What is the integral of 3e^{-4x}sin(e^{-4x}) ?

The integral of 3e^{-4x}sin(e^{-4x}) is 3/4 cos(e^{-4x})+C

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integral of 3e^{-4x}sin(e^{-4x})

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

\frac{3}{4} cos (e^{- 4 x}) + C

Solution steps

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

Take the constant out:

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

= 3 \cdot \int e^{- 4 x} sin (e^{- 4 x}) d x

Apply u-substitution

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

= 3 \cdot \int - \frac{1}{4} e^{u} sin (e^{u}) d u

Take the constant out:

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

= 3 (- \frac{1}{4} \cdot \int e^{u} sin (e^{u}) d u)

Apply u-substitution

: \int sin (v) d v

= 3 (- \frac{1}{4} \cdot \int sin (v) d v)

Use the common integral:

\int sin (v) d v = - cos (v)

= 3 (- \frac{1}{4} (- cos (v)))

Substitute back

= 3 (- \frac{1}{4} (- cos (e^{- 4 x})))

Simplify

3 (- \frac{1}{4} (- cos (e^{- 4 x}))) : \frac{3}{4} cos (e^{- 4 x})

= \frac{3}{4} cos (e^{- 4 x})

Add a constant to the solution

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

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What is the integral of 3e^{-4x}sin(e^{-4x}) ?
The integral of 3e^{-4x}sin(e^{-4x}) is 3/4 cos(e^{-4x})+C