Question

I can't seem to do this integral:
\[\int xarcsin(x^{2})dx\]

Answer 1

take x^2=sin(theta)

Answer 2

I don't see how that would help anything

Answer 3

I was thinking about integration by parts because I know the derivative for arcsin but i dont know

Answer 4

hmmm yeah by parts would help too! even if you substitute x^2 as sin0 it would ultimately be solved by parts. so yeah do it

Answer 5

so,
\[u=arcsin(x^{2})\rightarrow du=\frac{1}{\sqrt{1-(x^{2})^{2}}}dx\]
\[dv=xdx\rightarrow v=\frac{x^{2}}{2}\]
\[\frac{x^{2}}{2}arcsin(x^{2})-\frac{1}{2}\int \frac{x^{2}}{\sqrt{1-x^{4}}}dx\]

Answer 6

If I try to integrate by parts again I come back to the same answer as before.

Answer 7

how are you integrating that by parts?

Answer 8

It looks like a trig integral to me

Answer 9

it is a trig integral. i dont understand how to do it. so i broke it into part. arcsin and x

Answer 10

what about u substitution
\[\frac{1}{2}\int arcsin(u)dx \]
still I don't know the integral of arcsin

Answer 11

sorry, didn't notice this tab, okay let's see
\(u=x^2\implies \large \int \dfrac{u}{\sqrt{1-u^2}}du\\
s=1-u^2\\ds=-2udu\implies -\dfrac{1}{2}ds=udu
\\
\large -\dfrac{1}{2}\int \dfrac{1}{\sqrt{s}}ds=-\dfrac{1}{2}2\sqrt s=-\sqrt s= -\sqrt{1-u^2}=-\sqrt{1-x^4}
\\\large \frac{x^{2}}{2}\arcsin(x^{2})-\frac{1}{2}\int \frac{x^{2}}{\sqrt{1-x^{4}}}dx=\frac{x^{2}}{2}\arcsin(x^{2})-\dfrac{1}{2}*-\sqrt{1-x^4}
\\
\huge \dfrac{1}{2}({x^{2}}\arcsin(x^{2})+\sqrt{1-x^4})
\)

Answer 12

\(+ \; C\)

lol!!!!

Answer 13

@bibby question \[u=x^{2}\implies \large \int \frac{u}{\sqrt{1-u^2}}du\]
why not \[u=x^{2}\implies \large \frac{1}{2}\int \frac{u}{\sqrt{1-u^2}}\frac{du}{\sqrt{u}}\]