integral of sin^2(3xco)s^23x

Lösung

\int sin^{2} (3 x co) s^{2} 3 x d x

\frac{s ^{2}}{24 c ^{2} o ^{2}} (18 c^{2} o^{2} x^{2} - 6 co x sin (6 co x) - cos (6 co x)) + C

Schritte zur Lösung

\int sin^{2} (3 x co) s^{2} \cdot 3 x d x

Entferne die Konstante:

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

= 3 s^{2} \cdot \int sin^{2} (3 co x) x d x

Wende U-Substitution an

= 3 s^{2} \cdot \int \frac{u sin ^{2} ( u )}{9 c ^{2} o ^{2}} d u

Entferne die Konstante:

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

= 3 s^{2} \frac{1}{9 c ^{2} o ^{2}} \cdot \int u sin^{2} (u) d u

Wende die partielle Integration an

= 3 s^{2} \frac{1}{9 c ^{2} o ^{2}} (\frac{1}{2} u (u - \frac{1}{2} sin (2 u)) - \int \frac{1}{2} (u - \frac{1}{2} sin (2 u)) d u)

\int \frac{1}{2} (u - \frac{1}{2} sin (2 u)) d u = \frac{1}{2} (\frac{u ^{2}}{2} + \frac{1}{4} cos (2 u))

= 3 s^{2} \frac{1}{9 c ^{2} o ^{2}} (\frac{1}{2} u (u - \frac{1}{2} sin (2 u)) - \frac{1}{2} (\frac{u ^{2}}{2} + \frac{1}{4} cos (2 u)))

Setze in

u = 3 co x

ein

= 3 s^{2} \frac{1}{9 c ^{2} o ^{2}} (\frac{1}{2} \cdot 3 co x (3 co x - \frac{1}{2} sin (2 \cdot 3 co x)) - \frac{1}{2} (\frac{( 3 co x ) ^{2}}{2} + \frac{1}{4} cos (2 \cdot 3 co x)))

Vereinfache

3 s^{2} \frac{1}{9 c ^{2} o ^{2}} (\frac{1}{2} \cdot 3 co x (3 co x - \frac{1}{2} sin (2 \cdot 3 co x)) - \frac{1}{2} (\frac{( 3 co x ) ^{2}}{2} + \frac{1}{4} cos (2 \cdot 3 co x))) : \frac{s ^{2}}{24 c ^{2} o ^{2}} (18 c^{2} o^{2} x^{2} - 6 co x sin (6 co x) - cos (6 co x))

= \frac{s ^{2}}{24 c ^{2} o ^{2}} (18 c^{2} o^{2} x^{2} - 6 co x sin (6 co x) - cos (6 co x))

Füge eine Konstante zur Lösung hinzu

= \frac{s ^{2}}{24 c ^{2} o ^{2}} (18 c^{2} o^{2} x^{2} - 6 co x sin (6 co x) - cos (6 co x)) + C

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

\frac{d}{d v} (u sin (v))

\frac{\partial}{\partial y} (ln (5 + x^{2} y^{2}))

t an g e n t f (x) = 4 x - x^{2}, at x = 2

s l o p e s = t^{3} - t^{2}