What is the integral of 2e^{-6x}cos(e^{-6x}) ?

The integral of 2e^{-6x}cos(e^{-6x}) is -1/3 sin(e^{-6x})+C

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integral of 2e^{-6x}cos(e^{-6x})

\int 2 e^{- 6 x} cos (e^{- 6 x}) d x

- \frac{1}{3} sin (e^{- 6 x}) + C

Solution steps

\int 2 e^{- 6 x} cos (e^{- 6 x}) d x

Take the constant out:

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

= 2 \cdot \int e^{- 6 x} cos (e^{- 6 x}) d x

Apply u-substitution

: \int - \frac{1}{6} e^{u} cos (e^{u}) d u

= 2 \cdot \int - \frac{1}{6} 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

= 2 (- \frac{1}{6} \cdot \int e^{u} cos (e^{u}) d u)

Apply u-substitution

: \int cos (v) d v

= 2 (- \frac{1}{6} \cdot \int cos (v) d v)

Use the common integral:

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

= 2 (- \frac{1}{6} sin (v))

Substitute back

= 2 (- \frac{1}{6} sin (e^{- 6 x}))

Simplify

2 (- \frac{1}{6} sin (e^{- 6 x})) : - \frac{1}{3} sin (e^{- 6 x})

= - \frac{1}{3} sin (e^{- 6 x})

Add a constant to the solution

= - \frac{1}{3} sin (e^{- 6 x}) + C

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What is the integral of 2e^{-6x}cos(e^{-6x}) ?
The integral of 2e^{-6x}cos(e^{-6x}) is -1/3 sin(e^{-6x})+C