integrate (cos2x/sinx)
Answers
Answer:
Dec 24, 2016
=
cos
x
−
2
3
cos
3
x
+
C
Explanation:
Use the identity
cos
(
2
x
)
=
1
−
2
sin
2
x
.
=
∫
(
1
−
2
sin
2
x
)
sin
x
d
x
Multiply out.
=
∫
(
sin
x
−
2
sin
3
x
)
d
x
Separate using
∫
(
a
+
b
)
d
x
=
∫
a
d
x
+
∫
b
d
x
=
∫
(
sin
x
)
d
x
−
∫
(
2
sin
3
x
)
d
x
The antiderivative of
sin
x
is
−
cos
x
. Use the property of integrals that
∫
(
C
f
(
x
)
)
d
x
=
C
∫
f
(
x
)
where
C
is a constant. Note that
sin
3
x
can be factored as
sin
2
x
(
sin
x
)
, which can in turn be written as
(
1
−
cos
2
x
)
(
sin
x
)
by the identity
sin
2
x
+
cos
2
x
=
1
.
=
−
cos
x
−
2
∫
(
1
−
cos
2
x
)
sin
x
d
x
Let
u
=
cos
x
. Then
d
u
=
−
sin
x
d
x
→
d
x
=
−
d
u
sin
x
.
=
−
cos
x
−
2
∫
(
1
−
u
2
)
sin
x
⋅
−
d
u
sin
x
The sines under the integral cancel each other out.
=
−
cos
x
−
2
∫
(
1
−
u
2
)
⋅
−
(
d
u
)
Extract the negative
1
.
=
−
cos
x
+
2
∫
(
1
−
u
2
)
d
u
Separate the integrals.
=
−
cos
x
+
2
∫
1
d
u
−
2
∫
u
2
d
u
Integrate using the rule
∫
(
x
n
)
d
x
=
x
n
+
1
n
+
1
+
C
, where
C
is a constant.
=
−
cos
x
+
2
u
−
2
(
1
3
u
3
)
+
C
Resubstitute
u
=
cos
x
to define the function with respect to
x
.
=
−
cos
x
+
2
cos
x
−
2
3
cos
3
x
+
C
Finally, combine like terms.
=
cos
x
−
2
3
cos
3
x
+
C
Hopefully this helps!
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