integral from 2 to 1/x of tan^2(t)

解

\int_{2}^{\frac{1}{x}} tan^{2} (t) d t

- \frac{1}{x} + 2 + tan (\frac{1}{x}) - tan (2)

解答ステップ

\int_{2}^{\frac{1}{x}} tan^{2} (t) d t

三角関数の公式を使用して書き換える

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

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_{2}^{\frac{1}{x}} 1 d t + \int_{2}^{\frac{1}{x}} sec^{2} (t) d t

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

\int_{2}^{\frac{1}{x}} sec^{2} (t) d t = tan (\frac{1}{x}) - tan (2)

= - (\frac{1}{x} - 2) + tan (\frac{1}{x}) - tan (2)

簡素化

= - \frac{1}{x} + 2 + tan (\frac{1}{x}) - tan (2)