Integral of tan^3 2x dx
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Answered by
1
Answer:
Split up
tan
3
(
x
)
into
tan
2
(
x
)
tan
(
x
)
then rewrite
tan
2
(
x
)
using the identity
tan
2
(
θ
)
+
1
=
sec
2
(
θ
)
⇒
tan
2
(
θ
)
=
sec
2
(
θ
)
−
1
.
∫
tan
3
(
x
)
d
x
=
∫
tan
2
(
x
)
tan
(
x
)
d
x
=
∫
(
sec
2
(
x
)
−
1
)
tan
(
x
)
d
x
Distribute:
=
∫
sec
2
(
x
)
tan
(
x
)
d
x
−
∫
tan
(
x
)
d
x
For the first integral, apply the substitution
u
=
tan
(
x
)
⇒
d
u
=
sec
2
(
x
)
d
x
, both of which are already in the integral.
=
∫
u
.
d
u
−
∫
tan
(
x
)
d
x
=
u
2
2
−
∫
tan
(
x
)
d
x
=
tan
2
(
x
)
2
−
∫
tan
(
x
)
d
x
Now rewrite
tan
(
x
)
as
sin
(
x
)
cos
(
x
)
and apply the substitution
v
=
cos
(
x
)
⇒
d
v
=
−
sin
(
x
)
d
x
.
=
tan
2
(
x
)
2
−
∫
sin
(
x
)
cos
(
x
)
d
x
=
tan
2
(
x
)
2
+
∫
−
sin
(
x
)
cos
(
x
)
d
x
=
tan
2
(
x
)
2
+
∫
d
v
v
This is a common integral:
=
tan
2
(
x
)
2
+
ln
(
|
v
|
)
+
C
=
tan
2
(
x
)
2
+
ln
(
|
cos
(
x
)
|
)
+
C
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