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laplace transform sin^2(θ)

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(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

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laplace transform sin²(θ)

Solution

2s(s²+4)

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L{sin²(θ)}

Use the following identity: sin²(x)=12 −cos(2x)12

=L{12 −cos(2θ)12 }

Use the linearity property of Laplace Transform:
For functions f(t), g(t) and constants a, b: L{a·f(t)+b·g(t)}=a·L{f(t)}+b·L{g(t)}

=L{12 }−12 L{cos(2θ)}

L{12 }: 12s

L{cos(2θ)}: ss²+4

=12s −12 · ss²+4

Simplify 12s −12 · ss²+4 : 2s(s²+4)

=2s(s²+4)

Solve instead

inverse laplace sin²(θ)

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−▭▭	<	7	8	9	÷	`AC`
+▭▭	>	4	5	6	×	☐☐☐
×▭▭	(	1	2	3	−	`x`
▭ \|▭	)	.	0	=	+	`y`

laplace transform sin^2(θ)

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