integral from 0 to 1 of pi(arcsin(x))^2

解

\int_{0}^{1} π (arcsin (x))^{2} d x

π (\frac{π ^{2}}{4} - 2)

十進法表記

1.46838 \dots

解答ステップ

\int_{0}^{1} π (arcsin (x))^{2} d x

定数を除く:

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

= π \cdot \int_{0}^{1} (arcsin (x))^{2} d x

部分積分を適用する

= π [x arcsin^{2} (x) - \int \frac{2 x arcsin ( x )}{1 - x ^{2}} d x]_{0}^{1}

\int \frac{2 x arcsin ( x )}{1 - x ^{2}} d x = 2 (- arcsin (x) 1 - x^{2} + x)

= π [x arcsin^{2} (x) - 2 (- arcsin (x) 1 - x^{2} + x)]_{0}^{1}

限度を計算する:

\frac{π ^{2}}{4} - 2

= π (\frac{π ^{2}}{4} - 2)