integral from 0 to 1 of sin^4(x)

Lösung

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

- \frac{1}{4} sin^{3} (1) cos (1) + \frac{3}{8} (1 - \frac{1}{2} sin (2))

Dezimale

0.12402 \dots

Schritte zur Lösung

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

Wende integrale Reduktion an

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

\int_{0}^{1} sin^{2} (x) d x = \frac{1}{2} (1 - \frac{1}{2} sin (2))

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

Vereinfache

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

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

Berechne die Grenzen:

\frac{1}{4} sin^{3} (1) cos (1)

= - \frac{1}{4} sin^{3} (1) cos (1) + \frac{3}{8} (1 - \frac{1}{2} sin (2))

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