integral from 0 to pi of arctan(x)

解

\int_{0}^{π} arctan (x) d x

π arctan (π) - \frac{1}{2} ln (π^{2} + 1)

十進法表記

2.77367 \dots

解答ステップ

\int_{0}^{π} arctan (x) d x

部分積分を適用する

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

\int \frac{x}{x ^{2} + 1} d x = \frac{1}{2} ln x^{2} + 1

= [x arctan (x) - \frac{1}{2} ln x^{2} + 1]_{0}^{π}

簡素化

[x arctan (x) - \frac{1}{2} ln x^{2} + 1]_{0}^{π} : [\frac{1}{2} (2 x arctan (x) - ln x^{2} + 1)]_{0}^{π}

= [\frac{1}{2} (2 x arctan (x) - ln x^{2} + 1)]_{0}^{π}

限度を計算する:

π arctan (π) - \frac{1}{2} ln (π^{2} + 1)

= π arctan (π) - \frac{1}{2} ln (π^{2} + 1)