Question

Hi, does anyone knows how to integrate (sinx)/(2+x^2) dx from x=-1 to 1

Answer 1

\[\int\limits_{-1}^{1} \frac{ sinx }{ (2+x^2) }dx\]

Answer 2

This integral is obviously 0, but do you want to really solve it or just get to the answeR?

Answer 3

I have no idea how to start and pretty lost, so it'll be great ifyou can help me go through it steps by steps

Answer 4

And by obvious, I mean that since you have a sine function, it's an odd function, so any time you integrate it from some point that's symmetrical around 0, it's going to be 0.

Since (2+x^2) is also symmetrical for all values since it's even, it will give the same weight at every point on the sine function.

Answer 5

so even function cancel out odd function? as in sinx will cancel out the 2+x^2?

Answer 6

Yeah, so as for actually solving it, I really don't know how to do this one, and it would be painful to try! But once you understand that odd functions that are integrated from -a to +a is 0, it should make sense.

Imagine adding up the area from the integral of x^3 or just plain sin(x) from some random negative distance to the same positive distance. You'll get an equal negative area on the left and an equal positive area on the right that cancel out.

Answer 7

i'm going to try that now

Answer 8

The point of the even function on the bottom is that it won't change the left and right part's symmetry on the odd function. So now if you imagine cosine or x^2 then when you integrate from some random negative number to 0, it's equal to as if you had just integrated from 0 to that same number but positive now. So it won't affect how high and low the odd function is, so they'll still cancel out.

Answer 9

I'll be here doing some homework, think about it a bit and I'll check on you in a few minutes if you catch my drift! Good luck.

Answer 10

thanks man! you're a legend, i'll just reread what you said til i understand the concept

Answer 11

This one is hell to integrate by hand.. don't even know where to start

Answer 12

Try to draw a picture, or just calculate exact values of that integral between -1 and 1 at values like +1/2 and -1/2 to see that they will cancel out each other.

Answer 13

Does anyone know how to actually calculate the indefinite integral for this question

Answer 14

ahh i see it, i plugged in -1 and 1, since they're suppose to subtract each other, it will come out as 0!

Answer 15

nope, i dont know how to do it with indefinite integral either, good luck man !

Answer 16

@ECE I CAN integrate this, but no sane person would integrate this perfectly symmetrical integral that equals 0 lol.

Answer 17

In case you're curious:

http://www.wolframalpha.com/input/?i=sinx%2F(2%2Bx%5E2)dx&t=crmtb01

Answer 18

yeah I put it in wolfram too lol

Answer 19

thanks guys! I will close this question now

Answer 20

Looks like you would need IBP then a substitution?

Answer 21

Honestly no clue

Answer 22

\[u = sinx, du = cosxdx\]
\[dv = 1/2+x^2, v = \arctan(x/\sqrt{2})/\sqrt{2}\]

Answer 23

Ah forget it, would take forever to solve lol

Answer 24

who comes up with these crazy questions?

Answer 25

:(

Answer 26

show that ∫(z)=z^3 is analytical,,,,,can any one help me here?