# Untitled Page 34

- Page ID
- 148378

## 8.3 Partial fractions revisited

Suppose we want to evaluate the integral

by the method of partial fractions. The quadratic formula tells us that the
roots are *i* and -*i*, setting
1/(*x*^{2}+1)=*A*/(*x*+*i*)+*B*/(*x*-*i*) gives *A*=*i*/2 and *B*=-*i*/2, so

The attractive thing about this approach, compared with the method used on page 88, is that it doesn't require any tricks. If you came across this integral ten years from now, you could pull out your old calculus book, flip through it, and say, “Oh, here we go, there's a way to integrate one over a polynomial --- partial fractions.” On the other hand, it's odd that we started out trying to evaluate an integral that had nothing but real numbers, and came out with an answer that isn't even obviously a real number.

But what about that expression (*x*+*i*)/(*x*-*i*)? Let's give it a name,
*w*. The numerator and denominator are complex conjugates of one
another. Since they have the same magnitude, we must have |*w*|=1,
i.e., *w* is a complex number that lies on the unit circle, the
kind of complex number that Euler's formula refers to. The numerator
has an argument of tan^{-1}(1/*x*)=π/2-tan^{-1}*x*, and the denominator
has the same argument but with the opposite sign. Division means subtracting
arguments, so arg *w*=π-2tan^{-1}*x*. That means that the result can
be rewritten using Euler's formula as

In other words, it's the same result we found before, but found without the need for trickery.

##### Example 6

◊ Evaluate \int dx/sin x.

◊
This can be tackled by rewriting the sine function in terms of complex exponentials, changing variables
to *u*=*e*^{ix}, and then using partial fractions.