Given: The given function is: f(θ)= sec2θ

Calculation:

Let, [a,b]=[0,π/2]

The average of a given function is:

  faverage=1b−a∫abf(θ)dθ           =1(π2−0)∫0π/2sec2θ/2 dθ  ---(1)Let  sec2θ/2=tan2θ/2+1=sin2θ/2cos2θ/2+1substitute in equation (1),faverage=1(π2−0)∫0π/2sec2θ/2 dθ           =2π∫0π/2[sin2θ/2cos2θ/2+1]dθ= 2π∫0π/2sin2θ/2cos2θ/2dθ+2π∫0π/2(1)dθ    ----(1) 

     Now,  ∫ 0 π/2 sin 2 θ/2 cos 2 θ/2 dθ     =  ∫ 0 π/2 ( −sinθ/2 ) −sinθ/2 cos 2 θ/2  dθ  ----(2)

   use by parts formula, ∫ udv=uv− ∫ vdu let u=-sinθ/2 du=-(1/2)cosθ/2 dθ

   and  ∫ dv= ∫ −sinθ/2 cos 2 θ/2 dθ  let s = cosθ/2 ⇒ds=−(1/2)sinθ/2 dθ

   Then v = ∫ dv= ∫ −sinθ/2 cos 2 θ/2 dθ  ⇒ ∫ dv= ∫ 1 s 2 ds

   ⇒− 1 s =− 1 cosθ/2 Then equation (2) gives,

   ∫ 0 π/2 sin 2 θ/2 cos 2 θ/2  dθ= [[−(1/2)sinθ/2.(− 1 (1/2)cosθ/2 )]− ∫ (−(1/2)cosθ/2)(− 1 cosθ/2 )dθ ] 0 π/2

                                = [ sinθ/2 cosθ/2 − 1 2 (θ)] 0 π/2 = [tanθ/2− θ 2 ] 0 π/2 =[ (tan(π/4)−tan(0)− π 4 ]=1−0=1

   And,  2 π ∫ 0 π/2 (1) dθ= 2 π [ θ ] 0 π/2 = 2 π [ π 2 −0 ]=1 Then,  h average = 2 π ∫ 0 π/2 sec 2 θ  dθ=1+1=2 Hence, the average of the given equation is 2..

Hence, the average of a given function is: faverage=2