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人気のある微分積分 >

integral from 0 to pi/4 of tan(θ)

解

\int_{0}^{\frac{π}{4}} tan (θ) d θ

解

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

+1

十進法表記

0.34657 \dots

解答ステップ

\int_{0}^{\frac{π}{4}} tan (θ) d θ

三角関数の公式を使用して書き換える

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

u置換積分法を適用する

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

定数を除く:

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

= - (- \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 ∣)

= - (- [ln ∣ u ∣]_{\frac{2}{2}}^{1})

簡素化

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

= [ln ∣ u ∣]_{\frac{1}{2}}^{1}

限度を計算する:

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

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

グラフ

人気の例

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\int_{\frac{1}{2}}^{\frac{1}{3}} cos (π \cdot x) d x

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