1.3e

Calcule (de manera exacta) la siguiente integral:

∬

R

x
sen
⁡
(
x
y
)
d
A
,
# R

[

0
,

π
6

]

×

[

0
,

π
3

]

{\displaystyle \iint \limits \_{R}x\operatorname {sen}(x+y)dA,R=\left[0,{\frac {\pi }{6}}\right]\times \left[0,{\frac {\pi }{3}}\right]}

Usamos integración primero sobre

d
y

{\displaystyle dy}

 y después sobre

d
x

{\displaystyle dx}

, bajo el teorema de *Fubini*.

⇒

∫

0

π
6

∫

0

π
3

x
sen
⁡
(
x
y
)
d
y
d
x

=

∫

0

π
6

[
−
x
cos
⁡
(
x
y
)

0

π
3

]
d
x

⇒
−
x
cos
⁡
(
x

π
3

)
−
(
−
x
cos
⁡
(
x
0
)
)

=
−
x
cos
⁡
(
x

π
3

)
x
cos
⁡
(
x
)

=
−
x
(
cos
⁡
(
x
)
cos
⁡
(

π
3

)
−
sen
⁡
(
x
)
sen
⁡
(

π
3

)
)
x
cos
⁡
(
x
)

=
−
x
(
cos
⁡
(
x
)

1
2

−
sen
⁡
(
x
)

3

2

)
x
cos
⁡
(
x
)

=

−
x
cos
⁡
(
x
)
x

3

sen
⁡
(
x
)

2

x
cos
⁡
(
x
)

=

−
x
cos
⁡
(
x
)
x

3

sen
⁡
(
x
)
2
x
cos
⁡
(
x
)

2

=

x
cos
⁡
(
x
)
x

3

sen
⁡
(
x
)

2

=

x
(
cos
⁡
(
x
)

3

sen
⁡
(
x
)
)

2

=

x
(
cos
⁡
(
x
)

3

sen
⁡
(
x
)
)

2

{\displaystyle {\begin{aligned}&\Rightarrow \int \_{0}^{\frac {\pi }{6}}\int \_{0}^{\frac {\pi }{3}}x\operatorname {sen}(x+y)dydx\\&=\int \_{0}^{\frac {\pi }{6}}[-x\cos(x+y){\Big |}\_{0}^{\frac {\pi }{3}}]dx\\\\&\Rightarrow -x\cos(x+{\frac {\pi }{3}})-(-x\cos(x+0))\\\\&=-x\cos(x+{\frac {\pi }{3}})+x\cos(x)\\\\&=-x(\cos(x)\cos({\frac {\pi }{3}})-\operatorname {sen}(x)\operatorname {sen}({\frac {\pi }{3}}))+x\cos(x)\\\\&=-x(\cos(x){\frac {1}{2}}-\operatorname {sen}(x){\frac {\sqrt {3}}{2}})+x\cos(x)\\\\&={\frac {-x\cos(x)+x{\sqrt {3}}\operatorname {sen}(x)}{2}}+x\cos(x)\\\\&={\frac {-x\cos(x)+x{\sqrt {3}}\operatorname {sen}(x)+2x\cos(x)}{2}}\\\\&={\frac {x\cos(x)+x{\sqrt {3}}\operatorname {sen}(x)}{2}}\\\\&={\frac {x(\cos(x)+{\sqrt {3}}\operatorname {sen}(x))}{2}}\\\\&={\frac {x(\cos(x)+{\sqrt {3}}\operatorname {sen}(x))}{2}}\\\\\end{aligned}}}

Aquí agregamos un *Uno tramposo* en forma de

2
2

{\displaystyle \frac{2}{2}}

=

2
x
(

1
2

cos
⁡
(
x
)

3

2

sen
⁡
(
x
)
)

2

=

2
x
(
cos
⁡
(

π
3

)
cos
⁡
(
x
)
sen
⁡
(

π
3

)
sen
⁡
(
x
)
)

2

=
x
cos
⁡
(

π
3

−
x
)

{\displaystyle {\begin{aligned}&={\frac {2x({\frac {1}{2}}\cos(x)+{\frac {\sqrt {3}}{2}}\operatorname {sen}(x))}{2}}\\\\&={\frac {2x(\cos({\frac {\pi }{3}})\cos(x)+\operatorname {sen}({\frac {\pi }{3}})\operatorname {sen}(x))}{2}}\\\\&=x\cos({\frac {\pi }{3}}-x)\end{aligned}}}

Entonces regresamos a la integral:

⇒

∫

0

π
6

x
cos
⁡

(

π
3

−
x

)

d
x

{\displaystyle {\begin{aligned}&\Rightarrow \int \_{0}^{\frac {\pi }{6}}x\cos \left({\frac {\pi }{3}}-x\right)dx\end{aligned}}}

Realizamos integración por partes, tomando

# u
x

{\displaystyle u = x}

 y

d
# v
cos
⁡

(

π
3

−
x

)

{\displaystyle dv=\cos \left({\frac {\pi }{3}}-x\right)}

. Vemos que

u

{\displaystyle u}

 es facilmente derivable; veamos qué resulta de la integral de

d
v

{\displaystyle dv}

:

∫
cos
⁡
(

π
3

−
x
)
d
x
,
# t

π
3

−
x
⇒
d
# t
−
1
d
x
∴
d
# x

1

−
1

=
∫
cos
⁡
(

π
3

−
x
)

1

−
1

d
t

=
−
∫
cos
⁡
(
t
)
d
t

=
−
sen
⁡
(
t
)

=
−
sen
⁡
(

π
3

−
x
)
C

=
−
(
sen
⁡
(

π
3

)
cos
⁡
(
x
)
−
cos
⁡
(

π
3

)
sen
⁡
(
x
)
)
C

=
−
(

3

2

cos
⁡
(
x
)
−

1
2

sen
⁡
(
x
)
)
C

=

sen
⁡
(
x
)
−

3

cos
⁡
(
x
)

2

C

{\displaystyle {\begin{aligned}&\int \cos({\frac {\pi }{3}}-x)dx,t={\frac {\pi }{3}}-x\Rightarrow dt=-1dx\therefore dx={\frac {1}{-1}}\\\\&=\int \cos({\frac {\pi }{3}}-x){\frac {1}{-1}}dt\\\\&=-\int \cos(t)dt\\\\&=-\operatorname {sen}(t)\\&=-\operatorname {sen}({\frac {\pi }{3}}-x)+C\\&=-(\operatorname {sen}({\frac {\pi }{3}})\cos(x)-\cos({\frac {\pi }{3}})\operatorname {sen}(x))+C\\&=-({\frac {\sqrt {3}}{2}}\cos(x)-{\frac {1}{2}}\operatorname {sen}(x))+C\\\\&={\frac {\operatorname {sen}(x)-{\sqrt {3}}\cos(x)}{2}}+C\end{aligned}}}

Ahora, volviendo a nuestra integral principal y sustituyendo:

∫

0

π
6

x
cos
⁡
(

π
3

−
x
)
d
x

=
x

sen
⁡
(
x
)
−

3

cos
⁡
(
x
)

2

−
∫

sen
⁡
(
x
)
−

3

cos
⁡
(
x
)

2

d
x

=
x

sen
⁡
(
x
)
−

3

cos
⁡
(
x
)

2

−

1
2

∫
sen
⁡
(
x
)
−

3

cos
⁡
(
x
)
d
x

=
x

sen
⁡
(
x
)
−

3

cos
⁡
(
x
)

2

−

1
2

∫
2
(

1
2

sen
⁡
(
x
)
−

1
2

3

cos
⁡
(
x
)
)
d
x

=
x

sen
⁡
(
x
)
−

3

cos
⁡
(
x
)

2

−

1
2

∫
2
(

1
2

sen
⁡
(
x
)
−

3

2

cos
⁡
(
x
)
)
d
x

=
x

sen
⁡
(
x
)
−

3

cos
⁡
(
x
)

2

−

1
2

∫
2
(
cos
⁡
(

π
3

)
sen
⁡
(
x
)
−
sen
⁡
(

π
3

)
cos
⁡
(
x
)
)
d
x

=

x
sen
⁡
(
x
)
−

3

x
cos
⁡
(
x
)

2

−
∫
cos
⁡
(

π
3

)
sen
⁡
(
x
)
−
sen
⁡
(

π
3

)
cos
⁡
(
x
)
d
x

=

x
sen
⁡
(
x
)
−

3

x
cos
⁡
(
x
)

2

−
∫
sen
⁡
(
x
−

π
3

)
d
x

=

x
sen
⁡
(
x
)
−

3

x
cos
⁡
(
x
)

2

−
(
−
cos
⁡
(
x
−

π
3

)
)

=

x
sen
⁡
(
x
)
−

3

x
cos
⁡
(
x
)

2

cos
⁡
(
x
)
cos
⁡
(

π
3

)
sen
⁡
(
x
)
sen
⁡
(

π
3

)

=

x
sen
⁡
(
x
)
−

3

x
cos
⁡
(
x
)

2

cos
⁡
(
x
)

1
2

sen
⁡
(
x
)

3

2

=

x
sen
⁡
(
x
)
−

3

x
cos
⁡
(
x
)

2

cos
⁡
(
x
)

2

3

sen
⁡
(
x
)

2

=

x
sen
⁡
(
x
)
−

3

x
cos
⁡
(
x
)
cos
⁡
(
x
)

3

sen
⁡
(
x
)

2

C

=

x
sen
⁡
(
x
)
−

3

x
cos
⁡
(
x
)
cos
⁡
(
x
)

3

sen
⁡
(
x
)

2

0

π
6

=

π
6

sen
⁡
(

π
6

)
−

3

π
6

cos
⁡
(

π
6

)
cos
⁡
(

π
6

)

3

sen
⁡
(

π
6

)

2

−

0
sen
⁡
(
0
)
−

3

0
cos
⁡
(
0
)
cos
⁡
(
0
)

3

sen
⁡
(
0
)

2

=

π
6

1
2

−

3

π
6

3

2

3

2

3

1
2

2

−

1
2

=

π
12

−

π
4

3

2

3

2

2

−

1
2

=

−

π
6

2

3

2

2

−

1
2

=

−

π
6

3

2

−

1
2

=

−
π
6

3

6

2

−

1
2

=

−
π
6

3

12

−

1
2

≈
0.10422

{\displaystyle {\begin{aligned}&\int \_{0}^{\frac {\pi }{6}}x\cos({\frac {\pi }{3}}-x)dx\\\\&=x{\frac {\operatorname {sen}(x)-{\sqrt {3}}\cos(x)}{2}}-\int {\frac {\operatorname {sen}(x)-{\sqrt {3}}\cos(x)}{2}}dx\\\\&=x{\frac {\operatorname {sen}(x)-{\sqrt {3}}\cos(x)}{2}}-{\frac {1}{2}}\int \operatorname {sen}(x)-{\sqrt {3}}\cos(x)dx\\\\&=x{\frac {\operatorname {sen}(x)-{\sqrt {3}}\cos(x)}{2}}-{\frac {1}{2}}\int 2({\frac {1}{2}}\operatorname {sen}(x)-{\frac {1}{2}}{\sqrt {3}}\cos(x))dx\\\\&=x{\frac {\operatorname {sen}(x)-{\sqrt {3}}\cos(x)}{2}}-{\frac {1}{2}}\int 2({\frac {1}{2}}\operatorname {sen}(x)-{\frac {\sqrt {3}}{2}}\cos(x))dx\\\\&=x{\frac {\operatorname {sen}(x)-{\sqrt {3}}\cos(x)}{2}}-{\frac {1}{2}}\int 2(\cos({\frac {\pi }{3}})\operatorname {sen}(x)-\operatorname {sen}({\frac {\pi }{3}})\cos(x))dx\\\\&={\frac {x\operatorname {sen}(x)-{\sqrt {3}}x\cos(x)}{2}}-\int \cos({\frac {\pi }{3}})\operatorname {sen}(x)-\operatorname {sen}({\frac {\pi }{3}})\cos(x)dx\\\\&={\frac {x\operatorname {sen}(x)-{\sqrt {3}}x\cos(x)}{2}}-\int \operatorname {sen}(x-{\frac {\pi }{3}})dx\\\\&={\frac {x\operatorname {sen}(x)-{\sqrt {3}}x\cos(x)}{2}}-(-\cos(x-{\frac {\pi }{3}}))\\\\&={\frac {x\operatorname {sen}(x)-{\sqrt {3}}x\cos(x)}{2}}+\cos(x)\cos({\frac {\pi }{3}})+\operatorname {sen}(x)\operatorname {sen}({\frac {\pi }{3}})\\\\&={\frac {x\operatorname {sen}(x)-{\sqrt {3}}x\cos(x)}{2}}+\cos(x){\frac {1}{2}}+\operatorname {sen}(x){\frac {\sqrt {3}}{2}}\\\\&={\frac {x\operatorname {sen}(x)-{\sqrt {3}}x\cos(x)}{2}}+{\frac {\cos(x)}{2}}+{\frac {{\sqrt {3}}\operatorname {sen}(x)}{2}}\\\\&={\frac {x\operatorname {sen}(x)-{\sqrt {3}}x\cos(x)+\cos(x)+{\sqrt {3}}\operatorname {sen}(x)}{2}}+C\\\\&={\frac {x\operatorname {sen}(x)-{\sqrt {3}}x\cos(x)+\cos(x)+{\sqrt {3}}\operatorname {sen}(x)}{2}}{\Big |}\_{0}^{\frac {\pi }{6}}\\\\&={\frac {{\frac {\pi }{6}}\operatorname {sen}({\frac {\pi }{6}})-{\sqrt {3}}{\frac {\pi }{6}}\cos({\frac {\pi }{6}})+\cos({\frac {\pi }{6}})+{\sqrt {3}}\operatorname {sen}({\frac {\pi }{6}})}{2}}-{\frac {0\operatorname {sen}(0)-{\sqrt {3}}0\cos(0)+\cos(0)+{\sqrt {3}}\operatorname {sen}(0)}{2}}\\\\&={\frac {{\frac {\pi }{6}}{\frac {1}{2}}-{\sqrt {3}}{\frac {\pi }{6}}{\frac {\sqrt {3}}{2}}+{\frac {\sqrt {3}}{2}}+{\sqrt {3}}{\frac {1}{2}}}{2}}-{\frac {1}{2}}\\\\&={\frac {{\frac {\pi }{12}}-{\frac {\pi }{4}}+{\frac {\sqrt {3}}{2}}+{\frac {\sqrt {3}}{2}}}{2}}-{\frac {1}{2}}\\\\&={\frac {-{\frac {\pi }{6}}+{\frac {2{\sqrt {3}}}{2}}}{2}}-{\frac {1}{2}}\\\\&={\frac {-{\frac {\pi }{6}}+{\sqrt {3}}}{2}}-{\frac {1}{2}}\\\\&={\frac {\frac {-\pi +6{\sqrt {3}}}{6}}{2}}-{\frac {1}{2}}\\\\&={\frac {-\pi +6{\sqrt {3}}}{12}}-{\frac {1}{2}}\approx 0.10422\end{aligned}}}