cherrilyn
  • cherrilyn
∫dx/(x^2+6x+6)^2 Evaluate the integral by completing the square and using trig substitution. After completing the square I got (x+3)^2-3 but I don't know what to do next
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
amistre64
  • amistre64
your good with the completing the square part: [(x+3)^2 +6+9)]^2
amistre64
  • amistre64
you are adding +9 inside; so you add +9 outside as well....
amistre64
  • amistre64
wiat...-9 lol

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amistre64
  • amistre64
your right there...good job :)
anonymous
  • anonymous
lols amistre, "good job" :P
amistre64
  • amistre64
you need to substitute the trig..... x^2 -1 thing....
amistre64
  • amistre64
tan = x^2 + 1 sin = 1 - x^2 whats the other one? sec?
anonymous
  • anonymous
don't trig sub needed amistre what is with you and trig subbing!?!?! those are the worst!!! just to regular subbing
amistre64
  • amistre64
its in the prob itself...use trig sub lol :) not my fault this time lol
amistre64
  • amistre64
sec^2 - 1 = tan...i think thats the one...
amistre64
  • amistre64
x = sqrt(3) sec(t)
amistre64
  • amistre64
err.... (x+3) = sqrt(3)sec(t) 3sec^2(t) -3 = 3(sec^2(t)-1) [3tan^2(t)]^2 = 9 tan^4(t)
anonymous
  • anonymous
the integral of (x+3)^2-3 is \[\frac{1}{3}(x + 3)^3 - 3x\]
anonymous
  • anonymous
oy i should learn to read the question... amistre, i take everything back...
anonymous
  • anonymous
...but what i did was right...why do you need trig id?
amistre64
  • amistre64
d(sqrt(3) sec(t))/dt = ....everything? even the ceramic unicorn you gave me for my birthday?
amistre64
  • amistre64
d(sqrt(3) sec(t))/dt = d(x+3)/dx sqrt(3) sec(t)tan(t) dt = dx [sqrt(3) sec(t)tan(t) dt]/[9 tan^4(t)] is this looking right ;)
amistre64
  • amistre64
\[\int\limits_{} \frac{\sqrt{3} \sec(t) \tan(t)}{9 \tan^4 (t)}dt\]
amistre64
  • amistre64
we can ignore for the moment the sqrt(3)/9 reduce to: sec(t)/tan^3(t) = sin(t)/cos^4(t)
amistre64
  • amistre64
you sure you posted the question right?
cherrilyn
  • cherrilyn
yeah I just don't know how to add the division bar . lol . "Evaluate the integral by completing the square and using trigonometric substitution"
amistre64
  • amistre64
frac{top}{bottom} in the editor :)
amistre64
  • amistre64
any ideas how to integrate: sin(t) ----- ?? cos^4
amistre64
  • amistre64
sin(t) = sqrt(1-cos^2) sqrt(1-cos^2) ------------ doesnt seem to make it easier cos^4
amistre64
  • amistre64
cos^2 = 1-sin^2 (1-sin^2)(1-sin^2) = sin^4 -2sin^2 +1...... hmmm
amistre64
  • amistre64
D(tan^4) = 4 tan^3 sec^2....dont see that helping lol
cherrilyn
  • cherrilyn
No conclusion ? :(
amistre64
  • amistre64
i couldnt get to a result.....at leaast not without pencil and paper to do it with.... and a lot more braincells :)

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