# What is the integral of f(x) = (tan x) ^4?

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### 2 Answers

We have to find Int [f(x)] = Int [(tan x) ^4 dx]

=> Int [(tan x) ^2 * (tan x) ^2 dx]

=> Int [(tan x) ^2 * ((sec x) ^2 – 1) dx]

=> Int [(tan x) ^2 * (sec x) ^2 – (tan x) ^2 dx]

=> Int [(tan x) ^2 * (sec x) ^2 dx] – Int [(tan x) ^2 dx]

=> Int [(tan x) ^2 * (sec x) ^2 dx] – Int [((sec x) ^2- 1) dx]

=> Int [(tan x) ^2 * (sec x) ^2 dx] – Int [(sec x) ^2 dx] + Int [1 dx]

Let u = tan x, du/dx = (sec x) ^2

=> Int [(tan x) ^2 * (sec x) ^2 dx] – Int [(sec x) ^2 dx] + Int [1 dx]

=> Int [u^2 du] – tan x + x + C

=> u^3/ 3 – tan x + x + C

substitute u = tan x

=> (tan x) ^3 / 3 – tan x + x + C

**Therefore the integral of f(x) = (tan x) ^4 = (tan x) ^3 / 3 – tan x + x + C**

f(x) =( tanx)^4.

Therefore Inf(x) dx = int(tanx)^4 dx.

Put tanx = t. Differentiate tanx = t.

=> sec^2x dx = dt.

dx = dt/(sex)62 = dt/(1+t^2).

Therefore Int tan ^4 x dx = Int t^4* dt/(1+t^2)

Int tan ^4 x dx = Int {(t^2-1) +1/(t^2+1)}dt

Int tan ^4 x dx = t^3/3 - t +Int {dt/(1+t^2)}

Int tan ^4 x dx = t^3/3 -t + arc tan t. Replace t = tanx.

Int tan ^4 x dx = (tan^3 x)/3 - tanx +x +C.