# Integrate tan²x sec^4x dx

1. Feb 15, 2016

### hotjohn

1. The problem statement, all variables and given/known data
the correct solution is
∫ tan²x sec²x sec²x dx =

replace the first sec²x with (tan²x + 1):

∫ tan²x (tan²x + 1) sec²x dx =

expand it into:

∫ (tan^4x + tan²x) sec²x dx =

let tanx = u

differentiate both sides:

d(tanx) = du →

sec²x dx = du

substituting, you get:

∫ (tan^4x + tan²x) sec²x dx = ∫ (u^4 + u²) du =

break it up into:

∫ u^4 du + ∫ u² du =

[1/(4+1)] u^(4+1) + [1/(2+1)] u^(2+1) + c =

(1/5)u^5 + (1/3)u³ + c

(1/5)tan^5(x) + (1/3)tan³(x) + c
is it wrong to make in into ∫ tan²x ( (1 + tan²x )^2 ) dx
= ∫ tan²x ( 1 + 2tan²x + ((tanx)^4 ) ) dx ?
= (1/3)(tanx)^3 +(2/5)(tanx) ^5 + (1/7)(tan x ) ^7 ???

2. Relevant equations

3. The attempt at a solution

2. Feb 15, 2016

### Samy_A

You could try to do it that way (not sure it is easier than the solution you have), but
$\int \tan² x\ dx \neq \frac{1}{3} \tan³ x$ and likewise for the two other integrals.

3. Feb 15, 2016

### SammyS

Staff Emeritus
Thank you for typing this out.

Further improvements:
Always include the statement of the problem in the body of the Original Post no matter whether or not it's stated in the thread title.
Use the template provided for you.​

I've edited what you had in the above quoted material.