integral from 0 to (3pi)/2 of cos^3(5x)

Lösung

\int_{0}^{\frac{3 π}{2}} cos^{3} (5 x) d x

- \frac{2}{15}

Dezimale

- 0.13333 \dots

Schritte zur Lösung

\int_{0}^{\frac{3 π}{2}} cos^{3} (5 x) d x

Wende U-Substitution an

= \int_{0}^{\frac{15 π}{2}} cos^{3} (u) \frac{1}{5} d u

Entferne die Konstante:

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

= \frac{1}{5} \cdot \int_{0}^{\frac{15 π}{2}} cos^{3} (u) d u

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= \frac{1}{5} \cdot \int_{0}^{\frac{15 π}{2}} (1 - sin^{2} (u)) cos (u) d u

Wende U-Substitution an

= \frac{1}{5} \cdot \int_{0}^{- 1} 1 - v^{2} d v

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= \frac{1}{5} (- \int_{- 1}^{0} 1 - v^{2} d v)

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{5} (- (\int_{- 1}^{0} 1 d v - \int_{- 1}^{0} v^{2} d v))

\int_{- 1}^{0} 1 d v = 1

\int_{- 1}^{0} v^{2} d v = \frac{1}{3}

= \frac{1}{5} (- (1 - \frac{1}{3}))

Vereinfache

\frac{1}{5} (- (1 - \frac{1}{3})) : - \frac{2}{15}

= - \frac{2}{15}

\int_{0}^{3} x^{2} + 9 d x

t an g e n t y = 2 x^{3} + 3 x^{2} - 12 x + 2

\int x^{2} - x - 6 d x

\frac{d}{d x} (4 sin (4 x))

\frac{\partial}{\partial y} (x y + y^{2})