JR/3 sin(6) We can now use the substitution u = sin(0), so du = = sec(4), u = 15 4 15 4 Step 4 COS Cos(0) cos(0) de. Once more, we must also make a substitution for the limits of integration. When = 3 We have determined that if we let u = sin(6), then du = cos(6) de on the interval 15/4 sec JR/3 cos(0) sin2(0) de= Step 5 We can now evaluate the integral. 15/4 du = 3/2 142 15/4 3/2 du 15 Applying the substitution gives us the following result.