integral of 6sin^2(3x)cos(3x)ln(sin(3x))

Lösung

\int 6 sin^{2} (3 x) cos (3 x) ln (sin (3 x)) d x

2 (\frac{1}{3} sin^{3} (3 x) ln (sin (3 x)) - \frac{1}{9} sin^{3} (3 x)) + C

Schritte zur Lösung

\int 6 sin^{2} (3 x) cos (3 x) ln (sin (3 x)) d x

Entferne die Konstante:

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

= 6 \cdot \int sin^{2} (3 x) cos (3 x) ln (sin (3 x)) d x

Wende U-Substitution an

= 6 \cdot \int sin^{2} (u) cos (u) ln (sin (u)) \frac{1}{3} d u

Entferne die Konstante:

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

= 6 \cdot \frac{1}{3} \cdot \int sin^{2} (u) cos (u) ln (sin (u)) d u

Wende die partielle Integration an

= 6 \cdot \frac{1}{3} (\frac{1}{3} sin^{3} (u) ln (sin (u)) - \int \frac{1}{3} sin^{3} (u) cot (u) d u)

\int \frac{1}{3} sin^{3} (u) cot (u) d u = \frac{1}{9} sin^{3} (u)

= 6 \cdot \frac{1}{3} (\frac{1}{3} sin^{3} (u) ln (sin (u)) - \frac{1}{9} sin^{3} (u))

Setze in

u = 3 x

ein

= 6 \cdot \frac{1}{3} (\frac{1}{3} sin^{3} (3 x) ln (sin (3 x)) - \frac{1}{9} sin^{3} (3 x))

Vereinfache

6 \cdot \frac{1}{3} (\frac{1}{3} sin^{3} (3 x) ln (sin (3 x)) - \frac{1}{9} sin^{3} (3 x)) : 2 (\frac{1}{3} sin^{3} (3 x) ln (sin (3 x)) - \frac{1}{9} sin^{3} (3 x))

= 2 (\frac{1}{3} sin^{3} (3 x) ln (sin (3 x)) - \frac{1}{9} sin^{3} (3 x))

Füge eine Konstante zur Lösung hinzu

= 2 (\frac{1}{3} sin^{3} (3 x) ln (sin (3 x)) - \frac{1}{9} sin^{3} (3 x)) + C

\int_{0}^{4} 4 x - 2 x d x

\frac{d y}{d x} = e^{3 x} + 9 y

x \to - 2 lim (\frac{sin ( - π x )}{2 + x})

\int_{2}^{3} \frac{22}{3 - x} d x

t an g e n t f (x) = x^{3} - 4 x, at x = 0