Total power contained in 10.0[cos(160.7*pi*t)]^4 (Fourier Series)

Jd303
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Homework Statement



Compute the power contained in the periodic signal x(t) = 10.0[cos(160.7*pi*t)]^4

Homework Equations


The Attempt at a Solution



Hey guys,
I have just started Fourier Series and am struggling with this one. Without writing all my calculations, -I start with inverse Euler formula.
-Then integrate x(t)*e^(-j*ω*k*t) with respect to t. From 0 to (To)
-Then consider the value of the final exponentials when k is an odd and even number.

However i calculate that the answer for ak for any value of k to be 15/4?

Once again I have only started Fourier recently so any direction would be much appreciated.
 
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Please show your work.

Another approach you might try is using trig identities to find the Fourier series.
 
Provided are my calculations in the attachment
-The function has already been converted using the inverse Euler formula
- σ is equal to 160.7*pi*t
- The fundamental frequency is 2*σ/(2*π)

-Anyway to calculate this equation is fine, inverse euler was just given as a suggestion to begin.
 

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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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