That seems to be pretty much correct, but it's not obvious how to transform your answer to the given one. I would take the residues as , where k = 1, 3, 5, 7. Their negative-fifth powers are , where k = 1, 7, –3, –5. The sum of the first two of these is , and the sum of the other two is . So the sum of the four residues is (addition formula for trig functions!). Now use and to see that times the sum of the residues is .

That is the integral from to , so you need to divide by 2 to get the integral from 0 to .