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

The integral of 4e^{-4x}sin(4x) is 4(-(e^{-4x}cos(4x))/8-(e^{-4x}sin(4x))/8)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

4 (- \frac{e ^{- 4 x} cos ( 4 x )}{8} - \frac{e ^{- 4 x} sin ( 4 x )}{8}) + C

Solution steps

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

Apply Integration By Parts

: \frac{1}{4} e^{- 4 x} sin (4 x) - \int - e^{- 4 x} sin (4 x) d x

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

Take the constant out:

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

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

Therefore

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

Isolate

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

= 4 (- \frac{e ^{- 4 x} cos ( 4 x )}{8} - \frac{e ^{- 4 x} sin ( 4 x )}{8})

Add a constant to the solution

= 4 (- \frac{e ^{- 4 x} cos ( 4 x )}{8} - \frac{e ^{- 4 x} sin ( 4 x )}{8}) + C

\frac{d}{d x} (\frac{( 8 x - 1 ) ^{5}}{( 3 x - 1 ) ^{3}})

x \to 5 lim (\frac{∣ 5 - x ∣}{x - 5})

f (x) = 4 e^{x} cos (x)

\int_{2}^{3} t sin (π t) d t

\int \frac{( x ^{5} - x ^{3} + 2 x )}{x ^{4}} d x

What is the integral of 4e^{-4x}sin(4x) ?
The integral of 4e^{-4x}sin(4x) is 4(-(e^{-4x}cos(4x))/8-(e^{-4x}sin(4x))/8)+C