An electrical circuits problem involving integrals, all rules given, tell me if right

[Asphyxiated]

Homework Statement

A voltage of $$60cos(4 \pi t) V$$ appears across the terminals of a 3mF (milli-farad) capacitor. (which is equal to .003 F [farad]).

Calculate the current through the capacitor and the energy stored in it from t=0 to t=.125 s

Homework Equations

Current through an ideal linear capacitor is given by:

$$i = C \frac {dv}{dt}$$

where C is the capacitance in farads and dv/dt is the derivative of the voltage.

Power stored in a capacitor over time is given by:

$$w = C \int_{- \infty}^{t} v \frac {dv}{dt} dt = C \int_{v(t_{- \infty})}^{V(t)} v dv$$

The Attempt at a Solution

$$C= .003 F$$

$$v= 60cos(4 \pi t)$$

$$\frac {dv}{dt} = -60sin(4 \pi t) (4 \pi) =-240 \pi sin(4 \pi t)$$

so current is then:

$$i = C \frac {dv}{dt} = (.003) (-240 \pi sin(4 \pi t) = -.72 \pi sin(4 \pi t)$$

No questions there, now on to energy. I did this two different but equal ways and got the correct answer one way but not the other. Can someone help me see why? First the method leading to the correct answer:

$$w = C \int_{0}^{.125} v dv = \frac {C}{2} (v^{2}) = \frac {C}{2} (3600cos^{2} (4 \pi t))$$

evaluate v^2 at 0 and .125 and you get (note I am aware I could change the limits to V(t) but chose to change back to t):

$$\frac {C}{2} (0-3600) = \frac {.003*-3600}{2} = -5.4 J$$

so this is the correct answer, now if I do the same integral but sub in different (but equal!) statements I get the wrong answer:

$$w = C \int_{0}^{.125} v \frac {dv}{dt} dt = c \int_{0}^{.125} (60cos(4 \pi t))(-.72 \pi sin(4 \pi t)) dt$$

now all I did was sub in values found before for dv/dt and we were given v. Now if I integrate this with respect to t shouldn't this be the same as before?

Combine all constants and move them out of the integrand:

$$C \int_{0}^{.125} -43.2 \pi sin(4 \pi t)cos(4 \pi t) dt = -.1296 \pi \int_{0}^{.125} sin(4 \pi t) cos(4 \pi t) dt$$

note that on the second step here I used the value C=.003 so .003*-43.2=-.1296 and also moved the pi constant out. Now this can be solved with the identity:

$$\int sin(ax)cos(ax) dx = -\frac {1}{2a} cos^{2} (ax) +C$$

$$-.1296 \pi \int_{0}^{.125} sin(4 \pi t) cos(4 \pi t) dt = -.1296 \pi (-\frac {1}{8 \pi} cos^{2}(4 \pi t))$$

this needs to be evaluated, notice though that we get the cos^2 like before but the coefficient is all out of whack, its not even close to what it should be, evaluating this at 0 and .125 you get:

$$-.1296 \pi ( 0 + \frac {1}{8 \pi}) = \frac {-.1296 \pi}{8 \pi }= \frac {-.1296}{8} = - \frac {81}{500} = -.162J$$

whats up with that? Did I do something wrong here?

 

[Asphyxiated]

Nevermind,

I figured it out, I used the value found for i instead of the value for dv/dt. If i use the proper value all is well. Thanks though!