Kkutie7
  • Kkutie7
I can't seem to do this integral: \[\int xarcsin(x^{2})dx\]
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
random231
  • random231
take x^2=sin(theta)
Kkutie7
  • Kkutie7
I don't see how that would help anything
Kkutie7
  • Kkutie7
I was thinking about integration by parts because I know the derivative for arcsin but i dont know

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

random231
  • random231
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
Kkutie7
  • Kkutie7
so, \[u=arcsin(x^{2})\rightarrow du=\frac{1}{\sqrt{1-(x^{2})^{2}}}dx\] \[dv=xdx\rightarrow v=\frac{x^{2}}{2}\] \[\frac{x^{2}}{2}arcsin(x^{2})-\frac{1}{2}\int \frac{x^{2}}{\sqrt{1-x^{4}}}dx\]
Kkutie7
  • Kkutie7
If I try to integrate by parts again I come back to the same answer as before.
bibby
  • bibby
how are you integrating that by parts?
bibby
  • bibby
It looks like a trig integral to me
Kkutie7
  • Kkutie7
it is a trig integral. i dont understand how to do it. so i broke it into part. arcsin and x
Kkutie7
  • Kkutie7
what about u substitution \[\frac{1}{2}\int arcsin(u)dx \] still I don't know the integral of arcsin
bibby
  • bibby
sorry, didn't notice this tab, okay let's see \(u=x^2\implies \large \int \dfrac{u}{\sqrt{1-u^2}}du\\ s=1-u^2\\ds=-2udu\implies -\dfrac{1}{2}ds=udu \\ \large -\dfrac{1}{2}\int \dfrac{1}{\sqrt{s}}ds=-\dfrac{1}{2}2\sqrt s=-\sqrt s= -\sqrt{1-u^2}=-\sqrt{1-x^4} \\\large \frac{x^{2}}{2}\arcsin(x^{2})-\frac{1}{2}\int \frac{x^{2}}{\sqrt{1-x^{4}}}dx=\frac{x^{2}}{2}\arcsin(x^{2})-\dfrac{1}{2}*-\sqrt{1-x^4} \\ \huge \dfrac{1}{2}({x^{2}}\arcsin(x^{2})+\sqrt{1-x^4}) \)
IrishBoy123
  • IrishBoy123
\(+ \; C\) lol!!!!
Kkutie7
  • Kkutie7
@bibby question \[u=x^{2}\implies \large \int \frac{u}{\sqrt{1-u^2}}du\] why not \[u=x^{2}\implies \large \frac{1}{2}\int \frac{u}{\sqrt{1-u^2}}\frac{du}{\sqrt{u}}\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.