Question

Find the integrals of the function: $tan^3 2x\, sec 2x$

Answer 1

tan3 2x sec 2x = tan2 2x tan 2x sec 2x
=(sec2 2x-1)tan 2x sec 2x
=sec2 2x.tan 2x sec 2x-tan 2x sec 2x
∴ ∫tan3 2x sec 2x dx = ∫sec2 2x tan 2x sec 2x dx - ∫tan 2x sec 2x dx
= ∫sec2 2x tan 2x sec 2x dx-$\frac{sec2x}{2}$+C
Let sec 2x = t
∴ 2sec 2x tan 2x dx = dt
∴ ∫tan3 2x sec 2x dx = $\frac{1}{2}$ ∫t2dt-$\frac{sec2x}{2}$+C
$= (\frac{t^3}{6}) -\frac{sec2x}{2}+C$
$=\frac{(sec2x)^3}{6}- (\frac{sec2x}{2}) +C$