Question

get the lapace transform of f(X)=6sin2t-5cos2t​

Answer 1

Answer:

Laplace Transform of f(x) is $\dfrac{12-5S}{S^{2}+4 }$

Step-by-step explanation:

Given function as :

f(x) = 6 Sin 2 t - 5 Cos 2 t

Now, Laplace transform of given function as

$\zeta$ ( 6 Sin 2 t - 5 Cos 2 t ) = $\int_{0}^{\infty }$ $e^{-st}$ {  6 Sin 2 t - 5 Cos 2 t } dt

I =  $\int_{0}^{\infty }$ $e^{-st}$  Sin 2 t dt

Or I = [ $\dfrac{-1}{s}$$e^{-st}$Sin 2 t ] o to ∞

i.e I = 0 + [ - $\dfrac{2}{s^{2} }$ $e^{-st}$  Cos 2 t ] o to ∞ + $\int_{0}^{\infty }$ $\dfrac{-4}{S^{2} }$ Sin 2 t dt

Or, I = $\dfrac{2}{S^{2} }$ - $\dfrac{4}{S^{2} }$ I

Hence, I = $\dfrac{2}{S^{2}+4 }$

Similarly

$\int_{0}^{\infty }$ $e^{-st}$  Cos 2 t  dt   = $\dfrac{S}{S^{2}+4 }$

So, $\zeta$ ( 6 Sin 2 t - 5 Cos 2 t ) = 6 × ( $\dfrac{2}{S^{2}+4 }$ ) - 5 × ( $\dfrac{S}{S^{2}+4 }$ )

I.e Laplace Transform of f(x) = $\dfrac{12}{S^{2}+4 }$  - $\dfrac{5S}{S^{2}+4 }$

Hence,  Laplace Transform of f(x) is $\dfrac{12-5S}{S^{2}+4 }$ Answer