integral from 0 to 1/4 of sin^2(pix)

解

\int_{0}^{\frac{1}{4}} sin^{2} (π x) d x

\frac{π - 2}{8 π}

十進法表記

0.04542 \dots

解答ステップ

\int_{0}^{\frac{1}{4}} sin^{2} (π x) d x

三角関数の公式を使用して書き換える

= \int_{0}^{\frac{1}{4}} \frac{1 - cos ( 2 π x )}{2} d x

定数を除く:

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

= \frac{1}{2} \cdot \int_{0}^{\frac{1}{4}} 1 - cos (2 π x) d x

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

= \frac{1}{2} (\int_{0}^{\frac{1}{4}} 1 d x - \int_{0}^{\frac{1}{4}} cos (2 π x) d x)

\int_{0}^{\frac{1}{4}} 1 d x = \frac{1}{4}

\int_{0}^{\frac{1}{4}} cos (2 π x) d x = \frac{1}{2 π}

= \frac{1}{2} (\frac{1}{4} - \frac{1}{2 π})

簡素化

\frac{1}{2} (\frac{1}{4} - \frac{1}{2 π}) : \frac{π - 2}{8 π}

= \frac{π - 2}{8 π}