Find the Laplace of t cos 4t
Answers
Answered by
1
Answer:
Let,
I
=
∫
cos
4
t
⋅
sin
2
t
d
t
.
Observe that,
cos
and
sin
both have even Power. So, we have
to convert them into Multiple Angles, using the Identities,
(
1
)
:
cos
2
θ
=
1
+
cos
2
θ
2
,
(
2
)
:
sin
2
θ
=
1
−
cos
2
θ
2
,
(
3
)
:
2
cos
α
cos
β
=
cos
(
α
+
β
)
+
cos
(
α
−
β
)
.
Now,
cos
4
t
⋅
sin
2
t
=
1
4
(
4
cos
2
t
⋅
sin
2
t
)
(
cos
2
t
)
,
=
1
4
(
sin
2
(
2
t
)
)
(
cos
2
t
)
,
=
1
4
{
1
2
(
1
−
cos
4
t
)
}
{
1
2
(
1
+
cos
2
t
)
}
,
=
1
16
(
1
−
cos
4
t
+
cos
2
t
−
cos
4
t
cos
2
t
)
,
=
1
32
{
2
−
2
cos
4
t
+
2
cos
2
t
−
2
cos
4
t
cos
2
t
}
,
=
1
32
{
2
−
2
cos
4
t
+
2
cos
2
t
−
(
cos
6
t
+
cos
2
t
)
}
,
=
1
32
{
2
−
cos
6
t
−
2
cos
4
t
+
3
cos
2
t
}
.
Therefore,
I
=
∫
1
32
{
2
−
cos
6
t
−
2
cos
4
t
+
3
cos
2
t
}
d
t
,
=
1
32
(
2
t
−
1
6
sin
6
t
−
2
⋅
1
4
sin
4
t
+
3
⋅
1
2
sin
2
t
)
,
⇒
I
=
1
192
(
12
t
−
sin
6
t
−
3
sin
4
t
+
9
sin
2
t
)
+
C
.
Enjoy Maths.!
Answered by
0
Answer:
Find the Laplace transform of t cos 4t
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