What is the integral of (sin(e^{-x}))/(2e^{3x)} ?

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

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integral of (sin(e^{-x}))/(2e^{3x)}

\int \frac{sin ( e ^{- x} )}{2 e ^{3 x}} d x

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

Solution steps

\int \frac{sin ( e ^{- x} )}{2 e ^{3 x}} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{sin ( e ^{- x} )}{e ^{3 x}} d x

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

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

= \frac{1}{2} \cdot \int sin (e^{- x}) e^{- 3 x} d x

Apply u-substitution

= \frac{1}{2} \cdot \int - u^{2} sin (u) d u

Take the constant out:

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

= \frac{1}{2} (- \int u^{2} sin (u) d u)

Apply Integration By Parts

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

Take the constant out:

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

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

Apply Integration By Parts

= \frac{1}{2} (- (- u^{2} cos (u) - (- 2 (u sin (u) - \int sin (u) d u))))

Use the common integral:

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

= \frac{1}{2} (- (- u^{2} cos (u) - (- 2 (u sin (u) - (- cos (u))))))

Substitute back

u = e^{- x}

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

Simplify

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

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

Add a constant to the solution

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

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