Homework Help: Trig integral

1. Jul 3, 2009

nameVoid

1. The problem statement, all variables and given/known data

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
2. Relevant equations

3. The attempt at a solution

Last edited: Jul 3, 2009
2. Jul 3, 2009

3. Jul 3, 2009

nameVoid

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

4. Jul 3, 2009

g_edgar

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

5. Jul 3, 2009

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.