ksaimouli
  • ksaimouli
http://online.math.uh.edu/apcalculus/exams/images/AP_AB_version1__37.gif
Mathematics
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

ksaimouli
  • ksaimouli
@mathmate
ksaimouli
  • ksaimouli
i did this with u substituion it did not work
ksaimouli
  • ksaimouli
i am getting something like ln2/2

Looking for something else?

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

More answers

mathmate
  • mathmate
It looks to me more an atan(2x)
ksaimouli
  • ksaimouli
how
ksaimouli
  • ksaimouli
i took u=1+4t^2
ksaimouli
  • ksaimouli
du/8t=dt
mathmate
  • mathmate
The standard integral is \( \int \frac{dx}{1+x^2} = atan(x) + C \)
ksaimouli
  • ksaimouli
|dw:1358025284718:dw|
mathmate
  • mathmate
So you'd be integrating \( \int \frac{du}{8u(1+4t^2)} \)
ksaimouli
  • ksaimouli
use the equation button
mathmate
  • mathmate
The result is not ln(u)
ksaimouli
  • ksaimouli
ok i should remember that formula right?
mathmate
  • mathmate
I suggest you use y=2t
mathmate
  • mathmate
Yes, it's on the first page of standard integrals. Check it out.
ksaimouli
  • ksaimouli
ok
mathmate
  • mathmate
\( \large \int_0^{1/2} \frac{dt}{1+4t^2}= \int_0^{1} \frac{2dy}{1+y^2} = [atan(y)]_0^{1}= 2\pi/4=\pi/2 \)
ksaimouli
  • ksaimouli
@mathmate i found the formula to be \[\frac{ 1 }{ a }\tan^{-1} \frac{ x }{ a }+c\]
mathmate
  • mathmate
Right, so a=1/2.
ksaimouli
  • ksaimouli
what is x then
ksaimouli
  • ksaimouli
\[\int\limits_{}^{}\frac{ 1 }{ a^2+x^2}=\frac{ 1 }{ a }\tan^{-1} \frac{ x }{ a}+c\]
mathmate
  • mathmate
x is your t. you can replace the identity with \( \large \int \frac{dx}{a^2+x^2} = \int \frac{dt}{a^2(1+(t/a)^2)} \)
mathmate
  • mathmate
and a=1/2
ksaimouli
  • ksaimouli
can i do \[\frac{ 1 }{ 1 }\tan^{-1} \frac{ 2t }{ 1}\]
ksaimouli
  • ksaimouli
and t=1/2
mathmate
  • mathmate
t is a variable, and you cannot equate it to a constant. Use the formula you found, change t for x, and 1/2 for a, then the left-hand side is exactly your question. All you need to do after that is to evaluate the right-hand side.

Looking for something else?

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