Notice that the original integral fe39 sin(46) dB is once again present. In factr putting together all our work so far, if we let I
Notice that the original integral fe39 sin(46) dB is once again present. In factr putting together all our work so far, if we let I = [5.39 sin(49) d6: we have I = i639 sin(49) i[ie36 (205(46) + iI] + C 3 33 3 1' This can be rearranged to give us r= law we) E coma) :1 +
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