anonymous
  • anonymous
difficult integration question
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
1 Attachment
inkyvoyd
  • inkyvoyd
th;dr
kymber
  • kymber

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TuringTest
  • TuringTest
is this all about even-odd stuff maybe?
TuringTest
  • TuringTest
but these integrals are undefined at the boundary points, so... hm...
TuringTest
  • TuringTest
oh wait, no they're not
Zarkon
  • Zarkon
no..this is actually really trivial
TuringTest
  • TuringTest
of course... -.- for you
Zarkon
  • Zarkon
a simple subtitituon will give you the first part adding the two integrals ( the one with plus and the one with minus...will give the second part)
anonymous
  • anonymous
Here is the full question if you're interested
experimentX
  • experimentX
first seems true .. since cos is an even function, and sin is an odd function ... behaviour will be symmetric at 0
experimentX
  • experimentX
maybe redefining function or changing limits might give the first part
TuringTest
  • TuringTest
yeah but how do you write it? Zarkon says simple sub, but \(\theta=-\theta\) changes the bounds
TuringTest
  • TuringTest
you and I were thinking the same again ;)
Zarkon
  • Zarkon
TT that is what you want to do
Zarkon
  • Zarkon
dont use the same variable though..call it something different
TuringTest
  • TuringTest
huh, but it seems like I get the bounds reversed... I can't think it though clearly enough\[\phi=-\theta\]so the bounds change, does that not matter?
Zarkon
  • Zarkon
then use \[\int\limits_{a}^{b}f(x)dx=-\int\limits_{b}^{a}f(x)dx\]
experimentX
  • experimentX
i guess yes .... so does dQ which which will give the positive value
TuringTest
  • TuringTest
but that would be \(-I\)
Zarkon
  • Zarkon
no
experimentX
  • experimentX
rechanging our bounds results - and dQ yields - so +
TuringTest
  • TuringTest
aw duh!!! I told you I wasn't thinking it through clearly enough XD
anonymous
  • anonymous
Can anybody help me with the rest of the question? (I attached it earlier)
experimentX
  • experimentX
the second problem is quite easy ... it seeems
experimentX
  • experimentX
just add up two those different terms
experimentX
  • experimentX
i would like to do it here ... but very poor at latex
anonymous
  • anonymous
I mean part (ii) and (iii) and (iv) not the second problem in (i)
experimentX
  • experimentX
int fa + inf ga = int (fa+ga)
anonymous
  • anonymous
(ii) Find J.
experimentX
  • experimentX
set suppose tanQ = x, it should solve the question
anonymous
  • anonymous
but the boundaries would be invalid with tan(pi/2) and tan(-pi/2) since cos would be zero
experimentX
  • experimentX
it's on denominator .. which means function approaches 0 at those boundary
experimentX
  • experimentX
supposing tanQ = x also eliminates, SecQ^2 dQ on denominator
anonymous
  • anonymous
no i'm pretty sure it would be infinite if the denominator approaches zero lo
anonymous
  • anonymous
yes i know that, but i'm talking about the boundaries
anonymous
  • anonymous
they are pi/2 and -pi/2 so how do i evaluate them for my substitution?
anonymous
  • anonymous
tan(pi/2) is infinite
anonymous
  • anonymous
I don't understand whyy you are putting it like that. Aren't we using the sub u = tan(theta) so u = tan(pi/2) ?
experimentX
  • experimentX
still why not give a try ... int 1/(1+x^2a)dx
Zarkon
  • Zarkon
\[u=\tan(\theta)\] works you get \[\frac{\pi}{\sqrt{\cos^2(2\alpha)}}\]
anonymous
  • anonymous
Zarkon: I don't understand how I use \[u = \tan(\theta)\] when the limits are pi/2 and -pi/2 \[\tan(\pi/2) = \infty\]
Zarkon
  • Zarkon
you get new lmits from \(-\infty\) to \(\infty\)
anonymous
  • anonymous
Ok, thanks. I've never seen infinite limits before, so after resolving the integral do I just use the limit of the function as the variable approaches infinite?
TuringTest
  • TuringTest
basically yeah, it's called an improper integral
anonymous
  • anonymous
cheers
TuringTest
  • TuringTest
just so you know all the technicalities, here's how the whole idea works rather well-explained: http://tutorial.math.lamar.edu/Classes/CalcII/ImproperIntegrals.aspx

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