Step 5 Notice that the original integral [5.39 sin(49) d6 is once again present. In fact, putting together all our work so far, if we
Step 5 Notice that the original integral [5.39 sin(49) d6 is once again present. In fact, putting together all our work so far, if we let I = [5.39 sin(46) d6: we have I = lam sin(49) [%e36 cos(46) + 11*] + C1. 3 3 This can be re-arranged to give us I = gem? sin(46) m V I+C. 6'39 cos(49) " We can move all the I terms to one side and get :] I = e36 sin(49) 3633 cos(49) + Cl
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