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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

: \int cos (v) d v

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

Use the common integral:

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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