integral from 0 to pi/6 of (sec(x)+tan(x))^2

解

\int_{0}^{\frac{π}{6}} (sec (x) + tan (x))^{2} d x

\frac{- 3 π + 36}{6 3} - 2

十進法表記

0.94050 \dots

解答ステップ

\int_{0}^{\frac{π}{6}} (sec (x) + tan (x))^{2} d x

拡張

(sec (x) + tan (x))^{2} : sec^{2} (x) + 2 sec (x) tan (x) + tan^{2} (x)

= \int_{0}^{\frac{π}{6}} sec^{2} (x) + 2 sec (x) tan (x) + tan^{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_{0}^{\frac{π}{6}} sec^{2} (x) d x + \int_{0}^{\frac{π}{6}} 2 sec (x) tan (x) d x + \int_{0}^{\frac{π}{6}} tan^{2} (x) d x

\int_{0}^{\frac{π}{6}} sec^{2} (x) d x = \frac{1}{3}

\int_{0}^{\frac{π}{6}} 2 sec (x) tan (x) d x = 2 (\frac{2}{3} - 1)

\int_{0}^{\frac{π}{6}} tan^{2} (x) d x = - \frac{π}{6} + \frac{1}{3}

= \frac{1}{3} + 2 (\frac{2}{3} - 1) - \frac{π}{6} + \frac{1}{3}

簡素化

\frac{1}{3} + 2 (\frac{2}{3} - 1) - \frac{π}{6} + \frac{1}{3} : \frac{- 3 π + 36}{6 3} - 2

= \frac{- 3 π + 36}{6 3} - 2