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人気のある微分積分 >

integral from 0 to 1 of pi(tan(pi/4 y))^2

解

\int_{0}^{1} π (tan (\frac{π}{4} y))^{2} d y

解

- π + 4

+1

十進法表記

0.85840 \dots

解答ステップ

\int_{0}^{1} π (tan (\frac{π}{4} y))^{2} d y

定数を除く:

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

= π \cdot \int_{0}^{1} (tan (\frac{π}{4} y))^{2} d y

簡素化

= π \cdot \int_{0}^{1} tan^{2} (\frac{π}{4} y) d y

u置換積分法を適用する

= π \cdot \int_{0}^{\frac{π}{4}} tan^{2} (u) \frac{4}{π} d u

定数を除く:

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

= π \frac{4}{π} \cdot \int_{0}^{\frac{π}{4}} tan^{2} (u) d u

三角関数の公式を使用して書き換える

= π \frac{4}{π} \cdot \int_{0}^{\frac{π}{4}} - 1 + sec^{2} (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

= π \frac{4}{π} (- \int_{0}^{\frac{π}{4}} 1 d u + \int_{0}^{\frac{π}{4}} sec^{2} (u) d u)

\int_{0}^{\frac{π}{4}} 1 d u = \frac{π}{4}

\int_{0}^{\frac{π}{4}} sec^{2} (u) d u = 1

= π \frac{4}{π} (- \frac{π}{4} + 1)

簡素化

π \frac{4}{π} (- \frac{π}{4} + 1) : - π + 4

= - π + 4

グラフ

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