\begin{align} \int \frac{1}{\cosh x + 1} dx &= \int \frac{2}{e^x + e^{-x} + 2} dx \\ &= \int \frac{2}{t + \frac{1}{t} + 2} \frac{dt}{t} \ \ \ \ \ \ \ \ \left( t = e^x \right) \\ &= 2 \int \frac{dt}{(t+1)^2} \\ &= - \frac{2}{t+1} + C \ \ \ \ \ \ \ \ \left( C \text{ は積分定数 } \right) \\ &= - \frac{2}{e^x+1} + C \end{align}
\begin{align} \lim_{x \to 0} \frac{x \cos x - \log_e (1+x)}{x \sin x} &= \lim_{x \to 0} \frac{\cos x - x \sin x - \frac{1}{1+x}}{\sin x + x \cos x} \\ &= \lim_{x \to 0} \frac{- \sin x - \sin x - x \cos x + \frac{1}{(1+x)^2}} {\cos x + \cos x - x \sin x} \\ &= \lim_{x \to 0} \frac{- 2 \sin x - x \cos x + \frac{1}{(1+x)^2}} {2 \cos x - x \sin x} \\ &= \frac{1}{2} \end{align}
$x = r \cos \theta, y = r \sin \theta$ によって極座標 $r,\theta$ を導入して、次のように計算できる： \begin{align} \iint_D \sqrt{\frac{1-x^2-y^2}{1+x^2+y^2}} dx dy &= \int_0^{2 \pi} d \theta \int_0^1 dr r \sqrt{\frac{1-r^2}{1+r^2}} \\ &= 2 \pi \int_1^0 \frac{-2t dt}{(1+t^2)^2} \cdot t \ \ \ \ \ \ \ \ \left( t = \sqrt{\frac{1-r^2}{1+r^2}} \right) \\ &= 4 \pi \int_0^1 \frac{t^2}{(1+t^2)^2} dt \\ &= 4 \pi \int_0^\frac{\pi}{4} \frac{\tan^2 \theta}{(1+\tan^2 \theta)^2} \frac{d \theta}{\cos^2 \theta} \ \ \ \ \ \ \ \ \left( t = \tan \theta \right) \\ &= 4 \pi \int_0^\frac{\pi}{4} \sin^2 \theta d \theta \\ &= 2 \pi \int_0^\frac{\pi}{4} \left( 1 - \cos 2 \theta \right) d \theta \\ &= 2 \pi \left[ \theta - \frac{1}{2} \sin 2 \theta \right]_0^\frac{\pi}{4} \\ &= 2 \pi \left( \frac{\pi}{4} - \frac{1}{2} \right) \\ &= \frac{\pi (\pi - 2)}{2} \end{align}