How to Simplify an Integration Using Trigonometric Substitution?

[KFC]

Homework Statement

There is a integration need to be done

$$I = \int_{-\infty}^\infty \frac{1+x^2}{1+x^4}dx$$


2. The attempt at a solution
I use the following substitution

$$x=\tan \theta$$

such that

$$dx = \frac{d\theta}{\cos^2\theta}$$

Now the integration becomes
$$I = \int_{-\pi/2}^{\pi/2} \frac{1}{1-0.5\sin^2(2\theta)}d\theta$$

But I still stuck with the simplified integration. Any other way to do that?

 

[tiny-tim]

$$I = \int_{-\infty}^\infty \frac{1+x^2}{1+x^4}dx$$

Hi KFC!

Hae you tried factorising 1 + x⁴ and using partial fractions?

 

[KFC]

Hi KFC!

Hae you tried factorising 1 + x⁴ and using partial fractions?

Thanks a lot for your reply. I try to do something like that

$$1+x^4 = (1+x^2)(Ax^2 + Bx + C)$$

and figure out A, B and C by comparing it to 1+x^4, but it will gives a complex terms.

 

[Dick]

There are two quadratic real polynomials whose product is (x^4+1). To my mind the easiest what to find them is to factor (x^4+1)=(x-c1)(x-c2)(x-c3)(x-c4) over the complex numbers and then pick the pairs where ci and cj are complex conjugates and multiply them to get the quadratic factors. Can you do that? There might be tricks you can use to stay in the real numbers, but I don't know them offhand.

 

[tiny-tim]

Thanks a lot for your reply. I try to do something like that

$$1+x^4 = (1+x^2)(Ax^2 + Bx + C)$$

and figure out A, B and C by comparing it to 1+x^4, but it will gives a complex terms.

Nooo … 1+x² is not a factor of 1+x⁴.

As Dick says, roots come in complex conjugate pairs,

so x⁴ + 1 = (x² + Ax + B)(x² + Cx + D), with A B C and D real.

 

[TD]

As Dick mentioned, there's also a little trick possible if you notice that

x⁴+1 = (x²+1)²-2x²

Now you can just use the factorization of a difference of two squares.

 

[dextercioby]

The factoring is simpler to get if you add and subtract $2x^2$ to the denominator and you those 6th grade formulas for expanding the square of a sum and the difference of two squares...

 

[KFC]

Thanks all of you guys, it helps. I know how to do it now. :)