The result, when cleared down, is:
∫
ϵ
a
b
t
⋅
sin
2
π
n
t
⋅
d
t
=
ϵ
a
b
t
{
a
b
⋅
sin
2
π
n
t
−
2
π
n
b
2
⋅
cos
2
π
n
t
a
2
+
4
π
2
n
2
b
2
}
.
.
.
.
.
[D]
{\displaystyle {\begin{aligned}\int \epsilon ^{{\frac {a}{b}}t}\cdot \sin 2\pi nt\cdot dt\\&=\epsilon ^{{\frac {a}{b}}t}\left\{{\frac {ab\cdot \sin 2\pi nt-2\pi nb^{2}\cdot \cos 2\pi nt}{a^{2}+4\pi ^{2}n^{2}b^{2}}}\right\}.....{\text{[D]}}&\\\end{aligned}}}
Inserting this value in [A], we get
y
=
g
{
a
⋅
sin
2
π
n
t
−
2
π
n
b
⋅
cos
2
π
n
t
a
2
+
4
π
2
n
2
b
2
}
{\displaystyle y=g\left\{{\frac {a\cdot \sin 2\pi nt-2\pi nb\cdot \cos 2\pi nt}{a^{2}+4\pi ^{2}n^{2}b^{2}}}\right\}}
To simplify still further, let us imagine an angle
ϕ
{\displaystyle \phi }
tan
ϕ
=
2
π
n
b
a
{\displaystyle \tan \phi ={\dfrac {2\pi nb}{a}}}
sin
ϕ
=
2
π
n
b
a
2
+
4
π
2
n
2
b
2
{\displaystyle \sin \phi ={\frac {2\pi nb}{\sqrt {a^{2}+4\pi ^{2}n^{2}b^{2}}}}}
and
cos
ϕ
=
a
a
2
+
4
π
2
n
2
b
2
{\displaystyle \cos \phi ={\frac {a}{\sqrt {a^{2}+4\pi ^{2}n^{2}b^{2}}}}}
Substituting these, we get:
y
=
g
cos
ϕ
⋅
sin
2
π
n
t
−
sin
ϕ
⋅
cos
2
π
n
t
a
2
+
4
π
2
n
2
b
2
{\displaystyle y=g{\frac {\cos \phi \cdot \sin 2\pi nt-\sin \phi \cdot \cos 2\pi nt}{\sqrt {a^{2}+4\pi ^{2}n^{2}b^{2}}}}}
which may be written
y
=
g
sin
(
2
π
n
t
−
ϕ
)
a
2
+
4
π
2
n
2
b
2
{\displaystyle y=g{\frac {\sin(2\pi nt-\phi )}{\sqrt {a^{2}+4\pi ^{2}n^{2}b^{2}}}}}
which is the solution desired.
This is indeed none other than the equation of an alternating electric current, where
g
{\displaystyle g}
n
{\displaystyle n}
b
{\displaystyle b}
ϕ
{\displaystyle \phi }
![{\displaystyle {\begin{aligned}\int \epsilon ^{{\frac {a}{b}}t}\cdot \sin 2\pi nt\cdot dt\\&=\epsilon ^{{\frac {a}{b}}t}\left\{{\frac {ab\cdot \sin 2\pi nt-2\pi nb^{2}\cdot \cos 2\pi nt}{a^{2}+4\pi ^{2}n^{2}b^{2}}}\right\}.....{\text{[D]}}&\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/64c430af1330773ca75d9925961eaeddcf80d9e9)
