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人気のある微分積分 >

integral from 0 to (5pi)/4 of tan(x/5)

解

\int_{0}^{\frac{5 π}{4}} tan (\frac{x}{5}) d x

解

\frac{5}{2} ln (2)

+1

十進法表記

1.73286 \dots

解答ステップ

\int_{0}^{\frac{5 π}{4}} tan (\frac{x}{5}) d x

三角関数の公式を使用して書き換える

= \int_{0}^{\frac{5 π}{4}} \frac{sin ( \frac{x}{5} )}{cos ( \frac{x}{5} )} d x

u置換積分法を適用する

= \int_{1}^{\frac{2}{2}} - \frac{5}{u} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{\frac{2}{2}}^{1} - \frac{5}{u} d u

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - (- 5 \cdot \int_{\frac{2}{2}}^{1} \frac{1}{u} d u)

共通積分を使用する:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - (- 5 [ln ∣ u ∣]_{\frac{2}{2}}^{1})

簡素化

- (- 5 [ln ∣ u ∣]_{\frac{2}{2}}^{1}) : 5 [ln ∣ u ∣]_{\frac{1}{2}}^{1}

= 5 [ln ∣ u ∣]_{\frac{1}{2}}^{1}

限度を計算する:

\frac{1}{2} ln (2)

= 5 \cdot \frac{1}{2} ln (2)

簡素化

= \frac{5}{2} ln (2)

グラフ

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