Question

Find the integrals of the function:

tan⁴x

Answer 1

Let `I = int tan^4 x dx = int (sec^2 x - 1)^2  dx`

`= (sec^4 x - 2 sec^2 x + 1) dx`

`= int sec^4 x dx  - 2 int sec^2 x dx + int 1 dx`

`= int sec^4 x dx - 2 tan x + x + C_1`

⇒ `I = I_1 - 2 tan x + x + C_1`                 ...(i)

Where `I_1 = intsec^4 x dx`

Now, `I_1 =  int sec^4 x dx =  int sec^2 x * sec^2 x dx`

`= int (1 + tan^2 x) sec^2 x dx.`

Put tan x = t

⇒ sec² x dx  = dt

∴ `I_1 = int (1 + t^2) dt = t = t^3/3 + C_2`

`= tan x + 1/3 tan^3 x + C_2`              .....(ii)

From (i) and (iii), we have,

`I = tan x + 1/3 tan^3 x + C_2 - 2 tan x + x + C_1`

`= 1/3 tan^3 x - tan x + x + C`