integral from 0 to 1 of 4xarccos(x)

Lösung

\int_{0}^{1} 4 x arccos (x) d x

\frac{π}{2}

Dezimale

1.57079 \dots

Schritte zur Lösung

\int_{0}^{1} 4 x arccos (x) d x

Entferne die Konstante:

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

= 4 \cdot \int_{0}^{1} x arccos (x) d x

Wende die partielle Integration an

= 4 [\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)))

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

Vereinfache

= 4 [\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}

= 4 \cdot \frac{π}{8}

Vereinfache

= \frac{π}{2}

\int \frac{13 x ^{3}}{2 + x ^{2}} d x

d er i v a t i v e sin ((π x)^{3})

x^{2} y^{^{'}} - 2 x y = 5 y^{4}, y (1) = \frac{1}{4}

\int_{1}^{2} \frac{y ^{3}}{2} + \frac{1}{2 y ^{3}} d y

\int \frac{x ^{3} + 1}{x ^{3} - x ^{2} - 2 x} d x