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They are exactly the same!!!
You know the following trigonometric identity right?
cos a * cos b = 0.5*(cos(a+b) + cos(a-b))
In your case, a = y and b = wt - y.
"Good " <modeebuo@yahoo.co.uk> wrote in message <gv9t5p$qk1$1@fred.mathworks.com>...
> Hi Friends,
> This problem is to be solved by symbolic math but the solution it gives me does not tally with what the actual solution should be:
>
> The problem is to integrate a symbolic function of x, y, w and t with respect to x and evaluate the integrand from -pi/2 to pi/2. So the code below was written:
>
> syms N x y w t real
> A = N*cos(x-y)*cos(w*t-y);
> B = int(A, x, -pi/2, pi/2);
>
> The expected solution is:
>
> B should equal N*sin(x-y)*cos(w*t-y) evaluated at (x=pi/2) minus same expression evaluated at (x=-pi/2). Naturally, the final solution should be:
>
> B = N*(sin(pi/2-y) - sin(-pi/2-y))*cos(w*t-y) = 2*N*cos(y)*cos(w*t-y).
>
> Symbolic toolbox gives me:
>
> B = N*cos(-w*t+2*y)+N*cos(w*t)
>
> I cannot see how this solution is correct.
>
> Please I need this clarified so as to continue using this wonderful tool.
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