integral from 0 to pi/2 of arctan(2x)

解

\int_{0}^{\frac{π}{2}} arctan (2 x) d x

\frac{1}{4} (2 π arctan (π) - ln (π^{2} + 1))

十進法表記

1.38683 \dots

解答ステップ

\int_{0}^{\frac{π}{2}} arctan (2 x) d x

部分積分を適用する

= [x arctan (2 x) - \int \frac{2 x}{4 x ^{2} + 1} d x]_{0}^{\frac{π}{2}}

\int \frac{2 x}{4 x ^{2} + 1} d x = \frac{1}{4} ln 4 x^{2} + 1

= [x arctan (2 x) - \frac{1}{4} ln 4 x^{2} + 1]_{0}^{\frac{π}{2}}

簡素化

[x arctan (2 x) - \frac{1}{4} ln 4 x^{2} + 1]_{0}^{\frac{π}{2}} : [\frac{1}{4} (4 x arctan (2 x) - ln 4 x^{2} + 1)]_{0}^{\frac{π}{2}}

= [\frac{1}{4} (4 x arctan (2 x) - ln 4 x^{2} + 1)]_{0}^{\frac{π}{2}}

限度を計算する:

\frac{1}{4} (2 π arctan (π) - ln (π^{2} + 1))

= \frac{1}{4} (2 π arctan (π) - ln (π^{2} + 1))