Page:Calculus Made Easy.pdf/260

 Inserting these, the integral in question becomes:

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: {{c|$$\begin{align} &\left\{ \begin{aligned} u &= \sin 2 \pi n t ; \\ dv &= \epsilon^{\frac{a}{b} t} \cdot dt; \end{aligned} \right. \\[1ex] whence \quad \quad \quad &\left\{ \begin{aligned} du &= 2 \pi n \cdot \cos 2 \pi n t \cdot dt; \\ v &= \frac{b}{a} \epsilon ^{\frac{a}{b} t} \end{aligned} \right.\end{align}$$}} Inserting these, we get {{c|$$\begin{align} \int \epsilon^{\frac{a}{b} t} &{} \cdot \sin 2 \pi n t \cdot dt\\ &= \frac{b}{a} \cdot \epsilon^{\frac{a}{b} t} \cdot \sin 2 \pi n t - \frac{2 \pi n b}{a} \int \epsilon^{\frac{a}{b} t} \cdot \cos 2 \pi n t \cdot dt............\text{[C]} \end{align}$$}} 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. {{nop}}