integral from 0 to pi/8 of sec(2x)tan(2x)

解

\int_{0}^{\frac{π}{8}} sec (2 x) tan (2 x) d x

\frac{1}{2} - \frac{1}{2}

十進法表記

0.20710 \dots

解答ステップ

\int_{0}^{\frac{π}{8}} sec (2 x) tan (2 x) d x

三角関数の公式を使用して書き換える

= \int_{0}^{\frac{π}{8}} \frac{sin ( 2 x )}{cos ^{2} ( 2 x )} d x

部分積分を適用する

= [\frac{1}{2} (sin (2 x) tan (2 x) - 2 \cdot \int sin (2 x) d x)]_{0}^{\frac{π}{8}}

\int sin (2 x) d x = - \frac{1}{2} cos (2 x)

= [\frac{1}{2} (sin (2 x) tan (2 x) - 2 (- \frac{1}{2} cos (2 x)))]_{0}^{\frac{π}{8}}

簡素化

[\frac{1}{2} (sin (2 x) tan (2 x) - 2 (- \frac{1}{2} cos (2 x)))]_{0}^{\frac{π}{8}} : [\frac{1}{2} (sin (2 x) tan (2 x) + cos (2 x))]_{0}^{\frac{π}{8}}

= [\frac{1}{2} (sin (2 x) tan (2 x) + cos (2 x))]_{0}^{\frac{π}{8}}

限度を計算する:

\frac{1}{2} - \frac{1}{2}

= \frac{1}{2} - \frac{1}{2}