# Another Math Question

1. Nov 7, 2003

### mattmns

This question is ugly,

His hints were first use the substitution u=tan(&theta;) then use the substitution u=x^2

I let x=tan(&theta;) so that dx=sec^2(&theta;)d&theta;

And then got 2&int;tan^2(&theta;)d&theta;

which I got as 2tan(&theta;)-2&theta;+C, and d/dx of that is not &radic;tan(&theta;)

Any thoughts?

2. Nov 7, 2003

### Hurkyl

Staff Emeritus
It looks like you tried to do too much at once and got your dummy variables confused!

You need to complete all of the work for one substitution before you can move onto the next substitution. In particular, after making the substitution u = tan &theta;, you have to rewrite all occurences of &theta; in terms of u.

(hint: first write (sec &theta;)^2 in terms of tan &theta;)

3. Nov 8, 2003

### mattmns

ok I did that and ended up with

2&int;(x^2dx)/(1+x^4) and from there I am kinda stuck on what I want to do, should I let x^2=tan(&theta;)? But I think if I do that I end up with a 2x that I dont want...

4. Nov 8, 2003

### Hurkyl

Staff Emeritus
Hrm. Are you permitted to use complex numbers? The denominator of that is easy to factor over the complex domain, allowing you to use partial fractions.

5. Nov 8, 2003

### mattmns

hmmm complex numbers, not sure lol. This was in the chapter of integration techniques. Integration by parts, Substitution, Trig Substitution, Partial Fraction Decomposition, Improper Integrals and Hyperbolics. Not sure exactly what complex numbers are or if we can use them.

edit...

(x^4+1)= (x^2-&radic;(2)x+1)(x^2+&radic;(2)x+1) and then use partial fraction decomp?... lol maybe thats why he told us that factor

Last edited: Nov 8, 2003
6. Nov 8, 2003

### HallsofIvy

Yes, in terms of complex numbers, x4+1 factors as

Of course, the product of the first two of those is
x2-&radic(2)x+ 1 and the product of the last two is
x2+&radic(2)x+ 1 just as you have.

7. Nov 8, 2003

### mattmns

well hours later I finally solved it, and I came out with an insanely ugly answer. After I did the partial fraction decomp I completed the square twice and used some trig substitutions followed by another substitution.

this is what I ended up with, not sure if it will be easier to type it, or to scan it lol