integral from 0 to 2pi of sin^5(x)

Lösung

\int_{0}^{2 π} sin^{5} (x) d x

0

Schritte zur Lösung

\int_{0}^{2 π} sin^{5} (x) d x

Vereinfache

sin^{5} (x) : sin^{4} (x) sin (x)

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

Wende die partielle Integration an

= [- sin^{4} (x) cos (x) - \int - 4 sin^{3} (x) cos^{2} (x) d x]_{0}^{2 π}

\int - 4 sin^{3} (x) cos^{2} (x) d x = - 4 (- \frac{cos ^{3} ( x )}{3} + \frac{cos ^{5} ( x )}{5})

= [- sin^{4} (x) cos (x) - (- 4 (- \frac{cos ^{3} ( x )}{3} + \frac{cos ^{5} ( x )}{5}))]_{0}^{2 π}

Vereinfache

[- sin^{4} (x) cos (x) - (- 4 (- \frac{cos ^{3} ( x )}{3} + \frac{cos ^{5} ( x )}{5}))]_{0}^{2 π} : [- sin^{4} (x) cos (x) + 4 (- \frac{1}{3} cos^{3} (x) + \frac{1}{5} cos^{5} (x))]_{0}^{2 π}

= [- sin^{4} (x) cos (x) + 4 (- \frac{1}{3} cos^{3} (x) + \frac{1}{5} cos^{5} (x))]_{0}^{2 π}

Berechne die Grenzen:

0

= 0

\int_{1}^{5} x^{2} - x + 2 d x

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