We can now use the substitution u sin 8 so du 8 sec ec 4 u Step 4 sec 4 cos 0 sin 0 15 4 JAS 4 We have determined that if we let u sin 8 then du cos 8 de on the interval V15 4 V1 1 329 4 de cos 0 cos 0 du de Once more we must also make a substitution for the limits of integration When 0 V3 v15 2 14 Applying the substitution gives us the following result 15