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integral from+0 to pi of e^{cos(x)}sin(2x)
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∫
0
π
e
cos
(
x
)
sin
(
2
x
)
dx
=
4
e
Show Steps
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∫
0
π
e
cos
(
x
)
sin
(
2
x
)
dx
=
4
e
(
Decimal
:
1
.
4
7
1
5
1
…
)
steps
∫
0
π
e
cos
(
x
)
sin
(
2
x
)
dx
show steps
Rewrite
using
trig
identities
=
∫
0
π
e
cos
(
x
)
·
2
cos
(
x
)
sin
(
x
)
dx
Take
the
constant
out
:
∫
a
·
f
(
x
)
dx
=
a
·
∫
f
(
x
)
dx
=
2
·
∫
0
π
e
cos
(
x
)
cos
(
x
)
sin
(
x
)
dx
show steps
Apply
u
−
substitution
:
∫
1
−
1
−
e
u
udu
=
2
·
∫
1
−
1
−
e
u
udu
∫
a
b
f
(
x
)
dx
=
−
∫
b
a
f
(
x
)
dx
,
a
<
b
=
2
(
−
∫
−
1
1
−
e
u
udu
)
Take
the
constant
out
:
∫
a
·
f
(
x
)
dx
=
a
·
∫
f
(
x
)
dx
=
2
(
−
(
−
∫
−
1
1
e
u
udu
)
)
show steps
Apply
Integration
By
Parts
:
[
e
u
u
−
∫
e
u
du
]
−
1
1
=
2
(
−
(
−
[
e
u
u
−
∫
e
u
du
]
−
1
1
)
)
show steps
∫
e
u
du
=
e
u
=
2
(
−
(
−
[
e
u
u
−
e
u
]
−
1
1
)
)
Simplify
=
2
[
e
u
u
−
e
u
]
−
1
1
show steps
Compute
the
boundaries
:
2
e
=
2
·
2
e
Simplify
=
4
e
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