Step 3. integral, we put everything in terms of sin(e) and cos(0). Doing so gives the following result. sec (4) sec(0) tan(e) de= tan (0) sec (4) /3 sin? n (0) cos(8) de sin(e) So far we have the following. sec (4) cos(6) sin2(e) /3 de We can now use the substitution u = sin(6), so du= V15 = 0 sec (4), u= 4 15 4 Cos(8) cos(0) de. Once more, we must also make a substitution for the limits of integration. When & Step 4 We have determined that if we let u= sin(0), then du = cos(e) de on the interval Applying the substitution gives us the following result. 15/4 /3 sec (4) co cos(6) sin2(e) 2) de du de= /3/2