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

The integral of e^{-2t}sin(h)(3t) is 3/4 sin(h)(-2e^{-2t}t-e^{-2t})+C

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integral of e^{-2t}sin(h)(3t)

\int e^{- 2 t} sin (h) (3 t) d t

\frac{3}{4} sin (h) (- 2 e^{- 2 t} t - e^{- 2 t}) + C

Solution steps

\int e^{- 2 t} sin (h) \cdot 3 t d t

Take the constant out:

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

= 3 sin (h) \cdot \int e^{- 2 t} t d t

Apply u-substitution

: \int \frac{e ^{u} u}{4} d u

= 3 sin (h) \cdot \int \frac{e ^{u} u}{4} d u

Take the constant out:

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

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

Apply Integration By Parts

: e^{u} u - \int e^{u} d u

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

\int e^{u} d u = e^{u}

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

Substitute back

u = - 2 t

= 3 sin (h) \frac{1}{4} (e^{- 2 t} (- 2 t) - e^{- 2 t})

Simplify

3 sin (h) \frac{1}{4} (e^{- 2 t} (- 2 t) - e^{- 2 t}) : \frac{3}{4} sin (h) (- 2 e^{- 2 t} t - e^{- 2 t})

= \frac{3}{4} sin (h) (- 2 e^{- 2 t} t - e^{- 2 t})

Add a constant to the solution

= \frac{3}{4} sin (h) (- 2 e^{- 2 t} t - e^{- 2 t}) + C

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What is the integral of e^{-2t}sin(h)(3t) ?
The integral of e^{-2t}sin(h)(3t) is 3/4 sin(h)(-2e^{-2t}t-e^{-2t})+C