Page:Calculus Made Easy.pdf/260

Inserting these, the integral in question becomes:

${\begin{aligned}\int \epsilon ^{{\frac {a}{b}}t}&{}\cdot \sin 2\pi nt\cdot dt\\&=-{\frac {1}{2\pi n}}\cdot \epsilon ^{{\frac {a}{b}}t}\cdot \cos 2\pi nt-\int -{\frac {1}{2\pi n}}\cos 2\pi nt\cdot \epsilon ^{{\frac {a}{b}}t}\cdot {\frac {a}{b}}\,dt\\&=-{\frac {1}{2\pi n}}\epsilon ^{{\frac {a}{b}}t}\cos 2\pi nt+{\frac {a}{2\pi nb}}\int \epsilon ^{{\frac {a}{b}}t}\cdot \cos 2\pi nt\cdot dt............{\text{[B]}}\end{aligned}}$

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:

${\begin{aligned}&\left\{{\begin{aligned}u&=\sin 2\pi nt;\\dv&=\epsilon ^{{\frac {a}{b}}t}\cdot dt;\end{aligned}}\right.\\[1ex]whence\quad \quad \quad &\left\{{\begin{aligned}du&=2\pi n\cdot \cos 2\pi nt\cdot dt;\\v&={\frac {b}{a}}\epsilon ^{{\frac {a}{b}}t}\end{aligned}}\right.\end{aligned}}$

Inserting these, we get

${\begin{aligned}\int \epsilon ^{{\frac {a}{b}}t}&{}\cdot \sin 2\pi nt\cdot dt\\&={\frac {b}{a}}\cdot \epsilon ^{{\frac {a}{b}}t}\cdot \sin 2\pi nt-{\frac {2\pi nb}{a}}\int \epsilon ^{{\frac {a}{b}}t}\cdot \cos 2\pi nt\cdot dt............{\text{[C]}}\end{aligned}}$

Noting that the final intractable integral in [C] is the same as that in [B], we may eliminate it, by multiplying [B] by ${\dfrac {2\pi nb}{a}}$, and multiplying [C] by ${\dfrac {a}{2\pi nb}}$, and adding them.