integral from 0 to pi/2 of sin^4(2x)

Lösung

\int_{0}^{\frac{π}{2}} sin^{4} (2 x) d x

\frac{3 π}{16}

Dezimale

0.58904 \dots

Schritte zur Lösung

\int_{0}^{\frac{π}{2}} sin^{4} (2 x) d x

Wende U-Substitution an

= \int_{0}^{π} sin^{4} (u) \frac{1}{2} d u

Entferne die Konstante:

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

= \frac{1}{2} \cdot \int_{0}^{π} sin^{4} (u) d u

Wende integrale Reduktion an

= \frac{1}{2} (- [\frac{cos ( u ) sin ^{3} ( u )}{4}]_{0}^{π} + \frac{3}{4} \cdot \int_{0}^{π} sin^{2} (u) d u)

\int_{0}^{π} sin^{2} (u) d u = \frac{π}{2}

= \frac{1}{2} (- [\frac{cos ( u ) sin ^{3} ( u )}{4}]_{0}^{π} + \frac{3}{4} \cdot \frac{π}{2})

Vereinfache

\frac{1}{2} (- [\frac{cos ( u ) sin ^{3} ( u )}{4}]_{0}^{π} + \frac{3}{4} \cdot \frac{π}{2}) : \frac{1}{2} (- [\frac{1}{4} sin^{3} (u) cos (u)]_{0}^{π} + \frac{3 π}{8})

= \frac{1}{2} (- [\frac{1}{4} sin^{3} (u) cos (u)]_{0}^{π} + \frac{3 π}{8})

Berechne die Grenzen:

0

= \frac{1}{2} (- 0 + \frac{3 π}{8})

Vereinfache

= \frac{3 π}{16}

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