Question

Integration of (2x-3)sin^2x dx

Answer 1

Answer
∫
3
sin
(
2
x
)
d
x
=
−
3
cos
(
2
x
)
2
+
C
Explanation
You could solve this integral using
u
-substitution. Typically, we pick
u
to be our "inner" function, so here, let's denote
u
=
2
x
which is inside the
3
sin
(
2
x
)
=
3
sin
(
u
)
. Now, we can integrate, but we need to make sure all of our variables are in terms of
u
, so we need to find
d
x
in terms of
u
.
u
=
2
x

→
d
u
=
2
d
x

→

d
x
=
1
2
d
u
∫
3
sin
(
2
x
)
d
x
=
∫
3
sin
(
u
)
⋅
1
2
d
u
=
∫
3
2
sin
(
u
)
d
u
=
−
3
2
cos
(
u
)
+
C
Since the original question was in terms of
x
, we want to make sure our final answer is also in terms of
x
. Remember,
u
=
2
x
so we just sub back in:
−
3
2
cos
(
u
)
+
C
=
−
3
2
cos
(
2
x
)
+
C
Which you can rewrite as
−
3
cos
(
2
x
)
2
+
C
if you wanted to.
Alternatively...
You could also solve this integral through inspection, since we easily know the integral of
sin
(
x
)
alone is supposed to be
−
cos
(
x
)
+
C
. But we need the integral of
3
sin
(
2
x
)
. So we can pull out the constant 3 from the integral
∫
3
sin
(
2
x
)
d
x
=
3
∫
sin
(
2
x
)
d
x

Our lives would be easier if we had
2
sin
(
2
x
)
because that is easily integrated...and we can do that! As long as we multiply by 1/2 on the outside of the integral to cancel it out. So we end up with:
∫
3
sin
(
2
x
)
d
x
=
3
∫
sin
(
2
x
)
d
x
=
3
⋅
1
2
∫
2
sin
(
2
x
)
d
x
=
−
3
2
cos
(
2
x
)
+
C