integration

- anonymous

integration

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

\[\int\limits_{0}^{.5}8dx/(4x^2+1)^2\]

- amistre64

there it went .....thought maye you fell asleep on your keyboard ;)

- anonymous

sorry i am new took me a while to type the equation

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- amistre64

we can take out the constant "8" right?

- anonymous

yes

- amistre64

or do we want to use it later...
if u = 4x^2 +1 ; then du = 8x dx
dx = du/8x

- amistre64

and x would equal... sqrt(u-1)/2...not sure if that helps us tho...

- anonymous

well, this is supposed to be solved by using trig substitution. like,let 2x=tan\[\theta\]

- amistre64

\[\int\limits{} \frac{2}{8u^2 \sqrt{u-1}} du\]

- amistre64

we can try it by trig substitution....a little rusty, but why not :)

- amistre64

4x^2 = tan^2(t)
2x = tan(t) ok

- amistre64

tan^2 + 1 = sec^2
sec^2^2 = sec^4 right?

- amistre64

1/sec^4 = cos^4 right?

- anonymous

yes

- amistre64

then when we pull out the 8 we get:
[S] cos^4(t) dt to solve.... or did I miss somehting...

- amistre64

it looks good to me :)

- anonymous

wait, how did you get 4x^2 = tan^2t

- anonymous

what i have after solving is cos^2t

- amistre64

I simply defined it as such; since its a substitution, with the same value its good right?

- amistre64

tan(t) = 2x -> t = tan^-1(2x)

- anonymous

well, my teachet said
a^2-x^2 : let x=asint
a^2+x^2: x=atant
and so on

- amistre64

the value doesnt change, only the form it takes

- 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

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

makes sense?

- 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

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

i will show u what i have. i just have problem at last

- amistre64

so we end up with:
4 [S] cos^4 sec^2 dt

- amistre64

ok :)

- amistre64

4 [S] cos^2(t) dt is what it simplifies to.... im sure of it ;)

- anonymous

yes

- amistre64

cos^2 = 1 + sin^2..... and so on and so forth....

- amistre64

we can reduce the power by uing the cos(2t) equations...

- anonymous

yes what i cant do is do the final definite substitution

- amistre64

cos(2t) = 2cos^2 - 1
1 + cos(2t) = 2cos^2
(1+cos(2t))/2 = cos^2 right?

- amistre64

[S] 1/2 dt + [S] cos(2t)/2 dt .... looking more doable?

- amistre64

\[4 \int\limits_{} \frac{1}{2} dt + \int\limits_{} \frac{\cos(2t)}{2}dt\]

- amistre64

well that 4 times the whole lot of it :)

- amistre64

the left side goes to 4t/2 = 2t what we need to do is concentrate in the right side...

- amistre64

if we multiply it by 1 the value stays the same right?

- amistre64

\[\int\limits_{} \frac{2}{2} * \frac{\cos(2t)}{2} dt\]

- 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

this IS doable :)

- amistre64

D (sin(2t)) = 2 cos(2t) right?

- anonymous

##### 1 Attachment

- amistre64

very good :) make sure you keep track of the "4" we had and use it in both these integrals right?

- anonymous

yea. the problem with me is the final part where we substiture value. cant figure that out

- amistre64

so lets put up our solution as we have it so far...

- amistre64

r\[F(x) = 2t + \sin(2t) + C\]

- amistre64

you agree?

- anonymous

yes

- amistre64

this is our key then :)

##### 1 Attachment

- anonymous

wait, did we use pi/4 already?

- 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

i am confused

- 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

\[2 \tan^{-1}(2x) + \frac{4x}{4x^2+1}\]
from [0, .5]

- amistre64

if I did it right, my answer is 91 :)

- amistre64

my tan^-1(1) came back as 45 instead of pi/4..... I might have to adjust for that :)

- amistre64

(2+pi)/2 might be more appropriate....

- amistre64

2.57 is about my answer .... any way to check it?

- amistre64

2(pi/4) + sin(2pi/4) =
pi/2 + 1 = (2+pi)/2 same results as before

- anonymous

i will put that as answer. this was a test question . n way to check it.
thanks for your time

- amistre64

I hope it was helpful ;) thanx

Looking for something else?

Not the answer you are looking for? Search for more explanations.