integral of cos(2ln(x))

Lösung

\int cos (2 ln (x)) d x

\frac{2}{5} x sin (2 ln (x)) + \frac{1}{5} x cos (2 ln (x)) + C

Schritte zur Lösung

\int cos (2 ln (x)) d x

Wende U-Substitution an

= \int cos (2 u) e^{u} d u

Wende die partielle Integration an

= e^{u} \frac{1}{2} sin (2 u) - \int e^{u} \frac{1}{2} sin (2 u) d u

Entferne die Konstante:

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

= e^{u} \frac{1}{2} sin (2 u) - \frac{1}{2} \cdot \int e^{u} sin (2 u) d u

Wende die partielle Integration an

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

Entferne die Konstante:

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

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

Deshalb

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

Isoliere

\int e^{u} cos (2 u) d u

= \frac{2 e ^{u} sin ( 2 u )}{5} + \frac{e ^{u} cos ( 2 u )}{5}

Setze in

u = ln (x)

ein

= \frac{2 e ^{l n (x)} sin ( 2 ln ( x ) )}{5} + \frac{e ^{l n (x)} cos ( 2 ln ( x ) )}{5}

Vereinfache

\frac{2 e ^{l n (x)} sin ( 2 ln ( x ) )}{5} + \frac{e ^{l n (x)} cos ( 2 ln ( x ) )}{5} : \frac{2}{5} x sin (2 ln (x)) + \frac{1}{5} x cos (2 ln (x))

= \frac{2}{5} x sin (2 ln (x)) + \frac{1}{5} x cos (2 ln (x))

Füge eine Konstante zur Lösung hinzu

= \frac{2}{5} x sin (2 ln (x)) + \frac{1}{5} x cos (2 ln (x)) + C

\int (- \frac{2}{11 - 11 x ^{2}}) d x

\frac{d}{d x} (ln (x^{3}) + \frac{7}{x})

\int \frac{5}{x ^{2} + 6 x + 10} d x

\frac{d}{d x} (\frac{x - 4}{x ^{2} - 2 x + 8})

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