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

Lösung

\int_{0}^{π} t^{2} cos (3 t) d t

- \frac{2 π}{9}

Dezimale

- 0.69813 \dots

Schritte zur Lösung

\int_{0}^{π} t^{2} cos (3 t) d t

Wende U-Substitution an

= \int_{0}^{3 π} \frac{u ^{2} cos ( u )}{27} d u

Entferne die Konstante:

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

= \frac{1}{27} \cdot \int_{0}^{3 π} u^{2} cos (u) d u

Wende die partielle Integration an

= \frac{1}{27} [u^{2} sin (u) - \int 2 u sin (u) d u]_{0}^{3 π}

\int 2 u sin (u) d u = 2 (- u cos (u) + sin (u))

= \frac{1}{27} [u^{2} sin (u) - 2 (- u cos (u) + sin (u))]_{0}^{3 π}

Berechne die Grenzen:

- 6 π

= \frac{1}{27} (- 6 π)

Vereinfache

= - \frac{2 π}{9}

\int_{1}^{e} (12 + ln (x))^{2} d x

\int_{0}^{π} (x - sin^{2} (x)) d x

\int_{- 8}^{8} \frac{1}{x ^{\frac{1}{3}}} d x

\int_{0}^{1} x (x^{2} - 1)^{4} d x

\int_{4}^{5} (\frac{2}{x} - x) d x