integral of sin(x)sqrt(2+3cos(x))

Lösung

\int sin (x) 2 + 3 cos (x) d x

- \frac{2}{9} (2 + 3 cos (x))^{\frac{3}{2}} + C

Schritte zur Lösung

\int sin (x) 2 + 3 cos (x) d x

Wende U-Substitution an

= \int - 2 + 3 u d u

Entferne die Konstante:

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

= - \int 2 + 3 u d u

Wende U-Substitution an

= - \int \frac{v}{3} d v

Entferne die Konstante:

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

= - \frac{1}{3} \cdot \int v d v

Wende die Potenzregel an

= - \frac{1}{3} \cdot \frac{2}{3} v^{\frac{3}{2}}

Ersetze zurück

= - \frac{1}{3} \cdot \frac{2}{3} (2 + 3 cos (x))^{\frac{3}{2}}

Vereinfache

- \frac{1}{3} \cdot \frac{2}{3} (2 + 3 cos (x))^{\frac{3}{2}} : - \frac{2}{9} (2 + 3 cos (x))^{\frac{3}{2}}

= - \frac{2}{9} (2 + 3 cos (x))^{\frac{3}{2}}

Füge eine Konstante zur Lösung hinzu

= - \frac{2}{9} (2 + 3 cos (x))^{\frac{3}{2}} + C

d er i v a t i v e y = (3 x^{2} + 2) (2 x - 5)

d er i v a t i v e f (x) = \frac{36}{( 1 + 3 x ) ^{3}}

\int_{0}^{6 π} t^{2} sin (2 t) d t

\int 6 (x - 12)^{5} d x

t an g e n t f (x) = x^{2} - 6, at x = 3