To find the antiderivative of tan^2x +tan^4x.
Let y = tan^2x (1+tan^2x)
Then y = tan^2x (sec^2x).
Now let tan x = t.
Then d/x(tanx) = secx^2x.
So, dt = sec^2x dx
So Antiderevative tan^2x sec^2x dx = Int tan^2x (1+tan^2x) dx = t^2 dt
Int tan^2x(1+tan^2xdx) dx = t ^3/3 + constant, as Int x^n dx = (1/n+1) x^(n+1) + const.
Therefore Int tan^2x(1+tan^x) dx = (1/3)(tanx)^3 +C.
Therefore Int (tan^2x+tan^4x) dx = (1/3)(tanx)^3 +C.
To determine the antiderivative of the given function , we'll have to calculate the indefinite integral of the given function:
Int [(tan x)^4 + (tan x)^2]dx
We notice that we can factorize by (tan x)^2 and the integrand will become:
Int (tan x)^2*[(tan x)^2 + 1]dx
We'll solve the integral using substitution technique. We'll note tan x = t:
We'll differentiate both sides:
dx/(cos x)^2 = dt
We'll use the fundamental formula of trigonometry and we'll get:
(sin x)^2 + (cos x)^2 =1
We'll divide the relation by (cos x)^2:
(tan x)^2 + 1 = 1/(cos x)^2
We'll re-write the integral, substituting [(tan x)^2 + 1] by 1/(cos x)^2:
Int (tan x)^2*[(tan x)^2 + 1]dx = Int (tan x)^2*dx/(cos x)^2
Now, we'll re-write the integral replacing the variable x by t:
Int (tan x)^2*dx/(cos x)^2 = Int t^2*dt
Int t^2*dt = t^3/3 + C
We'll substitute t by tan x and we'll get the antiderivative of the function:
Int [(tan x)^4 + (tan x)^2]dx = (tan x)^3/3 + C