What is the integral of e^{-4x}sin(bx) ?

The integral of e^{-4x}sin(bx) is -(be^{-4x}cos(bx))/(b^2+16)-(4e^{-4x}sin(bx))/(b^2+16)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of e^{-4x}sin(bx)

\int e^{- 4 x} sin (b x) d x

- \frac{b e ^{- 4 x} cos ( b x )}{b ^{2} + 16} - \frac{4 e ^{- 4 x} sin ( b x )}{b ^{2} + 16} + C

Solution steps

\int e^{- 4 x} sin (b x) d x

Apply Integration By Parts

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

Take the constant out:

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

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

Apply Integration By Parts

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

Take the constant out:

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

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

Therefore

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

Isolate

\int e^{- 4 x} sin (b x) d x

= - \frac{b e ^{- 4 x} cos ( b x )}{b ^{2} + 16} - \frac{4 e ^{- 4 x} sin ( b x )}{b ^{2} + 16}

Add a constant to the solution

= - \frac{b e ^{- 4 x} cos ( b x )}{b ^{2} + 16} - \frac{4 e ^{- 4 x} sin ( b x )}{b ^{2} + 16} + C

What is the integral of e^{-4x}sin(bx) ?
The integral of e^{-4x}sin(bx) is -(be^{-4x}cos(bx))/(b^2+16)-(4e^{-4x}sin(bx))/(b^2+16)+C