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

⇒ .