Integrating with trig. substitution

[raptik]

Homework Statement

Integral of dx/[(x² - 2x + 2)²]

Homework Equations

Trig substitution rules:
for expression sqrt(a² - x²)
make x = asin(t) with -(pi/2) < t < (pi/2)

for sqrt(x² - a²)
make x = asec(t) with 0< t < (pi/2)

and
for sqrt(a² + x²)
make x = atan(t) with -(pi/2) < t < (pi/2)

The Attempt at a Solution

multiply out to get

integral dx/(x⁴ - 4x³ + 8x² - 8x + 4)

...I'm really at a loss for what to do next. Any help?

 

[Dick]

Expanding it is a bad idea. Complete the square the denominator. Then pick a trig substitution.

 

[raptik]

ok, so integral dx/[(x² - 2x + 2))²] = integral dx/[((x-1)² + 1)²]

t = x-1; dt = dx

integral dt/(t² + 1)²

t = tan(u); dt = sec²(u)du

= integral sec²(u)du/(tan²(u) + 1)²

= integral sec²(u)du/(sec²(u))² = integral du/sec²(u)

= integral cos²(u)du = 1/2 integral (1 + cos(2u))du

= (1/2)u + 1/4sin(2u) + c

...I hope I got all that right. But I still can't get it back in terms of x. I would use the triange knowing tan(u) = t/1 = opposite/adjacent, but the "2u" makes me a bit confused as to how to get back in terms of x.

 

[Dick]

sin(2u)=2*sin(u)*cos(u), it's a trig identity, yes? Try using that.

 

[raptik]

awsome! Figured it out. Thnx, Dick.