen x
ln 2
26. 27. 28. 29. 30. 31. 32. 33. 34. 35.
0
x
√ x − 1 dx
t2t3 − 1 dt
e4x
9 + e4x dx
x √ dx 1+x √ x 3 x + a dx x2 √ dx 1−x √ 4
3
4. 5. 6. 7. 8. 9.
x−2 dx (x2 − 4x + 3)3 x2 + x dx (4 − 3x2 − 2x3 )4 s √ ds 3 1 − 2s2 − (10t − 5t) dt √ ( z + 3)4 √dz z t4 t2
1 1/2 4
5
17. 18. 19. 20. 21. dt 22.
(ln x)2 dx x 1 √ dx x ln x ex dx x + 1) ln(ex + 1) (e tan−1 x dx 1 + x2 √ (1 + x)9 dx
ln(1/3)
x2 dx x−3 x3 + 1 x5 dx
3
3x2 (2x − 1)4dx sec v + π π tan v + 2 2 dv
1 1− t
5
1 t2
10.
1
√
ex sen(πex ) dx
0 π/4
x2 (x3 + 5)8 cos (x3 + 5)9 dx
1
5 − x dx 23.
11. 12. 13.
sen(2πθ − 4) dθ x cos(x + a) dx √ x...
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