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

The integral of (sin(6e^{-2x}))/(e^{2x)} is 1/12 cos(6e^{-2x})+C

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

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

\frac{1}{12} cos (6 e^{- 2 x}) + C

Solution steps

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

Apply u-substitution

: \int - \frac{sin ( u )}{12} d u

= \int - \frac{sin ( u )}{12} d u

Take the constant out:

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

= - \frac{1}{12} \cdot \int sin (u) d u

Use the common integral:

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

= - \frac{1}{12} (- cos (u))

Substitute back

u = 6 e^{- 2 x}

= - \frac{1}{12} (- cos (6 e^{- 2 x}))

Simplify

= \frac{1}{12} cos (6 e^{- 2 x})

Add a constant to the solution

= \frac{1}{12} cos (6 e^{- 2 x}) + C

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What is the integral of (sin(6e^{-2x}))/(e^{2x)} ?
The integral of (sin(6e^{-2x}))/(e^{2x)} is 1/12 cos(6e^{-2x})+C