Explained: "Find Laplace of given fn: f(t)=∫0tsin⁡(t−τ)cos⁡τdτ"

Name: Find Laplace of given fn: f(t)=∫0tsin⁡(t−τ)cos⁡τdτ
Uploaded: 2022-02-23T17:22:57+00:00
Description: Find Laplace of given fn: f(t)=∫0tsin⁡(t−τ)cos⁡τdτ

EunoR

Answered question

2020-11-27

Find Laplace of given fn:

f (t) = \int_{0}^{t} \sin (t - τ) \cos τ d τ

Answer & Explanation

toroztatG

Skilled2020-11-28Added 98 answers

Step 1
Given that

f (t) = \int_{0}^{t} \sin (t - τ) \cos τ d τ

The objective is to find the Laplace transform of f(t).

\sin (t - τ) = \sin t \cos τ - \cos t \sin t

So,

\sin (t - τ) \cos τ = \sin t \cos^{2} τ - \cos t \sin τ \cos τ

= \sin t (\frac{1 + \cos 2 τ}{2}) - \cos t (\frac{\sin 2 τ}{2})

Now,

f (t) = \frac{\sin t}{2} \int_{0}^{t} [1 + \cos 2 τ] d τ - \frac{\cos t}{2} \int_{0}^{t} \sin 2 τ d τ

= \frac{\sin t}{2} [t + \frac{\sin 2 t}{2}] - \frac{\cos t}{2} {[- \frac{\cos 2 t}{2}]}_{0}^{t}

= \frac{t \sin t}{2} + \frac{\sin 2 t \sin t}{4} - \frac{\cos t}{2} + \frac{\cos t \cos 2 t}{4}

= \frac{t \sin t}{2} - \frac{\cos t}{2} + \frac{1}{4} \cos (t)

= \frac{t \sin t}{2} - \frac{\cos t}{4}

Step 2
Now

f (t) = \frac{t \sin t}{2} - \frac{\cos t}{4}

L (f (t)) = L (\frac{t \sin t}{2} - \frac{\cos t}{4})

= L (\frac{t \sin t}{2}) + L (- \frac{\cos t}{4})

= \frac{1}{2} L (t \sin t) - \frac{1}{4} L (\cos t)

It is known that laplace transform of

\cos t

\frac{s}{(s^{2} + 1)}

Now let's find the

L (t \sin t),

L [t \sin t] = \frac{d}{(d s)} [L (\sin t)]

= \frac{d}{(d s)} [\frac{1}{s^{2} + 1}]

= \frac{- 2 s}{(s^{2} + 1)^{2}}

Hence the laplace transform for

f (t) = \int_{0}^{t} \sin (t - τ) \cos τ d τ

L (f (t)) = \frac{- s}{(s^{2} + 1)^{2}} - \frac{s}{4 (s^{2} + 1)}

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