Friday

August 1, 2014

August 1, 2014

Posted by **Bobby Smith** on Sunday, August 30, 2009 at 9:19pm.

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.

- Calculus -
**Count Iblis**, Monday, August 31, 2009 at 9:18amWrite 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)]

**Related Questions**

Integral - That's the same as the integral of sin^2 x dx. Use integration by ...

double integral - 1. Sketch the region of integration & reverse the order of ...

Calculus II - Integrate using integration by parts (integral) (5-x) e^3x u = 5-x...

calculus - 1. integral -oo, oo [(2x)/(x^2+1)^2] dx 2. integral 0, pi/2 cot(theta...

integration by parts - s- integral s ln (2x+1)dx ? = ln(2x+1)x - s x d( ln (2x+1...

calc - how do you start this problem: integral of xe^(-2x) There are two ways: 1...

Calculus - integral -oo, oo [(2x)/(x^2+1)^2] dx (a) state why the integral is ...

Integral Help - I need to find the integral of (sin x)/ cos^3 x I let u= cos x, ...

Calculus - Hello, I have some calculus homework that I can't seem to get started...

trig integration - s- integral endpoints are 0 and pi/2 i need to find the ...