integral of sin^3(5x)sqrt(cos(5x))

Lösung

\int sin^{3} (5 x) cos (5 x) d x

- \frac{1}{5} (\frac{2}{3} cos^{\frac{3}{2}} (5 x) - \frac{2}{7} cos^{\frac{7}{2}} (5 x)) + C

Schritte zur Lösung

\int sin^{3} (5 x) cos (5 x) d x

Wende U-Substitution an

= \int sin^{3} (u) cos (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 sin^{3} (u) cos (u) d u

Wende U-Substitution an

= \frac{1}{5} \cdot \int - v (1 - v^{2}) d v

Entferne die Konstante:

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

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

Multipliziere aus

v (1 - v^{2}) : v - v^{\frac{5}{2}}

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

\int v d v = \frac{2}{3} v^{\frac{3}{2}}

\int v^{\frac{5}{2}} d v = \frac{2}{7} v^{\frac{7}{2}}

= \frac{1}{5} (- (\frac{2}{3} v^{\frac{3}{2}} - \frac{2}{7} v^{\frac{7}{2}}))

Ersetze zurück

= \frac{1}{5} (- (\frac{2}{3} cos^{\frac{3}{2}} (5 x) - \frac{2}{7} cos^{\frac{7}{2}} (5 x)))

Vereinfache

= - \frac{1}{5} (\frac{2}{3} cos^{\frac{3}{2}} (5 x) - \frac{2}{7} cos^{\frac{7}{2}} (5 x))

Füge eine Konstante zur Lösung hinzu

= - \frac{1}{5} (\frac{2}{3} cos^{\frac{3}{2}} (5 x) - \frac{2}{7} cos^{\frac{7}{2}} (5 x)) + C

\int_{1}^{2} x 3^{(x^{2})} d x

(x^{2} + y^{2}) d x + x y d y = 0

\frac{\partial}{\partial x} (- y sin (x ln (y)) ln (y))

x \to e lim (- 11 - 12)

d er i v a t i v e y = x + 12