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Integrating with trig. substitution

  • Thread starter raptik
  • Start date
  • #1
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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?
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,258
618
Expanding it is a bad idea. Complete the square the denominator. Then pick a trig substitution.
 
  • #3
21
0
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.
 
  • #4
Dick
Science Advisor
Homework Helper
26,258
618
sin(2u)=2*sin(u)*cos(u), it's a trig identity, yes? Try using that.
 
  • #5
21
0
awsome! Figured it out. Thnx, Dick.
 

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