integral of e^{2x+3}*sin(4x)

Lösung

\int e^{2 x + 3} \cdot sin (4 x) d x

- \frac{e ^{2 x + 3} cos ( 4 x )}{5} + \frac{e ^{2 x + 3} sin ( 4 x )}{10} + C

Schritte zur Lösung

\int e^{2 x + 3} sin (4 x) d x

Wende die partielle Integration an

= - \frac{1}{4} e^{2 x + 3} cos (4 x) - \int - \frac{1}{2} e^{2 x + 3} cos (4 x) d x

Entferne die Konstante:

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

= - \frac{1}{4} e^{2 x + 3} cos (4 x) - (- \frac{1}{2} \cdot \int e^{2 x + 3} cos (4 x) d x)

Wende die partielle Integration an

= - \frac{1}{4} e^{2 x + 3} cos (4 x) - (- \frac{1}{2} (\frac{1}{4} e^{2 x + 3} sin (4 x) - \int \frac{1}{2} e^{2 x + 3} sin (4 x) d x))

Entferne die Konstante:

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

= - \frac{1}{4} e^{2 x + 3} cos (4 x) - (- \frac{1}{2} (\frac{1}{4} e^{2 x + 3} sin (4 x) - \frac{1}{2} \cdot \int e^{2 x + 3} sin (4 x) d x))

Deshalb

\int e^{2 x + 3} sin (4 x) d x = - \frac{1}{4} e^{2 x + 3} cos (4 x) - (- \frac{1}{2} (\frac{1}{4} e^{2 x + 3} sin (4 x) - \frac{1}{2} \cdot \int e^{2 x + 3} sin (4 x) d x))

Isoliere

\int e^{2 x + 3} sin (4 x) d x

= - \frac{e ^{2 x + 3} cos ( 4 x )}{5} + \frac{e ^{2 x + 3} sin ( 4 x )}{10}

Füge eine Konstante zur Lösung hinzu

= - \frac{e ^{2 x + 3} cos ( 4 x )}{5} + \frac{e ^{2 x + 3} sin ( 4 x )}{10} + C

\int x^{2} (x^{3} + 5)^{7} d x

d er i v a t i v e f (x) = e^{- 1}

f (x) = x^{8}

\int sin (- 4 x) cos (3 x) d x

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