anonymous
  • anonymous
integration
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[\int\limits_{0}^{.5}8dx/(4x^2+1)^2\]
amistre64
  • amistre64
there it went .....thought maye you fell asleep on your keyboard ;)
anonymous
  • anonymous
sorry i am new took me a while to type the equation

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amistre64
  • amistre64
we can take out the constant "8" right?
anonymous
  • anonymous
yes
amistre64
  • amistre64
or do we want to use it later... if u = 4x^2 +1 ; then du = 8x dx dx = du/8x
amistre64
  • amistre64
and x would equal... sqrt(u-1)/2...not sure if that helps us tho...
anonymous
  • anonymous
well, this is supposed to be solved by using trig substitution. like,let 2x=tan\[\theta\]
amistre64
  • amistre64
\[\int\limits{} \frac{2}{8u^2 \sqrt{u-1}} du\]
amistre64
  • amistre64
we can try it by trig substitution....a little rusty, but why not :)
amistre64
  • amistre64
4x^2 = tan^2(t) 2x = tan(t) ok
amistre64
  • amistre64
tan^2 + 1 = sec^2 sec^2^2 = sec^4 right?
amistre64
  • amistre64
1/sec^4 = cos^4 right?
anonymous
  • anonymous
yes
amistre64
  • amistre64
then when we pull out the 8 we get: [S] cos^4(t) dt to solve.... or did I miss somehting...
amistre64
  • amistre64
it looks good to me :)
anonymous
  • anonymous
wait, how did you get 4x^2 = tan^2t
anonymous
  • anonymous
what i have after solving is cos^2t
amistre64
  • amistre64
I simply defined it as such; since its a substitution, with the same value its good right?
amistre64
  • amistre64
tan(t) = 2x -> t = tan^-1(2x)
anonymous
  • anonymous
well, my teachet said a^2-x^2 : let x=asint a^2+x^2: x=atant and so on
amistre64
  • amistre64
the value doesnt change, only the form it takes
amistre64
  • amistre64
if we have something under a radical, then we want to substitute it with: asin(t) so that the "a" gets changed to a good value that can be taken away... there is no radical here to worry about so we dont have to worry about what "a" needs to be right?
amistre64
  • amistre64
what we need here is not to get rid of a radical sign; but to transform it from one form to another.... we need tan^2 to equal 4x^2, the only way thats possible is to have tan = 2x
amistre64
  • amistre64
makes sense?
amistre64
  • amistre64
but I tend to miss something whenever I do this...that dt in the problem is not simply "dt" it is dt = something dx....and thats what I need to find lol
amistre64
  • amistre64
tan(t) = 2x D (tan(t)) =D (2x) sec^2(t) dt = 2 dx dx = sec^2(t) dt/2 is what I should have... right?
anonymous
  • anonymous
i will show u what i have. i just have problem at last
amistre64
  • amistre64
so we end up with: 4 [S] cos^4 sec^2 dt
amistre64
  • amistre64
ok :)
amistre64
  • amistre64
4 [S] cos^2(t) dt is what it simplifies to.... im sure of it ;)
anonymous
  • anonymous
yes
amistre64
  • amistre64
cos^2 = 1 + sin^2..... and so on and so forth....
amistre64
  • amistre64
we can reduce the power by uing the cos(2t) equations...
anonymous
  • anonymous
yes what i cant do is do the final definite substitution
amistre64
  • amistre64
cos(2t) = 2cos^2 - 1 1 + cos(2t) = 2cos^2 (1+cos(2t))/2 = cos^2 right?
amistre64
  • amistre64
[S] 1/2 dt + [S] cos(2t)/2 dt .... looking more doable?
amistre64
  • amistre64
\[4 \int\limits_{} \frac{1}{2} dt + \int\limits_{} \frac{\cos(2t)}{2}dt\]
amistre64
  • amistre64
well that 4 times the whole lot of it :)
amistre64
  • amistre64
the left side goes to 4t/2 = 2t what we need to do is concentrate in the right side...
amistre64
  • amistre64
if we multiply it by 1 the value stays the same right?
amistre64
  • amistre64
\[\int\limits_{} \frac{2}{2} * \frac{\cos(2t)}{2} dt\]
amistre64
  • amistre64
lets use this to our advantage and take out the constants that dont matter.... ok?\[4* \frac{1}{4} \int\limits_{} 2 \cos(2t)dt\]
amistre64
  • amistre64
this IS doable :)
amistre64
  • amistre64
D (sin(2t)) = 2 cos(2t) right?
anonymous
  • anonymous
1 Attachment
amistre64
  • amistre64
very good :) make sure you keep track of the "4" we had and use it in both these integrals right?
anonymous
  • anonymous
yea. the problem with me is the final part where we substiture value. cant figure that out
amistre64
  • amistre64
so lets put up our solution as we have it so far...
amistre64
  • amistre64
r\[F(x) = 2t + \sin(2t) + C\]
amistre64
  • amistre64
you agree?
anonymous
  • anonymous
yes
amistre64
  • amistre64
this is our key then :)
1 Attachment
anonymous
  • anonymous
wait, did we use pi/4 already?
amistre64
  • amistre64
i didnt; I dont trust myself enough to change the original interval yet, so I just convert it back to values of x :)
anonymous
  • anonymous
i am confused
amistre64
  • amistre64
we can use this the way it is...minus the +C of course if we change the original interval to match oour substitution...which it looks like you might have down with that pi/4..... we can use that if your confident in it; but I prefer to undo the substitution of our trig with the key I provided...
amistre64
  • amistre64
\[2 \tan^{-1}(2x) + \frac{4x}{4x^2+1}\] from [0, .5]
amistre64
  • amistre64
if I did it right, my answer is 91 :)
amistre64
  • amistre64
my tan^-1(1) came back as 45 instead of pi/4..... I might have to adjust for that :)
amistre64
  • amistre64
(2+pi)/2 might be more appropriate....
amistre64
  • amistre64
2.57 is about my answer .... any way to check it?
amistre64
  • amistre64
2(pi/4) + sin(2pi/4) = pi/2 + 1 = (2+pi)/2 same results as before
anonymous
  • anonymous
i will put that as answer. this was a test question . n way to check it. thanks for your time
amistre64
  • amistre64
I hope it was helpful ;) thanx

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