integrate (x^3 - cos x + 1/x) dx
Answers
Step-by-step explanation:
x
3
sin
x
+
3
x
2
cos
x
−
6
x
sin
x
−
6
cos
x
+
C
Explanation:
I
=
∫
x
3
cos
x
d
x
This will require multiple iterations of integration by parts (IBP). Integration by parts takes the form
∫
u
d
v
=
u
v
−
∫
v
d
u
.
So, for
∫
x
3
cos
x
d
x
as
∫
u
d
v
, let:
{
u
=
x
3
d
v
=
cos
x
d
x
Now, differentiating
u
and integrating
d
v
, we see that:
{
u
=
x
3
==⇒
d
u
=
3
x
2
d
x
d
v
=
cos
x
d
x
==⇒
v
=
sin
x
Now, plugging this into the IBP formula:
I
=
u
v
−
∫
v
d
u
=
x
3
sin
x
−
∫
3
x
2
sin
x
d
x
Now, for
∫
3
x
2
sin
x
d
x
, perform IBP again:
{
u
=
3
x
2
==⇒
d
u
=
6
x
d
x
d
v
=
sin
x
d
x
==⇒
v
=
−
cos
x
Thus:
I
=
x
3
sin
x
−
[
3
x
2
(
−
cos
x
)
−
∫
6
x
(
−
cos
x
)
d
x
]
Pay close attention to sign:
I
=
x
3
sin
x
+
3
x
2
cos
x
−
∫
6
x
cos
x
d
x
IBP again on the integral:
{
u
=
6
x
==⇒
d
u
=
6
d
x
d
v
=
cos
x
d
x
==⇒
v
=
sin
x
So:
I
=
x
3
sin
x
+
3
x
2
cos
x
−
[
6
x
sin
x
−
∫
6
sin
x
d
x
]
Since
∫
sin
x
d
x
=
−
cos
x
:
I
=
x
3
sin
x
+
3
x
2
cos
x
−
[
6
x
sin
x
+
6
sin
x
]
I
=
x
3
sin
x
+
3
x
2
cos
x
−
6
x
sin
x
−
6
cos
x
+
C
_______________
Mark as Brainliest plz
