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

Hello Muhammad,

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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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**

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