How do I solve this trigonometric integral: I(tan^3x/cos^4x, x)?

[nameVoid]

Homework Statement

I(tan^3x/cos^4x,x)
I(tan^3x * sec^4x,x)
I((sec^2x-1)sec^4x*tanx,x)
u=secx du=secxtanx
I((u^2-1)u^3,u)
u^6/6-u^4/4+C
sec^6x/6-sec^4x/4+C

im new to this and my book is showing diffrent solutions i see nothing wrong here

Homework Equations

The Attempt at a Solution

 

[Bohrok]

Your answer is correct. What is the solution from your book?

 

[nameVoid]

tan^6x/6+tan^4x/4+C

 

[g_edgar]

Are you sure the two solutions are different? (By more than just adding a constant.)

 

[Bohrok]

That answer is also correct. You can get that answer from yours by using sec² = tan²x + 1, multiplying out the numerators, and simplifying.

The answer you got seems to be the easiest integral to find for the problem.