Question

hey james ishaan solve this
intergrate
(x)/(a^2cosx+b^2sinx)

Answer 1

Not elementary.
http://www.wolframalpha.com/input/?i=int+%28x%29%2F%28a%5E2cosx%2Bb%5E2sinx%29+dx

Answer 2

...which is to say, this integral is not the finite sum of elementary functions.

Answer 3

james it is , deduct pie from x everywhere in the eq

Answer 4

then u d get it

Answer 5

why don't you show us your solution wasiqss

Answer 6

I see. I stand corrected. Nonetheless, I second Z's motion.

Answer 7

and i third it

Answer 8

Let J be the original integral. Change variables, with u = x - pi. Then after a small amount of algebra,
\[ J = -J + \int \frac{\pi}{a^2\cos u + b^2\sin u} du \]
Hence
\[ J = \frac{\pi}{2} \int \csc(u + \phi) du \]
where
\[ \tan \phi = b^2/a^2 \]
and now that's straight-forward. Nice problem. I've not see it before.

Answer 9

I dropped a \[ \sqrt{a^4 + b^4} \] outside that last integral.

Answer 10

also, it's a nice example of where Wolfram isn't entirely reliable.

Answer 11

I came across a similar problem few months ago. I can't remember exactly what it was, but I'll try to find it and post it.

Answer 12

Something is bugging me about this result. Looking at the integrand in the last integral and the real function that comprises the anti-derivatve, both of those functions are periodic, with period 2pi.

But the _original_ integrand is not periodic with period 2pi. That is to say, if we evaluated the original integral from
x = 0 to pi/4 say,
that result should be a positive number, as the integrand is everywhere positive on that interval. Further, that result should be different from the integral from
x = 2pi to 2pi + pi/4 = 9pi/4,
because the integrand is clearly greater everywhere.

I hypothesize that the way we have to deal with that is by considering the ln function as the restriction to the real numbers of the complex valued ln function, which is a very non-standard thing to do.

Answer 13

ohhh so many replies :P, n yeh u know what this is a question of first semester in my uni :P

Answer 14

ohhh so many replies :P, n yeh u know what this is a question of first semester in my uni :P

Answer 15

i know this is very stupid kinda question , bt r teacher gave it :P

Answer 16

It's a very good question.

Answer 17

i am a bit confused so maybe you can help. liking to do things the simple way first i put
\[a=b=1\] and arrived at
\[\int\frac{x}{\sin(x)+\cos(x)}dx=\int\frac{x}{\sqrt{2}\sin(x+\frac{\pi}{4})}dx\]

Answer 18

now what can an anti- derivative of this thing look like? are we ignoring the domain of this thing?

Answer 19

then to cheat i graphed it, and i cannot even begin to imagine a closed form of this integral.

Answer 20

This is how I think the complex branching gets introduced.

Using the change of variables u = x - pi, we clearly cross a point of discontinuity for the function, but that's ok under the rules of substitution. Put that into your integral, call it K, then you see that
\[ K = \int \frac{x- \pi}{\sqrt{2} \sin(x + \pi/4 + \pi/2) } dx = \int \frac{-(x-\pi)}{\sqrt{2} \sin(x + \pi/4)} = -K + \int \frac{\pi}{\sqrt{2} \sin(x + \pi/4)} \ \]

Answer 21

So yes, there are some heavy duty complications. But we do arrive at a closed form.

Answer 22

\[K = \int \frac{x- \pi}{\sqrt{2} \sin(x + \pi/4 + \pi/2) } dx \]
\[= \int \frac{-(x-\pi)}{\sqrt{2} \sin(x + \pi/4)} = -K + \int \frac{\pi}{\sqrt{2} \sin(x + \pi/4)} \] i just rewrote so it would fit on my page

Answer 23

right. Sorry about that.

Answer 24

BUT, and this only something you can appreciate if you've taken complex analysis, in crossing over onto another branch of the function, the K on the LHS--the original integral--is actually not necessarily the same thing as the K on the RHS.

This is why the answer we arrive at using naive integration is periodic, even though the original integrand is not.

I've never seen this in this form before.

Answer 25

ok so i can follow the algebra no problem, but i still have a problem with the solution. of course maybe that is just my problem

Answer 26

ahhhhh ok, i am beginning to understand

Answer 27

which (correct me if i am wrong) means the elementary algebra from now on is not necessarily correct. if
\[K\neq K\] you cannot add to both sides, divide by two etc.

Answer 28

Yes, I think that's right.

Answer 29

It's K and K', and they're not quite the same.

Answer 30

ok
in any case if we get the integral it should be
\[-\frac{\pi}{\sqrt{2}}\ln(\cot(x+\frac{\pi}{4})+\csc(x+\frac{\pi}{4}))\]
so i guess we are leaving real numbers and entering the world of complex in any case

Answer 31

Yes. Now as my original 'complaint' about this suggested, there's something odd going on here.

Evaluate that from x=0 to pi/4 and you get the same answer as though you evaluated it from x = 2pi to 2pi+pi/4. Yet the fact that those two integral are equal is clearly false, as the integrand in the second is always strictly greater than in the first.

Answer 32

or to avoid improper integrals, even x = 0 to pi/16 and x = 2pi to 2pi + pi/16

Answer 33

so maybe wolfram is not as dumb as it first appears

Answer 34

Perhaps not. And perhaps my original intuition about this was in fact correct.

I'm now dismissing my earlier hypothesis about how to 'fix' this with the (infinitely-branched) log function; that's not it.