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Math  /  Calculus

QuestionThe integral ∫0∞e−ste3tcos⁡(2t)dt=\int_{0}^{\infty} e^{-s t} e^{3 t} \cos (2 t) d t=
Select one: 3!s2−9\frac{3!}{s^{2}-9} s−3(s−3)2+4\frac{s-3}{(s-3)^{2}+4} s(s+3)2−4\frac{s}{(s+3)^{2}-4} ss2+4\frac{s}{s^{2}+4}

Studdy Solution

STEP 1

1. The integral is of the form ∫0∞e(a+bi)tdt\int_{0}^{\infty} e^{(a + bi)t} dt, where aa and bb are real numbers.
2. The integral involves an exponential decay term e−ste^{-st} and a growth term e3te^{3t}.
3. The integral also involves a trigonometric function cos⁡(2t)\cos(2t).
4. We will use the formula for the Laplace transform of a cosine function.

STEP 2

1. Simplify the integrand.
2. Identify the Laplace transform form.
3. Apply the formula for the Laplace transform of cos⁡(bt)\cos(bt).
4. Solve for the integral.

STEP 3

First, simplify the integrand by combining the exponential terms:
e−ste3t=e(3−s)te^{-st} e^{3t} = e^{(3-s)t}
Thus, the integral becomes:
∫0∞e(3−s)tcos⁡(2t) dt\int_{0}^{\infty} e^{(3-s)t} \cos(2t) \, dt

STEP 4

Recognize that this integral is in the form of the Laplace transform of cos⁡(bt)\cos(bt), which is:
L{cos⁡(bt)}=ss2+b2\mathcal{L}\{\cos(bt)\} = \frac{s}{s^2 + b^2}
where ss is replaced by 3−s3-s and b=2b = 2.

STEP 5

Apply the Laplace transform formula for cos⁡(bt)\cos(bt):
The Laplace transform of cos⁡(2t)\cos(2t) with the exponential term e(3−s)te^{(3-s)t} is given by:
3−s(3−s)2+22\frac{3-s}{(3-s)^2 + 2^2}

STEP 6

Simplify the expression:
3−s(3−s)2+4=s−3(s−3)2+4\frac{3-s}{(3-s)^2 + 4} = \frac{s-3}{(s-3)^2 + 4}
This matches the option:
s−3(s−3)2+4\boxed{\frac{s-3}{(s-3)^2 + 4}}
The value of the integral is:
s−3(s−3)2+4\boxed{\frac{s-3}{(s-3)^2 + 4}}

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