Question

alenahelenash · Accepted Answer

We want to find the Laplace transform of

\sin (3 t)

. The Laplace transform of a function

f (t)

is defined as:

ℒ {f (t)} (s) = \int_{0}^{\infty} e^{- s t} f (t) d t

Using this definition, we have:

ℒ {\sin (3 t)} (s) = \int_{0}^{\infty} e^{- s t} \sin (3 t) d t

We can use integration by parts to evaluate this integral. Specifically, if we let

u = \sin (3 t)

and

d v = e^{- s t} d t

, then

d u = 3 \cos (3 t) d t

and

v = - \frac{1}{s} e^{- s t}

. Using the integration by parts formula, we have:

\int_{0}^{\infty} e^{- s t} \sin (3 t) d t = {[- \frac{1}{s} \sin (3 t) e^{- s t}]}_{0}^{\infty} - \int_{0}^{\infty} - \frac{1}{s} e^{- s t} (3 \cos (3 t)) d t

The first term evaluates to zero since

\sin (3 t) e^{- s t}

goes to zero as

t \to \infty

. Simplifying the second term, we have:

\int_{0}^{\infty} - \frac{1}{s} e^{- s t} (3 \cos (3 t)) d t = \frac{3}{s} \int_{0}^{\infty} e^{- s t} \cos (3 t) d t

We can use integration by parts again, this time letting

u = \cos (3 t)

and

d v = e^{- s t} d t

. Then,

d u = - 3 \sin (3 t) d t

and

v = - \frac{1}{s} e^{- s t}

, and we have:

\int_{0}^{\infty} e^{- s t} \cos (3 t) d t = {[- \frac{1}{s} \cos (3 t) e^{- s t}]}_{0}^{\infty} - \int_{0}^{\infty} - \frac{1}{s} e^{- s t} (- 3 \sin (3 t)) d t

Again, the first term evaluates to zero. Simplifying the second term, we have:

\int_{0}^{\infty} - \frac{1}{s} e^{- s t} (- 3 \sin (3 t)) d t = \frac{3}{s} \int_{0}^{\infty} e^{- s t} \sin (3 t) d t

Thus, we have:

\int_{0}^{\infty} e^{- s t} \sin (3 t) d t = - \frac{3}{s} \int_{0}^{\infty} e^{- s t} \cos (3 t) d t

Substituting this result back into the original integral, we have:

ℒ {\sin (3 t)} (s) = - \frac{3}{s} \int_{0}^{\infty} e^{- s t} \cos (3 t) d t

We can use the same integration by parts approach as before to evaluate this integral, giving us:

\int_{0}^{\infty} e^{- s t} \cos (3 t) d t = \frac{s}{s^{2} + 9}

Answered question