# Integral of arcsin (x^2)

1. Sep 10, 2012

### Kiwiro0ls

Just got out of a Calculus 2/ Lineal Algebra exam and im still wondering about the first question:
∫arcsin(2×2)dx

I used integration by parts letting u=arcsin(2×2) and so du=dx/sqrt[1-4x4], v=x

=xarcsin(2×2) - ∫[x/sqrt[1-4x4]dx

this second integral is what I can´t solve.

I tried doing trigonometric substitution where sin∅=2x2, and then later used a trig identity to integrate... but I really think I messed up somewhere.

If anyone out there has any clue, its much appreciated! :)

Last edited: Sep 10, 2012
2. Sep 10, 2012

### susskind_leon

If the second integral WERE ∫[x/sqrt[1-4x^4]dx, you could just substitute w=x^2 which would then give you arcsin again.
However, du=dx 4x/sqrt[1-4x4] (chain rule) so that the integral you actually have to solve becomes ∫[4x^2/sqrt[1-4x^4]dx which, according to wolfram alpha has a complicated solution involving elliptic integrals.
Maybe you are missing a factor in the question?

3. Sep 10, 2012

### Kiwiro0ls

No im very very very sure this is the question,
theres some talk amongst the class that it was typed wrong or something, apparently no one got those 15pts...
anyway, im still out of ideas and trying at it!

4. Sep 10, 2012

### SammyS

Staff Emeritus
Are you sure it wasn't something like $\displaystyle \int x\ \arcsin(2x^2)\,dx\ ?$

5. Sep 10, 2012

### Kiwiro0ls

Nope D: im 100% sure it was ∫arcsin (2x^2)dx

6. Sep 10, 2012

### Zondrina

Err I'm surprised a question like that would show up, but you can use the fact that :

$\int_{}^{}arcsin(2x^2)dx = \int_{}^{}\frac{\pi}{2} - arccosxdx = \frac{\pi}{2}x - xarccosx + \sqrt{1-x^2} + c$

7. Sep 10, 2012

### LCKurtz

Wouldn't it be $\int \arcsin(2x^2)\, dx = \int\frac \pi 2 -\arccos(2x^2)\, dx$?

8. Sep 11, 2012

### Kiwiro0ls

So
=xarcsin(2x2)-4∫[x2dx/sqrt(1-4x4)]
and thats what im trying to solve

9. Sep 11, 2012

### hogrampage

Looks like a valid problem for trigonometric substitution, particularly the 1/sqrt(...).

10. Sep 11, 2012

### LCKurtz

Beats me why this thread is still alive. As has already been pointed out, the solution involves elliptic functions and no "standard" calculus technique is going to work it. It is pretty obvious that if that was intended, the problem has a typo.

11. Sep 11, 2012

oh well :´(