Question

evaluate the indefinite integra1 as an infinite series : int sin(x^2)dx

Answer 1

\[\sin(x) = \sum_{0}^{\infty} \frac{ (-1)^{n} x^{2n+1}}{ (2n+1)! }\]

sin(x^2) --> plug x = (x^2). find integral in the fashion of \[\int\limits_{}^{} x^{n} = \frac{ x^{n+1} }{ n+1 } +C\]

Answer 2

∫x^n=(x^n+1)/(n+1)+C this is what you evaluated?

Answer 3

how did you get that series so easily?

Answer 4

wait wait, so i need to plug x with x^2?

Answer 5

i know it by memory :P.

it is not what i evaluated. that's the general rule for integrating function that look like power series (polynomials).

and yes you plug x^2 everywhere there is x in the power series of sinx

Answer 6

but there's only one x exists in the power series of sinx ?

Answer 7

it sounds so simple now then???

Answer 8

so same series but x^2^(2n+1) ???????????

Answer 9

the proper way of doing it would be x^(2(2n+1))

Answer 10

but yes. based on the same series

Answer 11

was this that easy one???????

Answer 12

idk what you mean

Answer 13

this could be done easily if i knew power series of sinx???

Answer 14

yes. that is the only way to do it. to base it on the power series of sinx. finding the power series of sin(x^2) right away would be very long

Answer 15

okay thank you!

Answer 16

hi unkle

Answer 17

hey unkle can you look over it? if you agree or not?

Answer 18

( i am trying a substitution method , ill post it when i get it to work )

Answer 19

okay thank you! :)

Answer 20

*if

Answer 21

[i cannot help here]

Answer 22

simpler than you'd think:

sin(x^2) = \[\int\limits_{}^{} \sum_{0}^{\infty} \frac{ (-1)^{n} x^{2(2n+1)}}{ (2n + 1)! } = \int\limits_{}^{} \sum_{0}^{\infty} \frac{ (-1)^{n} x^{4n+2}}{ (2n + 1)! } \]
=\[\sum_{0}^{\infty} \frac{ (-1)^{n} x^{4n+3}}{ (2n + 1)!*(4n +3)} \]

Answer 23

first line of my last post should read integral of sin(x^2), not just sin(x^2)