Question

integration

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

we can take out the constant "8" right?

Answer 5

yes

Answer 6

or do we want to use it later...

if u = 4x^2 +1 ; then du = 8x dx

dx = du/8x

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

4x^2 = tan^2(t)

2x = tan(t) ok

Answer 12

tan^2 + 1 = sec^2

sec^2^2 = sec^4 right?

Answer 13

1/sec^4 = cos^4 right?

Answer 14

yes

Answer 15

then when we pull out the 8 we get:

[S] cos^4(t) dt to solve.... or did I miss somehting...

Answer 16

it looks good to me :)

Answer 17

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

Answer 18

what i have after solving is cos^2t

Answer 19

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

Answer 20

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

Answer 21

well, my teachet said

a^2-x^2 : let x=asint
a^2+x^2: x=atant
and so on

Answer 22

the value doesnt change, only the form it takes

Answer 23

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?

Answer 24

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

Answer 25

makes sense?

Answer 26

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

Answer 27

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?

Answer 28

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

Answer 29

so we end up with:

4 [S] cos^4 sec^2 dt

Answer 30

ok :)

Answer 31

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

Answer 32

yes

Answer 33

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

Answer 34

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

Answer 35

yes what i cant do is do the final definite substitution

Answer 36

cos(2t) = 2cos^2 - 1

1 + cos(2t) = 2cos^2

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

Answer 37

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

Answer 38

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

Answer 39

well that 4 times the whole lot of it :)

Answer 40

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

Answer 41

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

Answer 42

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

Answer 43

lets use this to our advantage and take out the constants that dont matter.... ok?\[4* \frac{1}{4} \int\limits_{} 2 \cos(2t)dt\]

Answer 44

this IS doable :)

Answer 45

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

Answer 46

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

Answer 47

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

Answer 48

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

Answer 49

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

Answer 50

you agree?

Answer 51

yes

Answer 52

this is our key then :)

Answer 53

wait, did we use pi/4 already?

Answer 54

i didnt; I dont trust myself enough to change the original interval yet, so I just convert it back to values of x :)

Answer 55

i am confused

Answer 56

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...

Answer 57

\[2 \tan^{-1}(2x) + \frac{4x}{4x^2+1}\]

from [0, .5]

Answer 58

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

Answer 59

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

Answer 60

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

Answer 61

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

Answer 62

2(pi/4) + sin(2pi/4) =

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

Answer 63

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

Answer 64

I hope it was helpful ;) thanx