pls-help-Find-L-cos-2-t-

Question Number 46542 by Umar last updated on 28/Oct/18

p l s h e l p

F i n d L ({c o s}^{2} t)

Commented by Umar last updated on 28/Oct/18

L a p l a c e t r a n s f o r m

Commented by maxmathsup by imad last updated on 28/Oct/18

w e h a v e g e n e r a l l y L (f (x)) = \int_{0}^{\infty} f (t) e^{- t x} d t \Rightarrow

L ({c o s}^{2} x) = \int_{0}^{\infty} {c o s}^{2} t e^{- t x} d t = \frac{1}{2} \int_{0}^{\infty} (1 + c o s (2 t)) e^{- t x} d t

= \frac{1}{2} \int_{0}^{\infty} e^{- t x} d t + \frac{1}{2} \int_{0}^{\infty} e^{- t x} c o s (2 t) d t b u t

\int_{0}^{\infty} e^{- t x} d t = {[- \frac{1}{x} e^{- t x}]}_{t = 0}^{\infty} = \frac{1}{x}

\int_{0}^{\infty} e^{- x t} c o s (2 t) d t = R e (\int_{0}^{\infty} e^{- x t + i 2 t} d t) a n d

\int_{0}^{\infty} e^{(- x + 2 i) t} d t = {[\frac{1}{- x + 2 i} e^{(- x + 2 i) t}]}_{t = 0}^{\infty} = \frac{- 1}{- x + 2 i} = \frac{1}{x - 2 i} = \frac{x + 2 i}{x^{2} + 4} \Rightarrow

\int_{0}^{\infty} e^{- x t} c o s (2 t) d t = \frac{x}{x^{2} + 4} \Rightarrow L ({c o s}^{2} x) = \frac{1}{2 x} + \frac{x}{2 (x^{2} + 4)}

Commented by maxmathsup by imad last updated on 28/Oct/18

a n o t h e r m e t h o d b y u s i n g l i n e a r i t y o f L a n d t a b l e

L ({c o s}^{2} x) = L (\frac{1 + c o s (2 x)}{2}) = \frac{1}{2} L (1) + \frac{1}{2} L (c o s (2 x))

= \frac{1}{2} \frac{1}{x} + \frac{1}{2} \frac{x}{x^{2} + 4} = \frac{1}{2 x} + \frac{x}{2 x^{2} + 8} .

Commented by Umar last updated on 28/Oct/18

t h a n k s

Commented by maxmathsup by imad last updated on 28/Oct/18

y o u a r e w e l c m e .