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

The integral of (sin(4e^{-3x}))/(e^{3x)} is 1/12 cos(4e^{-3x})+C

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

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

\frac{1}{12} cos (4 e^{- 3 x}) + C

Solution steps

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

= - \frac{1}{12} (- cos (4 e^{- 3 x}))

Simplify

= \frac{1}{12} cos (4 e^{- 3 x})

Add a constant to the solution

= \frac{1}{12} cos (4 e^{- 3 x}) + C

\int \frac{sec ^{2} ( θ )}{1 + 2 tan ( θ )} d θ

f (x) = x^{4}, at x = 6.87

\int x^{3} + 2 x + 1 d x

f (x) = \frac{2 x ^{2}}{x + 3}

(2 x - 1)^{2}

What is the integral of (sin(4e^{-3x}))/(e^{3x)} ?
The integral of (sin(4e^{-3x}))/(e^{3x)} is 1/12 cos(4e^{-3x})+C