integral from 0 to 1 of arcsin(sqrt(x))

Lösung

\int_{0}^{1} arcsin (x) d x

\frac{π}{4}

Dezimale

0.78539 \dots

Schritte zur Lösung

\int_{0}^{1} arcsin (x) d x

Wende U-Substitution an

= \int_{0}^{1} arcsin (u) \cdot 2 u d u

Entferne die Konstante:

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

= 2 \cdot \int_{0}^{1} arcsin (u) u d u

Wende die partielle Integration an

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

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

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

Berechne die Grenzen:

\frac{π}{8}

= 2 \cdot \frac{π}{8}

Vereinfache

= \frac{π}{4}

\int_{0}^{1} 4 π x d x

\int_{0}^{2} 1 + x^{2} d x

\int_{- \frac{π}{4}}^{\frac{π}{4}} tan^{2} (x) d x

\int_{- 1}^{1} (3 + 1 - x^{2}) d x

\int_{0}^{1} (x + 4 1 - x^{2}) d x