To find the Laplace transform of the given function f(t) = sin(t) + sin(2t), we can use the following formula:ℒ{f(t)}=∫0∞e−stf(t)dtwhere ℒ{f(t)} denotes the Laplace transform of f(t), and s is the Laplace variable.So, let's apply this formula to the given function:ℒ{sin(t)+sin(2t)}=∫0∞e−st(sin(t)+sin(2t))dtWe can simplify this integral using the following trigonometric identity:sin(2t)=2sin(t)cos(t)Substituting this into the above integral, we get:ℒ{sin(t)+sin(2t)}=∫0∞e−st(sin(t)+2sin(t)cos(t))dtFactoring out sin(t), we have:ℒ{sin(t)+sin(2t)}=∫0∞e−stsin(t)(1+2cos(t))dtUsing the Laplace transform of sin(t) and cos(t), we have:ℒ{sin(t)}=1s2+1andℒ{cos(t)}=ss2+1Substituting these into the integral above, we get:ℒ{sin(t)+sin(2t)}&amp;=∫0∞e−stsin(t)(1+2cos(t))dt&amp;=∫0∞e−stsin(t)dt+2∫0∞e−stsin(t)cos(t)dt&amp;=1s2+1+2·s(s2+1)2Therefore, the Laplace transform of f(t)=sin(t)+sin(2t) is:ℒ{sin(t)+sin(2t)}=1s2+1+2s(s2+1)2And we are done.

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2022-05-11

Answer & Explanation

Andre BalkonE

Skilled2023-05-06Added 110 answers

To find the Laplace transform of the given function f(t) = sin(t) + sin(2t), we can use the following formula:

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

where

ℒ {f (t)}

denotes the Laplace transform of

f (t)

, and

s

is the Laplace variable.
So, let's apply this formula to the given function:

ℒ {\sin (t) + \sin (2 t)} = \int_{0}^{\infty} e^{- s t} (\sin (t) + \sin (2 t)) d t

We can simplify this integral using the following trigonometric identity:

\sin (2 t) = 2 \sin (t) \cos (t)

Substituting this into the above integral, we get:

ℒ {\sin (t) + \sin (2 t)} = \int_{0}^{\infty} e^{- s t} (\sin (t) + 2 \sin (t) \cos (t)) d t

Factoring out

\sin (t)

, we have:

ℒ {\sin (t) + \sin (2 t)} = \int_{0}^{\infty} e^{- s t} \sin (t) (1 + 2 \cos (t)) d t

Using the Laplace transform of

\sin (t)

and

\cos (t)

, we have:

ℒ {\sin (t)} = \frac{1}{s^{2} + 1} and ℒ {\cos (t)} = \frac{s}{s^{2} + 1}

Substituting these into the integral above, we get:

ℒ {\sin (t) + \sin (2 t)} & = \int_{0}^{\infty} e^{- s t} \sin (t) (1 + 2 \cos (t)) d t & = \int_{0}^{\infty} e^{- s t} \sin (t) d t + 2 \int_{0}^{\infty} e^{- s t} \sin (t) \cos (t) d t & = \frac{1}{s^{2} + 1} + 2 \cdot \frac{s}{(s^{2} + 1)^{2}}

Therefore, the Laplace transform of

f (t) = \sin (t) + \sin (2 t)

is:

ℒ {\sin (t) + \sin (2 t)} = \frac{1}{s^{2} + 1} + \frac{2 s}{(s^{2} + 1)^{2}}

And we are done.

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