[Fourier transform] Convolution product with sin and cos

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I'm asked to transform y(t) = x(t)*x(t) (where * is the convolution product) and x(t)= sinc(t)cos(2π10t) ( sinc(t)= sin(πt)/(πt) ).The attempt at a solution
Clearly everything is simple if you know X(f), because y(t)=InverseFourier{ X(f)2 }. The problem is that I can't find X(f). By the way, because x(t) is even, I can use the easier version of the transform, but it doesn't help so much, in the end I have to solve:
W4RJlCC.jpg

I tried in vain with some trigonometric formula, but I still have product of sin/cos in the end.

Trying with the convolution product doesn't help either, because I get this:
iRsJv0m.jpg

And I can't see anything that may help me...Wolfram says this is the solution:
1/4 sqrt(pi/2) (sgn(-w-20pi+1)+sgn(-w+20pi+1)+sgn(w-20pi+1)+sgn(w+20pi+1))

EDIT: wait a minute! I'm dumb! because I can use the modulation theorem...
 
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You may put all of this stuff into the exponential form.. :)
 
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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