integral from 1/4 to 1 of arcsin(x)

解

\int_{\frac{1}{4}}^{1} arcsin (x) d x

\frac{π}{2} - \frac{arcsin ( \frac{1}{4} ) + 15}{4}

十進法表記

0.53938 \dots

解答ステップ

\int_{\frac{1}{4}}^{1} arcsin (x) d x

部分積分を適用する

= [x arcsin (x) - \int \frac{x}{1 - x ^{2}} d x]_{\frac{1}{4}}^{1}

\int \frac{x}{1 - x ^{2}} d x = - 1 - x^{2}

= [x arcsin (x) - (- 1 - x^{2})]_{\frac{1}{4}}^{1}

簡素化

= [x arcsin (x) + 1 - x^{2}]_{\frac{1}{4}}^{1}

限度を計算する:

\frac{π}{2} - \frac{arcsin ( \frac{1}{4} ) + 15}{4}

= \frac{π}{2} - \frac{arcsin ( \frac{1}{4} ) + 15}{4}