1. Aug 10, 2008

Davidk1

Hi, I posted a thread earlier with

int/ 1/(1-x^2)^3

I was able to solve it w/ partial fractions but it took 4 pages...
Can you do this problem w/o partial fractions

2. Aug 10, 2008

HallsofIvy

Staff Emeritus
No, partial fractions is the right way to do it.

3. Aug 10, 2008

rootX

Try using substitution
x = sin t ?

I got int {1/cos^2 t}.dt
There might be a shortcut to solve this in one line ...

4. Aug 10, 2008

Chrisas

I get that there would be a cos upstairs and a cos^6 downstairs so it becomes 1/Cos^5, or sec^5. Not sure if that would be any easier than partial fractions.

5. Aug 10, 2008

rootX

let's see:
=1/(1-x^2)^3 .dx

x = sin (t)
dx = cos(t).dt

subs:
=cos/(1-sin^2)^3 .dt

=1/cos^5 dt

Yes, you are right, I forgot (^3).

It seems messier

>> syms x;
>> f = sec(x)^5

f =

sec(x)^5

>> int(f,x)

ans =

1/4/cos(x)^4*sin(x)+3/8/cos(x)^2*sin(x)+3/8*log(sec(x)+tan(x))