integral from 0 to 1 of tan^3(x)

解

\int_{0}^{1} tan^{3} (x) d x

- ln (sec (1)) + \frac{sec ^{2} ( 1 ) - 1}{2}

十進法表記

0.59713 \dots

解答ステップ

\int_{0}^{1} tan^{3} (x) d x

三角関数の公式を使用して書き換える

= \int_{0}^{1} (- 1 + sec^{2} (x)) tan (x) d x

u置換積分法を適用する

= \int_{1}^{s e c (1)} \frac{- 1 + u ^{2}}{u} d u

拡張

\frac{- 1 + u ^{2}}{u} : - \frac{1}{u} + u

= \int_{1}^{s e c (1)} - \frac{1}{u} + u d u

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_{1}^{s e c (1)} \frac{1}{u} d u + \int_{1}^{s e c (1)} u d u

\int_{1}^{s e c (1)} \frac{1}{u} d u = ln (sec (1))

\int_{1}^{s e c (1)} u d u = \frac{sec ^{2} ( 1 ) - 1}{2}

= - ln (sec (1)) + \frac{sec ^{2} ( 1 ) - 1}{2}