Question

is this problem solvable via integration by parts?

(sin(x))/(x^2+1)

Answer 1

\[\int\limits_{}^{}\frac{ \sin x }{ x^2 + 1} = ?\]

Answer 2

i presume that
u = sin (x)
dv = \[\frac{ 1 }{ x^2+1 }dx\]

Answer 3

or is this not solvable at all by that method? it tends to give a zero eh...

Answer 4

\[\int\limits_{}^{}\frac{ \sin x }{ x^2+1 } ={ \sin x \times \tanh^{-1} x } + \int\limits_{}^{} \tanh^{-1}x \times \cos x\]

Answer 5

tanh?

Answer 6

\[\int \frac{dx}{x^2 + 1} = \tan^{-1} x\]
right?

Answer 7

yes sir.

Answer 8

oh sorry. it should be tan -1

Answer 9

right

Answer 10

let me rewrite

\[\int\limits_{}^{}\frac{ \sin x }{ x^2 + 1} = \sin x \times \tan^{-1} x - \int\limits_{}^{} \tan^{-1} x \times \cos x\]

Answer 11

right.. now integrate by parts again

Answer 12

then lhs=rhs...i mean...same integrand will come on RHS

Answer 13

dv = \[\frac{ 1 }{ x^2 + 1 }\]
u = cos x

so, \[\int\limits_{}^{}\frac{ \sin x }{ x^2 + 1 } = \sin x \times \tan^{-1}x - (\cos x \times \tan^{-1} + \int\limits_{}^{} \tan^{-1} x \times \sin x)\]

is that it akash, or i had it wrong?

Answer 14

should be \[\int \frac{\sin x}{x^2 + 1}\]
because it's vdu

Answer 15

sorry if i'm a bit confused... hahaha

Answer 16

dv = tan^-1 x

v = 1/x^2 + 1 right?

Answer 17

or no?

Answer 18

oh yeah. you're right LGS. i was confused. hahahahaha.

Answer 19

i remembered i isolated it because of LIATE. =)))

Answer 20

y =tan^-1 x

tan y = x

sec^2 y = 1

y = 1/sec^2 y

y = 1/x^2 + 1

yup that's right

Answer 21

so you should have \[\int \frac{\sin x}{x^2 + 1} = \sin x \tan^{-1} x - \frac{\cos x}{x^2 + 1} - \int \frac{\sin x}{x^2 + 1}\]
right?

Answer 22

oh... and then move the integral on the other side, divide by to and eat your bacon? =)))

Answer 23

remember to eat the fats first

Answer 24

yessir. but, wolfram alpha has a weird answer. =))) ((-1 + E^2)*CosIntegral[I - x] + (-1 + E^2)*CosIntegral[I + x] + I*(1 + E^2)*(SinIntegral[I - x] + SinIntegral[I + x]))/(4*E)

Answer 25

huh

Answer 26

that's interesting...

Answer 27

it has imaginary numbers. that looked like some f-d up s-it to me. =))

Answer 28

\[\int \frac{\sin x}{x^2 + 1}\]
\[u = \frac{1}{x^2 + 1} \implies du = -\frac{2x}{(x^2 + 1)^2}\]
\[dv = \sin x \implies v = -\cos x\]
\[\int \frac{\sin x}{x^2 + 1} = -\frac{\cos x}{x^2 + 1} - 2\int \frac{x\cos x}{x^2 + 1}\]
yes.. i think we made a mistake a while ago...this is complex...very complex

Answer 29

i saw our mistake earlier...we made wrong substitution in the second IBP

Answer 30

ah yes. we forgot to apply LIATE immediately. instead of LIATE, ETAIL happened. =)))

Answer 31

well..that wasnt the problem i was talking about...

Answer 32

order of LIATE doesn't really matter

Answer 33

however.... you let \[u = \sin x \implies du = \cos x\]
\[dv = \frac{1}{x^2 + 1} \implies v = \tan^{-1}x\]
so... \[\int \frac{\sin x}{x^2 + 1} = \sin x \tan^{-1} - \int \tan^{-1} \cos x dx\]
see th mistake?

Answer 34

what we did earlier was + instead of -

Answer 35

yeap. that should have been a negative. =)))

Answer 36

yes. so now, this is equally difficult and complex

Answer 37

unfortunately there is No closed form for this integral in terms of elementary functions

Answer 38

indeed

Answer 39

sorry to have dragged it. i thought there was hope

Answer 40

no problem. My professor and I have been working on it for a few days already. Looks like it needs something called, "taylor", is it? =)))

Answer 41

i hear it's a very swift method