Question

The integral ∫(sec²x/(secx + tanx)^9/2)dx  equals (for some arbitrary constant K)

 

Answer 1

Answer is (C)

Let secx + tanx = t. Then

secx - tanx = 1/t

Now,

(secxtanx + sec²x)dx = dt² ⇒ secx(secx + tanx)dx = dt