How Do You Integrate tan²x sec^4x dx?

[hotjohn]

Homework Statement

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 ?

Homework Equations

The Attempt at a Solution

 

[Samy_A]

Homework Statement

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 ?

Homework Equations

The Attempt at a Solution

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.

 

[SammyS]

Homework Statement

Integrate tan²x sec⁴x dx

the correct solution is
∫ tan²x sec²x sec²x dx =

Homework Equations

The Attempt at a Solution

Replace the first sec²x with (tan²x + 1):

∫ tan²x (tan²x + 1) sec²x dx =

expand it into:
...

Thank you for typing this out.

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