Question

Find the integrals of the function: $tan^4x$

Answer 1

tan4x
=tan2 x. tan2 x
=(sec2 x-1)tan2 x
=sec2 x tan2 x-tan2 x
=sec2 x tan2 x-(sec2 x-1)
= sec2 x tan2 x-sec2 x+1

∴ ∫tan4 x dx = ∫sec2 xtan2 x dx- ∫sec2 x dx+ ∫1.dx
= ∫sec2 x tan2 x dx-tan x +x+C ...(1)

Consider ∫sec2 x tan2 x dx
Let tan x = t ⇒ sec2 x dx = dt
⇒ ∫sec2 x tan2 xdx = ∫t2dt $= \frac{t^3}{3} = \frac{tan^3x}{3}$
From equation (1), we obtain
$∫tan^4 x dx = \frac{1}{3} tan^3 x-tan x +x+C$