# 6. Laplace Transforms of Integrals

We first saw the following properties in the Table of Laplace Transforms.

1. If G(s)= Lap{g(t)}, then Lap{int_0^tg(t)dt}=(G(s))/s.

2. For the general integral, if

[intg(t)dt]_(t=0)

is the value of the integral when t=0, then:

Lap{intg(t)dt} =(G(s))/s+1/s[intg(t)dt]_(t=0)

### Examples

Use the above information and the Table of Laplace Transforms to find the Laplace transforms of the following integrals:

(a) int_0^tcos\ at\ dt

In this example, g(t) = cos at and from the Table of Laplace Transforms, we have:

G(s)= Lap{cosat} =s/((s^2+a^2))

Now, applying the first rule above, we have:

Lap{int_0^tcosat\ dt}=1/sxxs/(s^2+a^2)

=1/(s^2+a^2)

(b) int_0^te^(at)cos\ bt\ dt

This is similar to example (a). We find the transform of the function g(t) = eatcos bt, then divide by s, since we are finding the Laplace transform of the integral of g(t) evaluated from 0 to t.

Lap{int_0^te^(at)cosbt\ dt}=1/sxx(s-a)/((s-a)^2+b^2)

=(s-a)/(s((s-a)^2+b^2)

(c) int_0^t te^(-3t) dt

This follows the same process as examples (a) and (b).

Find the Laplace transform of the function g(t)=te^(-3t) then divide by s.

Lap{int_0^t te^(-3t)dt}=1/sxx1/((s+3)^2)

=1/(s(s+3)^2)

(d) int_0^tsin\ at\ cos\ at\ dt

Recall from the Double Angle Formula that

sin 2α = 2\ sin α\ cos α

We can use this to re-express our integrand (the part we are integrating):

sin at\ cos at=1/2 sin 2at

So the Laplace Transform of the integral becomes:

Lap{int_0^t\ sin at\ cos at\ dt}=1/2 Lap{int_0^t\ sin 2at\ dt}

=1/2(2a)/(s(s^2+4a^2))

=a/(s(s^2+4a^2))

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