Find the integral of sin^3x cos^5x
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Answer:
Recall that through the Pythagorean Identity
sin
2
(
x
)
=
1
−
cos
2
(
x
)
.
Thus,
sin
3
(
x
)
=
sin
(
x
)
sin
2
(
x
)
=
sin
(
x
)
(
1
−
cos
2
(
x
)
)
. Substituting this into the integral we see:
∫
sin
3
(
x
)
cos
5
(
x
)
d
x
=
∫
sin
(
x
)
(
1
−
cos
2
(
x
)
)
cos
5
(
x
)
d
x
Distributing just the cosines, this becomes
=
∫
(
cos
5
(
x
)
−
cos
7
(
x
)
)
sin
(
x
)
d
x
Now use the substitution:
u
=
cos
(
x
)
⇒
d
u
=
−
sin
(
x
)
d
x
Noting that
sin
(
x
)
d
x
=
−
d
u
, the integral becomes:
=
−
∫
(
u
5
−
u
7
)
d
u
Integrating, this becomes
=
−
(
u
6
6
−
u
8
8
)
+
C
Reordering and back-substituting with
u
=
cos
(
x
)
:
=
cos
8
(
x
)
8
−
cos
6
(
x
)
6
+
C
Note that this integration could have also been done my modifying the cosines like:
cos
5
(
x
)
=
cos
(
x
)
(
cos
2
(
x
)
)
2
=
cos
(
x
)
(
1
−
sin
2
(
x
)
)
2
And then proceeding by expanding and letting
u
=
sin
(
x
)
.
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