Find the indefinite integral of the function y = tan^4x + tan^2x.

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Math

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hala718's profile pic

Posted on

y= tan^4 x + tan^2 x

First let us simplify;

We will factor tan^2 x

==> y= tan^2 x ( tan^2 x + 1)

==> intg y = intg (tan^2 x ( tan^2 +1) ) dx

We know that : tanx = sinx/cosx

==> intg y= intg (sin^2 x/cos^2 x) ( sin^2 x + cos^2 x/cos^2 x) dx

               = intg ( sin/cos)^2 ( 1/cos^2)   dx

Now let t= tanx= sinx/cosx

==> dt = 1/cos^2 x  dx

==> dx = cos^2 x dt

Now subsitute in y:

intg y= intg ( t^2 ( 1/cos^2x) * cos^2 x  dt

           = intg (t^2 ) dt

           = t^3/3 + C

==> intg y = (tanx)^3 /3   + C

neela's profile pic

Posted on

To find the integral of tan^4x +tan^2x .

tan^4x +tan^2x = tan^2x (tan^2x+1).

Put tanx = t.

d/dx(tanx) = sec^2x.

dt= sec^2x dx.

Therefore Integral tan^2 (1+tan^2)  dx = Integral t^2 dt

Integral t^2 dt = t^3/3 +C.

Therefore Integral (tan^4 x+ tan^2 x) dx = (tanx)^3/3 + C.

giorgiana1976's profile pic

Posted on

The indefinite integral of the given function is written:

Int [(tan x)^4 + (tan x)^2]dx

We'll factorize by (tan x)^2:

Int (tan x)^2*[(tan x)^2 + 1]dx

We'll solve the integral using substitution technique:

tan x = t

We'll differentiate both sides:

dx/(cos x)^2 = dt

We'll write 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 result of the indefinite integral of the function:

Int [(tan x)^4 + (tan x)^2]dx = (tan x)^3/3 + C

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