∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ u(t) ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫

Derivadasa
d(c)
dt = 0
du(t)
d cos u(t)
= − sen u(t) dt
dt
d(cu(t))
du(t)
= c dt
dt
d[u(t)]n
dt
d tan u(t)
du(t)
= sec2 u(t) dt
dt
du(t)
= n[u(t)]n−1 dt
u(t)
du(t)
dv(t)
v(t) dt − u(t) dt
v(t)
=
dt
[v(t)]2
la funciones exponencial y logaritmo
natural. Nótese como una función es
d ln u(t)
du(t)
= 1 dt
dt
u(t)
el reflejo de la otra a través de la recta
y = x.
f (k) (h) k
f 00 (h) 2
x + ... +
x + ...
2
k!
2
4
6
cos x = 1 − x2! + x4! − x6! + . . .
2
3
4
ex = 1 + x + x2! + x3! + x4! + . . .
Ejemplos
du
R
z }| {
du(t)
dt =
dt
u(t)n+1
[u(t)]n du = n + 1 + C
R
(3t + 5)3 dt = 31
Rz
du(t)
dt
dt
u(t)
4
(3t + 5)
(3t + 5)3 3dt = 13
+C
4
}| {
z}|{
du(t)
dt
dt
du
R
1 du = ln |u(t)| + C
u(t)
R
Rz
}|
{
cos 2t(2dt)
cot 2tdt = 12
= 12 ln | sen 2t| + C
sen
| {z2t}
u(t)
R
[c1 f1 (t) + . . . + cn fn (t)] dt =
R
[v0 + at] dt = v0 t + 21 at2 + C
R
c1 f1 (t)dt + . . . + cn fn (t)dt
du
R
z }| {
du(t)
sen u(t)
dt =
dt
R
sen u(t)du = − cos u(t)+C
R
t sen t2 dt
= 12
R
u(t)
z }| {
cos u(t)
du(t)
dt =
dt
R
cos u(t)du = sen u(t) + C
du
R
z }| {
eu(t)
a
du(t)
dt =
dt
R
1
cos ωtdt = ω
R
sen t
eu(t) du = eu(t) + C
c es una constante
R
dt
2tdt = − 12 cos t2 + C
2
u(t)
du(t)
dt
dt
1 sen ωt + C
cos ωt ωdt = ω
z}|{ z}|{
u(t)
R
du(t)
dt
z}|{ z}|{
du
R
6
Figura 1. Representación gráfica de
Integralesa
1 du(t) dt =
u(t) dt
4
-2
deu(t) = eu(t) du(t)
dt
dt
Por ejemplo:
3
5
7
sen x = x − x3! + x5! − x7! + . . .
z }| {
2
t
c es una constante
[u(t)]n
0
du(t)
d csc u(t)
= − csc u(t) cot u(t) dt
dt
f (x + h) = f (h) + f 0 (h)x +
R
ln x
-2
Series de Taylor
R
2
d sec u(t)
du(t)
= sec u(t) cot u(t) dt
dt
d sen u(t)
du(t)
= cos u(t) dt
dt
R
y=x
0
dv(t)
du(t)
d(u(t)v(t))
= u(t) dt + v(t) dt
dt
a
ex
4
du(t)
d cot u(t)
= − csc2 u(t) dt
dt
d(u(t) + v(t))
du(t) dv(t)
= dt + dt
dt
d
6
R
z }| {
du(t)
dt
dt
2 z }| {
2
1 e−at −2atdt = − 1 e−at2 + C
te−at dt = − 2a
2a