dx Sen2Cos2(x)

1
Metodo de
R sustitución
dx
Evaluar Sen2 Cos
2 (x) tenemos que
Z
Z
Z
dx
Sen2 (x) + Cos2 (x)dx
dx
dx
=
=
+
=
2
2
2
2
2
Sen Cos (x)
Sen Cos (x)
Cos (x)
Sen2 (x)
Z
Z
2
Sec (x)dx + Csc2 (x)dx = T an(x) − Cot(x) + k
Z
Evaluar
Z
R
(x+1)dx
√
x x−2
(x + 1)dx
√
x x−2
| {z }
x−2=t2
tenemos que
(t2 + 2 + 1)2tdt
=2
(t2 + 2)t
Z
=
Z
t2 + 2
dt + 2
t2 + 2
Z
t2
1
dt =
+2
dx=2tdt
√
√
√
√
1
t
x−2
2
dt = 2t+ 2Arctan( √ )+k = 2 x − 2+ 2Arctan( √ )+k
2
1+t
2
2
R Cos(√x)
√
Evalua
dx tenbemos que
x
Z
Z
dt+
Z
√
Z
√
Cos( x)
√
dx = 2 Cos(t)dt = 2Sen(t) + k = 2Sen( x) + k
x
|
{z
}
√
x=t
Evaluar
Z
R
dx
√ =dt
2 x
Sen2 (x)Cos(x)
dx
1+Sen2 (x)
Sen2 (x)Cos(x)
dx =
1 + Sen2 (x)
|
{z
}
Sen(x)=t
Z
tenemos que
t2 dt
=
1 + t2
Z
t2 + 1 − 1dt
=
1 + t2
Z
1−
1
dt = t−Arctan(t)+k
1 + t2
Cos(x)dx=dt
= Sen(x) + Arctan(Sen(x)) + k
Evaluar
Z
R
5
√ x
dx
a3 −x3
x5
√
dx
a3 − x 3
|
{z
}
a3 −x3 =t2
tenemos que
Z
=
(a3 − t2 )(− 32 )tdt
2
=−
t
3
Z
2
t3
(a3 −t2 )dt = − (a3 t− )+k =
3
3
−3x2 dx=2tdt
2 √
1p 3
− ( 3 a3 − x 3 −
(a − x3 )3 ) + k
3
3
2
Metodo por RPartes
Evaluar ArcSen(x)dx tenemos que
Z
Z
xdx
√
ArcSen(x)dx = xArcSen(x)−
{z
}
|
1 − x2
| {z }
u=ArcSen(x)
dv=dx
du= √1dx
1−x2
v=x
u=1−x2
1
= xArcSen(x)+
2
Z
du
√
u
du=−2xdx
√
√
= xArcSen(x) + u + k = xArcSen(x) + 1 − x2 + k
R
Evaluar Arctan(x)dx tenemos que
Z
Z
Z
xdx
1
du
Arctan(x)dx = xArctan(x)−
=
xArctan(x)−
2
{z
}
|
2
u
{zx }
|1 +
u=arctan(x)
dv=dx
du=
1
1+x2
v=x
u=1+x2
du=2xdx
1
1
= xArctan(x) − Ln(u) + k = xArctan(x) − Ln(1 + x2 ) + k
2
2
R
dx
Calcular In = (x2 +a2 )n n = 1, 2, 3... tenemos que
Z
Z
dx
x
2nx2 dx
In =
=
+
(x2 + a2 )n
(x2 + a2 )n+1
(x2 + a2 )n+1
|
{z
}
1
dv=dx
(x2 +a2 )n
du=− 2 2nx
v=x
(x +a2 )n+1
u=
Z
Z 2
x
x
x2 dx
x + a2 − a2 dx
+
2n
=
+
2n
= 2
(x + a2 )n+1
(x2 + a2 )n+1
(x2 + a2 )n+1
(x2 + a2 )n+1
Z
Z
x
x
dx
dx
= 2
+2n(
−a2
)= 2
+2nIn −2a2 nIn+1
2
n+1
2
2
n
2
2
n+1
(x + a )
(x + a )
(x + a )
(x + a2 )n+1
Despejando In+1 obtenemos
1
x
2n − 1
In+1 =
+
In
2
2
2
n+1
2na (x + a )
2na2
tenemos que
Z
dx
1
x
I1 =
= Arctan( ) + k
2
2
a +x
a
a
usando la fórmula In+1 hallamos
Z
dx
1
x
1
x
= I2 = I1+1 = 2 2
+ 3 Arctan( ) + k
2
2
2
2
(x + a )
2a x + a
2a
a
Z
dx
1
x
3 1
x
1
x
=
I
=
I
=
+
(
+
Arctan(
))+k
3
2+1
(x2 + a2 )3
4a2 (x2 + a2 )2 4a2 2a2 a2 + x2 2a3
a