Asked by | 18th Jan, 2010, 08:37: AM
The integral can be evaluated by substituting, cosx = (eix + e-ix)/2, and of the difficulty level which is not expected.
The solution involves polylogarithmic function, Li(s,z).
The solution is,
x2i/2 - xlog(1+e2ix) + xlog cosx + i Li(2,-e2ix) + C
where i = (-1)
Answered by | 18th Jan, 2010, 08:49: PM
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