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Find the Laplace transform of f(t)=integral of (t-torque)^2 cos 2*torque d*torque from 0 to t.
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Does "torque" stand for tau, \(\tau\) ?
Yes.
\[f(t)=\int_0^t (t-\tau)^2\cos2\tau~d\tau\] Apply the convolution theorem: \[\mathcal{L}\left\{\int_0^t f(t-\tau)g(\tau)~d\tau\right\}=F(s)G(s)\] Here, \(f(t)=t^2\) and \(g(t)=\cos(2t)\). So, the Laplace transform of the integral is \[\mathcal{L}\left\{t^2\right\}\mathcal{L}\left\{\cos(2t)\right\}\]
Hmm, maybe that's not the convolution theorem... well, whatever it's called, use that formula.
Sorry, but I don't understand what you wrote above. The processing math is 0%.
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