Evaluate:∫x sin²x dx
Answers
Answer:
The solution is really simple if you do it by parties.
We start by integrating the function
sin
2
(
x
)
so
∫
sin
2
(
x
)
d
x
=
∫
1
2
(
1
−
cos
(
2
x
)
)
=
1
2
(
x
−
sin
(
2
x
)
2
)
+
C
so then you do the original integral by party.
you take:
u
=
x
⇒
.
u
=
1
.
v
=
sin
2
x
⇒
v
=
1
2
(
x
−
sin
(
2
x
)
2
)
and we know that
∫
u
.
v
=
u
v
−
∫
.
u
v
and so
∫
x
sin
2
x
d
x
=
x
2
(
x
−
sin
(
2
x
)
2
)
−
∫
1
2
(
x
−
sin
(
2
x
)
2
)
=
x
2
(
x
−
sin
(
2
x
)
2
)
−
1
2
(
x
2
+
cos
(
2
x
)
4
)
Which then can be simplified to
x
2
4
−
x
sin
(
2
x
)
4
−
cos
(
2
x
)
8
Answer link
Answer:
The solution is really simple if you do it by parties.
We start by integrating the function
sin
2
(
x
)
so
∫
sin
2
(
x
)
d
x
=
∫
1
2
(
1
−
cos
(
2
x
)
)
=
1
2
(
x
−
sin
(
2
x
)
2
)
+
C
so then you do the original integral by party.
you take:
u
=
x
⇒
.
u
=
1
.
v
=
sin
2
x
⇒
v
=
1
2
(
x
−
sin
(
2
x
)
2
)
and we know that
∫
u
.
v
=
u
v
−
∫
.
u
v
and so
∫
x
sin
2
x
d
x
=
x
2
(
x
−
sin
(
2
x
)
2
)
−
∫
1
2
(
x
−
sin
(
2
x
)
2
)
=
x
2
(
x
−
sin
(
2
x
)
2
)
−
1
2
(
x
2
+
cos
(
2
x
)
4
)
Which then can be simplified to
x
2
4
−
x
sin
(
2
x
)
4
−
cos
(
2
x
)
8
The solution is really simple if you do it by parties.
We start by integrating the function
sin
2
(
x
)
so
∫
sin
2
(
x
)
d
x
=
∫
1
2
(
1
−
cos
(
2
x
)
)
=
1
2
(
x
−
sin
(
2
x
)
2
)
+
C
so then you do the original integral by party.
you take:
u
=
x
⇒
.
u
=
1
.
v
=
sin
2
x
⇒
v
=
1
2
(
x
−
sin
(
2
x
)
2
)
and we know that
∫
u
.
v
=
u
v
−
∫
.
u
v
and so
∫
x
sin
2
x
d
x
=
x
2
(
x
−
sin
(
2
x
)
2
)
−
∫
1
2
(
x
−
sin
(
2
x
)
2
)
=
x
2
(
x
−
sin
(
2
x
)
2
)
−
1
2
(
x
2
+
cos
(
2
x
)
4
)
Which then can be simplified to
x
2
4
−
x
sin
(
2
x
)
4
−
cos
(
2
x
)
8
The solution is really simple if you do it by parties.
We start by integrating the function
sin
2
(
x
)
so
∫
sin
2
(
x
)
d
x
=
∫
1
2
(
1
−
cos
(
2
x
)
)
=
1
2
(
x
−
sin
(
2
x
)
2
)
+
C
so then you do the original integral by party.
you take:
u
=
x
⇒
.
u
=
1
.
v
=
sin
2
x
⇒
v
=
1
2
(
x
−
sin
(
2
x
)
2
)
and we know that
∫
u
.
v
=
u
v
−
∫
.
u
v
and so
∫
x
sin
2
x
d
x
=
x
2
(
x
−
sin
(
2
x
)
2
)
−
∫
1
2
(
x
−
sin
(
2
x
)
2
)
=
x
2
(
x
−
sin
(
2
x
)
2
)
−
1
2
(
x
2
+
cos
(
2
x
)
4
)
Which then can be simplified to
x
2
4
−
x
sin
(
2
x
)
4
−
cos
(
2
x
)
8