L-sin-2-t-

Question Number 5629 by LMTV last updated on 23/May/16

L (\sin^{2} t) = ?

Commented by FilupSmith last updated on 23/May/16

Laplace transform ?

Commented by LMTV last updated on 23/May/16

y e s > ⌣ <

Commented by FilupSmith last updated on 23/May/16

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

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

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

unsure how to solve

Commented by Yozzii last updated on 23/May/16

e^{(a + b i) x} = e^{a x} e^{b i x} = e^{a x} (c o s b x + i s i n b x)

\Rightarrow R e \int e^{(a + b i) x} d x = \int e^{a x} c o s b x d x

∴ \int e^{a x} c o s b x d x = R e (\frac{1}{a + b i} e^{(a + b i) x}) + C

\int e^{a x} c o s b x d x = R e (\frac{(a - b i) e^{(a + b i) x}}{a^{2} + b^{2}}) + C

\int e^{a x} c o s b x d x = \frac{e^{a x} (a c o s b x + b s i n b x)}{a^{2} + b^{2}} + C

L e t a = - s a n d b = 2 .

∴ \int e^{- s x} c o s 2 x d x = \frac{e^{- s x} (2 s i n 2 x - s c o s 2 x)}{s^{2} + 4} + C

∴ L ({s i n}^{2} t) = \frac{1}{2} \lim_{m \to \infty} (\frac{- e^{- s t}}{s} + \frac{e^{- s t} (s c o s 2 t - 2 s i n 2 t)}{s^{2} + 4}) ∣_{0}^{m}

L ({s i n}^{2} t) = 0.5 \lim_{m \to \infty} (e^{- s m} (\frac{s c o s 2 m - 2 s i n 2 m}{s^{2} + 4} - \frac{1}{s}) + \frac{1}{s} - \frac{s}{s^{2} + 4})

L ({s i n}^{2} t) = 0.5 (\frac{s^{2} + 4 - s^{2}}{s (s^{2} + 4)})

L ({s i n}^{2} t) = \frac{2}{s (s^{2} + 4)}

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