Evaluate the following integrals:
∫ tan32x sec 2x dx
tan32x. sec 2x = tan22x. tan2x.sec2x.dx
tan22x = sec22x - 1
⇒ tan22x. tan2x.sec2x.dx = (sec22x - 1). tan2x.sec2x.dx
⇒ sec22x tan2x.sec2xdx - tan2x.sec2xdx
∴
⇒
Assume sec2x = t
d(sec2x) = dt
sec2x.tan2x.dx = dt
⇒
⇒
But t = sec2x
⇒ .