integral from-pi/4 to pi/4 of tan^2(x)

解

\int_{- \frac{π}{4}}^{\frac{π}{4}} tan^{2} (x) d x

- \frac{π}{2} + 2

十進法表記

0.42920 \dots

解答ステップ

\int_{- \frac{π}{4}}^{\frac{π}{4}} tan^{2} (x) d x

三角関数の公式を使用して書き換える

= \int_{- \frac{π}{4}}^{\frac{π}{4}} - 1 + sec^{2} (x) d x

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int_{- \frac{π}{4}}^{\frac{π}{4}} 1 d x + \int_{- \frac{π}{4}}^{\frac{π}{4}} sec^{2} (x) d x

\int_{- \frac{π}{4}}^{\frac{π}{4}} 1 d x = \frac{π}{2}

\int_{- \frac{π}{4}}^{\frac{π}{4}} sec^{2} (x) d x = 2

= - \frac{π}{2} + 2