\begin{align} f(x) &= \left( 1 + 2x + \frac{(2x)^2}{2} + \frac{(2x)^3}{6} + \cdots \right) \left( 1 - \frac{x^2}{2} + \frac{x^4}{24} + \cdots \right) \\ &= 1 + 2x + \frac{3}{2} x^2 + \frac{1}{3} x^3 + \cdots \end{align}
$t = x + \sqrt{a^2 + x^2}$ とおくと、 \begin{align} \frac{dt}{dx} &= 1 + \frac{x}{\sqrt{a^2 + x^2}} \\ &= \frac{\sqrt{a^2 + x^2} + x}{\sqrt{a^2 + x^2}} \\ &= \frac{t}{\sqrt{a^2 + x^2}} \\ \therefore \ \ \frac{dx}{\sqrt{a^2 + x^2}} &= \frac{dt}{t} \end{align} なので、 \begin{align} \int \frac{dx}{\sqrt{a^2 + x^2}} &= \int \frac{dt}{t} \\ &= \log \left| t \right| + C \\ &= \log \left| x + \sqrt{a^2 + x^2} \right| + C \ \ \ \ \ \ \ \ ( C \text{ は積分定数 } ) \end{align} である。
\begin{align} \iint_D \sin (x+y) \ dx dy &= \int_0^\pi dx \int_0^{\pi - x} dy \sin (x+y) \\ &= \int_0^\pi dx \left[ - \cos (x+y) \right]_{y=0}^{y = \pi - x} \\ &= \int_0^\pi dx \left( 1 + \cos x \right) \\ &= \left[ x + \sin x \right]_0^\pi \\ &= \pi \end{align}