Integration using complex analysis

gipc
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I have to integrate S cos^8 (t) dt from 0 to 2 pi, presumably using complex analysis

I got to S [(e^(it) + e(-it))/2]^8 dt from 0 ti 2pi

How do I take it from here?

I have a hint- use binomial theorem.
 
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Use the binomial theorem on [(e^(it) + e(-it))/2]^8, then write the answers back to cosinus form and integrate.
 
gipc said:
How do I take it from here?

I have a hint- use binomial theorem.

At the risk of being snarky, have you tried using the binomial theorem? If you have not, you really should have tried it before coming here. If you have, then you should describe why you couldn't continue from there to get the answer.
 
well, the problem is i didn't study the binomial theorem (obviously with taking the complex analysis course and not the discrete math one). and this example is a little tricky to begin with :(
 
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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