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# Physics/formula

Question

numerical
from image,can you plz tell me what happened to the step just next to the form. 2sin^2(theta)=1-cos2(theta),

Since the sin^2 term was not 2*sin^2(pi*x/L), the identity used was
sin^2(pi*x/L) = 1/2*(1 - cos(2*pi*x/L))

So 1/2 - (1/2)cos(2*pi*x/L) replaced the sin^2(pi*x/L) in the integral.

The result of that integration was
(1/2)*x - (1/2)*(L/(2*pi))*sin(2*pi*x/L) with limits L and 0, as shown. Plugging the limits in for x gives you
L/2 - (L/(4*pi))*sin(2*pi) which is simply L/2 since sin(2*pi)=0.

I hope this helps,
Steve

Physics

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#### Steve Johnson

##### Expertise

I would be delighted to help with questions up through the first year of college Physics. Particularly Electricity, Electronics and Newtonian Mechanics (motion, acceleration etc.). I decline questions on relativity and Atomic Physics. I also could discuss the Space Shuttle and space flight in general.

##### Experience

I have a BS in Physics and an MS in Electrical Engineering. I am retired now. My professional career was in Electrical Engineering with considerable time spent working with accelerometers, gyroscopes and flight dynamics (Physics related topics) while working on the Space Shuttle. I gave formal classroom lessons to technical co-workers periodically over a several year period.

Education/Credentials
BS Physics, North Dakota State University
MS Electrical Engineering, North Dakota State University