I need help with this integral.
w= the integral from 0 to 5
24e^-6t cos(2t) dt.
i found the the integration in the integral table.
(e^ax/a^2 + b^2) (a cos bx + b sin bx)
im having trouble finishing the problem from here.
Write cos(x) as the real part of
exp(ix)
The integral of exp(ax)cos(bx) is then the real part of the integral of
exp[(a + ib)x]
The integral is:
1/(a+ib) exp[(a+ib)x] =
(a-ib)/(a^2 + b^2) exp(ax)*
[cos(bx) + i sin(bx)]
The real part of this is:
exp(ax)/(a^2 + b^2)
[a cos(bx) + b sin(bx)]
To solve the integral, you correctly used the formula from the integral table:
∫ e^(ax) * cos(bx) dx = [(e^(ax) * (a * cos(bx) + b * sin(bx)))] / (a^2 + b^2)
In your case, a = -6 and b = 2. So let's substitute these values into the formula:
∫ 24e^(-6t) * cos(2t) dt = [24 * e^(-6t) * (-6 * cos(2t) + 2 * sin(2t))] / (36 + 4)
Simplifying further:
= (-24/40) * e^(-6t) * (-6 * cos(2t) + 2 * sin(2t))
= (-3/5) * e^(-6t) * (-6 * cos(2t) + 2 * sin(2t))
Now you can integrate term by term:
∫ (-3/5) * e^(-6t) * (-6 * cos(2t) + 2 * sin(2t)) dt
To integrate each of the terms, you can use integration by parts for the term involving e^(-6t), and the formula for integrating cos(2t) and sin(2t).
After integrating, you can evaluate the integral from the given limits of 0 to 5.