Inserting these, the integral in question becomes:
∫
ϵ
a
b
t
⋅
sin
2
π
n
t
⋅
d
t
=
−
1
2
π
n
⋅
ϵ
a
b
t
⋅
cos
2
π
n
t
−
∫
−
1
2
π
n
cos
2
π
n
t
⋅
ϵ
a
b
t
⋅
a
b
d
t
=
−
1
2
π
n
ϵ
a
b
t
cos
2
π
n
t
+
a
2
π
n
b
∫
ϵ
a
b
t
⋅
cos
2
π
n
t
⋅
d
t
.
.
.
.
.
.
.
.
.
.
.
.
[B]
{\displaystyle {\begin{aligned}\int \epsilon ^{{\frac {a}{b}}t}&{}\cdot \sin 2\pi nt\cdot dt\\&=-{\frac {1}{2\pi n}}\cdot \epsilon ^{{\frac {a}{b}}t}\cdot \cos 2\pi nt-\int -{\frac {1}{2\pi n}}\cos 2\pi nt\cdot \epsilon ^{{\frac {a}{b}}t}\cdot {\frac {a}{b}}\,dt\\&=-{\frac {1}{2\pi n}}\epsilon ^{{\frac {a}{b}}t}\cos 2\pi nt+{\frac {a}{2\pi nb}}\int \epsilon ^{{\frac {a}{b}}t}\cdot \cos 2\pi nt\cdot dt............{\text{[B]}}\end{aligned}}}
The last integral is still irreducible. To evade the difficulty, repeat the integration by parts of the left side, but treating it in the reverse way by writing:
{
u
=
sin
2
π
n
t
;
d
v
=
ϵ
a
b
t
⋅
d
t
;
w
h
e
n
c
e
{
d
u
=
2
π
n
⋅
cos
2
π
n
t
⋅
d
t
;
v
=
b
a
ϵ
a
b
t
{\displaystyle {\begin{aligned}&\left\{{\begin{aligned}u&=\sin 2\pi nt;\\dv&=\epsilon ^{{\frac {a}{b}}t}\cdot dt;\end{aligned}}\right.\\[1ex]whence\quad \quad \quad &\left\{{\begin{aligned}du&=2\pi n\cdot \cos 2\pi nt\cdot dt;\\v&={\frac {b}{a}}\epsilon ^{{\frac {a}{b}}t}\end{aligned}}\right.\end{aligned}}}
Inserting these, we get
∫
ϵ
a
b
t
⋅
sin
2
π
n
t
⋅
d
t
=
b
a
⋅
ϵ
a
b
t
⋅
sin
2
π
n
t
−
2
π
n
b
a
∫
ϵ
a
b
t
⋅
cos
2
π
n
t
⋅
d
t
.
.
.
.
.
.
.
.
.
.
.
.
[C]
{\displaystyle {\begin{aligned}\int \epsilon ^{{\frac {a}{b}}t}&{}\cdot \sin 2\pi nt\cdot dt\\&={\frac {b}{a}}\cdot \epsilon ^{{\frac {a}{b}}t}\cdot \sin 2\pi nt-{\frac {2\pi nb}{a}}\int \epsilon ^{{\frac {a}{b}}t}\cdot \cos 2\pi nt\cdot dt............{\text{[C]}}\end{aligned}}}
Noting that the final intractable integral in [C] is the same as that in [B], we may eliminate it, by multiplying [B] by
2
π
n
b
a
{\displaystyle {\dfrac {2\pi nb}{a}}}
, and multiplying [C] by
a
2
π
n
b
{\displaystyle {\dfrac {a}{2\pi nb}}}
, and adding them.