integral(sin^2xcos^4x dx)
Answers
Answered by
0
Step-by-step explanation:
∫
s
i
n
2
(
x
)
c
o
s
4
(
x
)
d
x
∫sin2(x)cos4(x)dx
=
1
8
∫
(
1
−
c
o
s
(
2
x
)
)
(
1
+
c
o
s
(
2
x
)
)
2
d
x
=18∫(1−cos(2x))(1+cos(2x))2dx
=
1
8
∫
(
1
−
c
o
s
2
(
2
x
)
)
(
1
+
c
o
s
(
2
x
)
)
d
x
=18∫(1−cos2(2x))(1+cos(2x))dx
=
1
8
∫
s
i
n
2
(
2
x
)
d
x
+
1
8
∫
s
i
n
2
(
2
x
)
c
o
s
(
2
x
)
d
x
=18∫sin2(2x)dx+18∫sin2(2x)cos(2x)dx
=
1
16
∫
(
1
−
c
o
s
(
4
x
)
)
d
x
+
1
48
s
i
n
3
(
2
x
)
=116∫(1−cos(4x))dx+148sin3(2x)
=
1
16
x
−
1
64
s
i
n
(
4
x
)
+
1
48
s
i
n
3
(
2
x
)
+
C
=116x−164sin(4x)+148sin3(2x)+C
Answered by
12
Answer:
hope it's correct i m not sure
Step-by-step explanation:
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