integral from 0 to 1 of 2xarccos(x)

Lösung

\int_{0}^{1} 2 x arccos (x) d x

\frac{π}{4}

Dezimale

0.78539 \dots

Schritte zur Lösung

\int_{0}^{1} 2 x arccos (x) d x

Entferne die Konstante:

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

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

Wende die partielle Integration an

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

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

= 2 [\frac{1}{2} x^{2} arccos (x) - (- \frac{1}{4} (arcsin (x) - \frac{1}{2} sin (2 arcsin (x))))]_{0}^{1}

Vereinfache

= 2 [\frac{1}{2} x^{2} arccos (x) + \frac{1}{4} (arcsin (x) - \frac{1}{2} sin (2 arcsin (x)))]_{0}^{1}

Berechne die Grenzen:

\frac{π}{8}

= 2 \cdot \frac{π}{8}

Vereinfache

= \frac{π}{4}

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