Integration for radian value (answer = 18.264 watts)

[Hancko]

Dear Tutors,

Given equations:

i = 5cos(60(pi)t)
p = vi
v = (6+10(integration of idt)

I have encountered the following problem considering integration and am unable to complete this sum:



This is my attempt:

 

[skeeter]

[MATH]i = 5\cos(60\pi t)[/MATH]
[MATH]v = 6 + \frac{5}{6\pi} \sin(60 \pi t)[/MATH]
I get [MATH]p=iv \approx 18.3 \, W [/MATH] at t = 5 msec.

 

[Hancko]

[MATH]i = 5\cos(60\pi t)[/MATH]
[MATH]v = 6 + \frac{5}{6\pi} \sin(60 \pi t)[/MATH]
I get [MATH]p=iv \approx 18.3 \, W [/MATH] at t = 5 msec.

I do not understand how you got 5/6(pi)

 

[skeeter]

the antiderivative of [MATH]\cos(60 \pi t)[/MATH] is [MATH]\frac{\sin(60 \pi t)}{60\pi}[/MATH]
[MATH]6 + 10 \int 5\cos(60 \pi t) \, dt = 6 + 50 \int \cos(60 \pi t) \, dt = 6 + \frac{50\sin(60 \pi t)}{60 \pi}[/MATH]
reduce the fraction