anonymous
  • anonymous
antiderivative of x^2/ sqr(1+x^2)
Mathematics
schrodinger
  • schrodinger
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Mr.Math
  • Mr.Math
Is it \(\frac{x^2}{\sqrt{1+x^2}}?\)
anonymous
  • anonymous
\( \int \frac{x^2 dx}{\sqrt{1+x^2}} \)
anonymous
  • anonymous
yes

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Mr.Math
  • Mr.Math
You know that the most general anti-derivative of it is: \[\int\limits {x^2 \over \sqrt{1+x^2}}dx\] The best way to evaluate this integral is by using trig substitution, \(x=\tan(z)\).
anonymous
  • anonymous
oh, I'm so sorry, insted of 1+x^2 is 1-x^2
Mr.Math
  • Mr.Math
\(x=\tan(z) \implies dx=\sec^2(z)dz\), the integral becomes: \[\int\limits {\tan^2(z)\sec^2(z) dz \over \sec(z)}=\int\limits \tan^2(z)\sec(z)dz\]
Mr.Math
  • Mr.Math
Oh man!
anonymous
  • anonymous
if he at least gives me any idea to solve is already worth it :D
Mr.Math
  • Mr.Math
You will use the same method, but substitute \(x=\sin(z)\) instead!
anonymous
  • anonymous
\[\frac{x^2+ 1-1 }{\sqrt{1-x^2}}\] \[-\sqrt{1-x^2} + \frac{1}{\sqrt{1-x^2}}\] \[x = \sin\theta \] \[dx = \cos\theta d\theta\] \[-\cos^2\theta + 1\]
anonymous
  • anonymous
that's what i did but i'm kind of dyslexic so i must have mixed something :D i'll check again to see what's wrong

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