Question

difficult integration question

Answer 1

th;dr

Answer 2

is this all about even-odd stuff maybe?

Answer 3

but these integrals are undefined at the boundary points, so... hm...

Answer 4

oh wait, no they're not

Answer 5

no..this is actually really trivial

Answer 6

of course... -.-
for you

Answer 7

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)

Answer 8

Here is the full question if you're interested

Answer 9

first seems true .. since cos is an even function, and sin is an odd function ... behaviour will be symmetric at 0

Answer 10

maybe redefining function or changing limits might give the first part

Answer 11

yeah but how do you write it?
Zarkon says simple sub, but \(\theta=-\theta\) changes the bounds

Answer 12

you and I were thinking the same again ;)

Answer 13

TT that is what you want to do

Answer 14

dont use the same variable though..call it something different

Answer 15

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?

Answer 16

then use

\[\int\limits_{a}^{b}f(x)dx=-\int\limits_{b}^{a}f(x)dx\]

Answer 17

i guess yes .... so does dQ which which will give the positive value

Answer 18

but that would be \(-I\)

Answer 19

no

Answer 20

rechanging our bounds results -
and dQ yields -
so +

Answer 21

aw duh!!!
I told you I wasn't thinking it through clearly enough XD

Answer 22

Can anybody help me with the rest of the question? (I attached it earlier)

Answer 23

the second problem is quite easy ... it seeems

Answer 24

just add up two those different terms

Answer 25

i would like to do it here ... but very poor at latex

Answer 26

I mean part (ii) and (iii) and (iv) not the second problem in (i)

Answer 27

int fa + inf ga = int (fa+ga)

Answer 28

(ii) Find J.

Answer 29

set suppose tanQ = x, it should solve the question

Answer 30

but the boundaries would be invalid with tan(pi/2) and tan(-pi/2) since cos would be zero

Answer 31

it's on denominator .. which means function approaches 0 at those boundary

Answer 32

supposing tanQ = x also eliminates, SecQ^2 dQ on denominator

Answer 33

no i'm pretty sure it would be infinite if the denominator approaches zero lo

Answer 34

yes i know that, but i'm talking about the boundaries

Answer 35

they are pi/2 and -pi/2 so how do i evaluate them for my substitution?

Answer 36

tan(pi/2) is infinite

Answer 37

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) ?

Answer 38

still why not give a try ... int 1/(1+x^2a)dx

Answer 39

\[u=\tan(\theta)\]

works

you get \[\frac{\pi}{\sqrt{\cos^2(2\alpha)}}\]

Answer 40

Zarkon: I don't understand how I use \[u = \tan(\theta)\] when the limits are pi/2 and -pi/2 \[\tan(\pi/2) = \infty\]

Answer 41

you get new lmits from \(-\infty\) to \(\infty\)

Answer 42

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?

Answer 43

basically yeah, it's called an improper integral

Answer 44

cheers

Answer 45

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