integral from 0 to pi/2 of (x+1)cos(2x)

Lösung

\int_{0}^{\frac{π}{2}} (x + 1) cos (2 x) d x

- \frac{1}{2}

Dezimale

- 0.5

Schritte zur Lösung

\int_{0}^{\frac{π}{2}} (x + 1) cos (2 x) d x

Wende die partielle Integration an

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

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

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

Vereinfache

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

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

Berechne die Grenzen:

- \frac{1}{2}

= - \frac{1}{2}

\int_{0}^{2 π} 1 + cos (x) d x

\int_{0}^{1} 3^{x} e^{x} sin (x) d x

\int_{3}^{5} - x^{2} + 7 x - 10 d x

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

\int_{- π}^{π} e^{x} cos (x) d x