integral from 0 to pi of x^3cos(2x)

Lösung

\int_{0}^{π} x^{3} cos (2 x) d x

\frac{3 π ^{2}}{4}

Dezimale

7.40220 \dots

Schritte zur Lösung

\int_{0}^{π} x^{3} cos (2 x) d x

Wende die partielle Integration an

= [\frac{1}{2} (x^{3} sin (2 x) - 2 \cdot \int \frac{3}{2} x^{2} sin (2 x) d x)]_{0}^{π}

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

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

Vereinfache

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

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

Berechne die Grenzen:

- \frac{3 ( 1 - 2 π ^{2} )}{8} + \frac{3}{8}

= - \frac{3 ( 1 - 2 π ^{2} )}{8} + \frac{3}{8}

Vereinfache

= \frac{3 π ^{2}}{4}

\int_{2}^{\infty} x^{- \frac{1}{3}} d x

\int_{- 1}^{2} (- x^{4} - x^{2} + 6 x + 8) d x

\int_{0}^{2 π} sin (2 x) sin (3 x) d x

\int_{0}^{2} 7 x^{4} d x

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