integral from 0 to pi of arctan(x)

Lösung

\int_{0}^{π} arctan (x) d x

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

Dezimale

2.77367 \dots

Schritte zur Lösung

\int_{0}^{π} arctan (x) d x

Wende die partielle Integration an

= [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}

Vereinfache

[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}

Berechne die Grenzen:

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

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

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