Question

what is the anti derivative of 4/(1+4t^2)

Answer 1

you can use trigonometric substitution for this

Answer 2

didnt recognize that , can you show me?

Answer 3

this lists the three cases
https://en.wikipedia.org/wiki/Trigonometric_substitution

In this case you would use tan(x)

Answer 4

great more formulas to remember before the test wednesday :)

Answer 5

could i also use u substitution for this?

Answer 6

so first factor out 4

4/4(1/4 + t^(2))
=
1/(1/4 + t^(2))

now using the identity tan^(2)(theta) + 1 = sec^2(theta)
so we have that
t = (1/4)^(1/2)tan(theta)
t = (1/2)tan(theta)
Now solve for theta
theta = arctan(2t)

Now you need to right the integral it in terms of theta

so

d(theta) = (1/2)sec^(2)(theta)dx

Answer 7

no you cannot use u substitution for this

Answer 8

oh god we havent done much trig substiution if any..

Answer 9

that 1/4 has to be 1 + tan^(2)(theta) crap

Answer 10

is there any other way to do this problem?

Answer 11

lol this only looks hard it really isn't and no this is a special case

Answer 12

I guess you could use partial fractions

Answer 13

Really you should just watch a video on trig sub, try khan academy he explains it really well

Answer 14

thanks will do

Answer 15

u = 1+4t^2
du = 8tdx
du/8t = dx

\[\int\limits_{}^{} 4du/(u)8t\]


yeah this will get ugly if we use u sub

as

(u - 1)/4 =t^(2)
((u-1)/4)^(1/2) = t

Answer 16

I messed up on my algebra and my brain is too dead I wil lget someone else to help you

Answer 17

This video might apply to oyur problem

http://www.khanacademy.org/math/calculus/v/integrals--trig-substitution-2

Answer 18

I think I made a mistake factoring the problem initially let me try again

Answer 19

ok

Answer 20

so

\[\int\limits_{}^{}4dt/(1 + 4t^{2})\]

so set

2t = tan(theta)
t = tan(theta)/2
theta = arctan(2t)

Now write it in terms of d(theta)

d(theta) = sec^(2)(theta)/2 dt

2d(theta)/sec^(2)(theta) = dt

\[\int\limits_{}^{}d(\theta)8/\sec^{2}(\theta)(1+\tan^{2}(\theta)) = 8\int\limits_{}^{}d(\theta)/\sec^{4}(\theta)\]

Answer 21

\[8\int\limits_{}^{}d(\theta)/(1/\cos(\theta)^{4}) = 8\int\limits_{}^{}\cos^{4}(\theta)d(\theta)\]

Answer 22

this can be solved pretty easily using identities now

Answer 23

thanks think i got it now

Answer 24

\[8 \int\limits_{}^{} ((1+\cos(2\theta))/2)^{2}d(\theta)\]

trigonometric functions to the even power we use this identity 1 + cos(2theta)/2

Now

\[2 \int\limits_{}^{} (1 + 2\cos(2\theta) + \cos^{2}(2\theta))d(\theta)\]

Now we can just split the integral

\[2( \int\limits_{}^{} 1d(\theta) + \int\limits_{}^{} 2\cos(2\theta)d(\theta) + \int\limits_{}^{} \cos^{2}(2\theta)d(\theta) )\]

Answer 25

u = 2(theta)
du/2 = d(theta)

\[2\int\limits_{}^{}\cos(u)du \]

\[2\int\limits_{}^{}\cos^{2}(2\theta)d(\theta) = \int\limits_{}^{}\cos^{2}(u)du = (1/2) \int\limits_{}^{} 1 du + (1/2)\int\limits_{}^{}\cos(2u)du\]

=

2theta - 2sin(2theta) + (theta) - sin(4theta)/2 + c

Answer 26

and I'm still not done we need to write it in terms of t. I probably made a mistake along the way it is almost certain

2theta - 2sin(2theta) + (theta) - sin(4theta)/2 + c

2t = tan(theta)
t = tan(theta)/2
theta = arctan(2t)
sec^(2) = (1+4t^2)

2arctan(2t) - 2sin(2arctan(2t)) + arctan(2t) - sin(4arctan(2t))/2 + c
=
3arctan(2t) - 2sin(2arctan(2t)) - sin(4arctan(2t))/2 + c

is the final answer

Answer 27

integral of du/(a^2+u^2)=(1/a)arc tan u/a +C
let a=1, u=2t so that
du=2 dt thus
integral of du/(a^2+u^2)=int of 2*2dt/(1+4t^2)
=2arc tan 2t +C ......ans