integral of sin(ax)e^{bx+c}

Lösung

\int sin (a x) e^{b x + c} d x

- \frac{a e ^{b x + c} cos ( a x )}{a ^{2} + b ^{2}} + \frac{b e ^{b x + c} sin ( a x )}{a ^{2} + b ^{2}} + C

Schritte zur Lösung

\int sin (a x) e^{b x + c} d x

Wende die partielle Integration an

= - e^{b x + c} \frac{1}{a} cos (a x) - \int - b e^{b x + c} \frac{1}{a} cos (a x) d x

Entferne die Konstante:

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

= - e^{b x + c} \frac{1}{a} cos (a x) - (- b \frac{1}{a} \cdot \int e^{b x + c} cos (a x) d x)

Vereinfache

- b \frac{1}{a} : - \frac{b}{a}

= - e^{b x + c} \frac{1}{a} cos (a x) - (- \frac{b}{a} \cdot \int e^{b x + c} cos (a x) d x)

Wende die partielle Integration an

= - e^{b x + c} \frac{1}{a} cos (a x) - (- \frac{b}{a} (e^{b x + c} \frac{1}{a} sin (a x) - \int e^{b x + c} b \frac{1}{a} sin (a x) d x))

Entferne die Konstante:

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

= - e^{b x + c} \frac{1}{a} cos (a x) - (- \frac{b}{a} (e^{b x + c} \frac{1}{a} sin (a x) - b \frac{1}{a} \cdot \int e^{b x + c} sin (a x) d x))

Vereinfache

b \frac{1}{a} : \frac{b}{a}

= - e^{b x + c} \frac{1}{a} cos (a x) - (- \frac{b}{a} (e^{b x + c} \frac{1}{a} sin (a x) - \frac{b}{a} \cdot \int e^{b x + c} sin (a x) d x))

Deshalb

\int e^{b x + c} sin (a x) d x = - e^{b x + c} \frac{1}{a} cos (a x) - (- \frac{b}{a} (e^{b x + c} \frac{1}{a} sin (a x) - \frac{b}{a} \cdot \int e^{b x + c} sin (a x) d x))

Isoliere

\int e^{b x + c} sin (a x) d x

= - \frac{a e ^{b x + c} cos ( a x )}{a ^{2} + b ^{2}} + \frac{b e ^{b x + c} sin ( a x )}{a ^{2} + b ^{2}}

Füge eine Konstante zur Lösung hinzu

= - \frac{a e ^{b x + c} cos ( a x )}{a ^{2} + b ^{2}} + \frac{b e ^{b x + c} sin ( a x )}{a ^{2} + b ^{2}} + C