# Thread: Trig Substitution

1. ## Trig Substitution

These are 2 problems that I'm stumped on and can't get...we are supposed to evaluate the integral using trig substitution.

x^2/(3+4x-4x^2)

(x^2+1)/(x^2-2x+2)

2. Originally Posted by sfgiants13
These are 2 problems that I'm stumped on and can't get...we are supposed to evaluate the integral using trig substitution.

x^2/(3+4x-4x^2)

[snip]
$\displaystyle \frac{x^2}{3 + 4x - 4x^2} = -\frac{1}{4} + \frac{x + \frac{3}{4}}{-4x^2 + 4x + 3}$

A trig substitution is not needed here because the quadratic is not irreducible. use a partial fraction decomposition

3. Originally Posted by sfgiants13
[snip]
(x^2+1)/(x^2-2x+2)
$\displaystyle \frac{x^2 + 1}{x^2 - 2x + 2} = \frac{(x^2 - 2x + 2) + 2x - 1}{x^2 - 2x + 2} = 1 + \frac{2x - 1}{x^2 - 2x + 2}$

$\displaystyle = 1 + \frac{(2x - 2) + 1}{x^2 - 2x + 2} = 1 + \frac{2x - 2}{x^2 - 2x + 2} + \frac{1}{x^2 - 2x + 2}$

$\displaystyle 1 + \frac{2x - 2}{x^2 - 2x + 2} + \frac{1}{(x - 1)^2 + 1}$.

The first two terms are trivial to integrate. To integrate the last term make the substitution $\displaystyle x-1 = tan \theta$ (or just recognise the standard form).