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Find the indefinite integral of the function y = tan^4x + tan^2x.
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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
Posted by giorgiana1976 on November 9, 2010 at 1:48 AM (Answer #1)
High School Teacher
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
Posted by hala718 on November 9, 2010 at 1:55 AM (Answer #2)
High School Teacher
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.
Posted by neela on November 9, 2010 at 2:04 AM (Answer #3)
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