What is the integral of+e^{9x}sin(2x) ?

The integral of+e^{9x}sin(2x) is -(2e^{9x}cos(2x))/(85)+(9e^{9x}sin(2x))/(85)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of+e^{9x}sin(2x)

\int + e^{9 x} sin (2 x) d x

- \frac{2 e ^{9 x} cos ( 2 x )}{85} + \frac{9 e ^{9 x} sin ( 2 x )}{85} + C

Solution steps

\int e^{9 x} sin (2 x) d x

Apply Integration By Parts

= - \frac{1}{2} e^{9 x} cos (2 x) - \int - \frac{9}{2} e^{9 x} cos (2 x) d x

Take the constant out:

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

= - \frac{1}{2} e^{9 x} cos (2 x) - (- \frac{9}{2} \cdot \int e^{9 x} cos (2 x) d x)

Apply Integration By Parts

= - \frac{1}{2} e^{9 x} cos (2 x) - (- \frac{9}{2} (\frac{1}{2} e^{9 x} sin (2 x) - \int \frac{9}{2} e^{9 x} sin (2 x) d x))

Take the constant out:

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

= - \frac{1}{2} e^{9 x} cos (2 x) - (- \frac{9}{2} (\frac{1}{2} e^{9 x} sin (2 x) - \frac{9}{2} \cdot \int e^{9 x} sin (2 x) d x))

Therefore

\int e^{9 x} sin (2 x) d x = - \frac{1}{2} e^{9 x} cos (2 x) - (- \frac{9}{2} (\frac{1}{2} e^{9 x} sin (2 x) - \frac{9}{2} \cdot \int e^{9 x} sin (2 x) d x))

Isolate

\int e^{9 x} sin (2 x) d x

= - \frac{2 e ^{9 x} cos ( 2 x )}{85} + \frac{9 e ^{9 x} sin ( 2 x )}{85}

Add a constant to the solution

= - \frac{2 e ^{9 x} cos ( 2 x )}{85} + \frac{9 e ^{9 x} sin ( 2 x )}{85} + C

What is the integral of+e^{9x}sin(2x) ?
The integral of+e^{9x}sin(2x) is -(2e^{9x}cos(2x))/(85)+(9e^{9x}sin(2x))/(85)+C