i really need to get this integration ....it is,...[1/(2*pi/q) integral (vm^2 cos^2 t dt)]^1/2?

Question

the limits are 0 to 2 pi/q . the answer that comes out should be...

vm[(q/2*pi){(pi/q)+(1/2) sin(2*pi/q)}]^1/2

thnx in advance.

Answer 1

You haven't stated the question properly -- there's a mistake somewhere if you want that answer to come out. However, I can give you the essential trick you need to solve this. The integral you wish to evaluate is of the form

∫ cos² t dt

over the interval from 0 to 2π/q. First observe that

2 cos² t = 1 + cos 2t

(you should have seen this trig indentity at some point). So we can rewrite the integral as

∫ cos² t dt = ½ ∫ (1 + cos 2t)dt

= ½ [1 + ½ sin 2t ] (evaluated between said limits),

= ½ [1 + ½ sin (4π/q) - 0 - 0]

= ½ + ¼ sin (4π/q).

Now, if you put in the corrections you need to make to your question, I might be willing to go through more of the calculation. But that depends on you.