Integral by substitution

**sinewave85** · August 26th 2009, 11:41 AM

I am stuck!

$\int\frac{dx}{\sqrt{1-2x^{2}}}$

$\mbox{Let }u=1-2x^{2}, du=-4xdx$

$\int\frac{dx}{\sqrt{1-2x^{2}}}=\frac{-1}{4}\int\frac{u^{-1/2}du}{x}$

$\int\frac{dx}{\sqrt{1-2x^{2}}}=\frac{-1}{4}\int\frac{du}{u^{1/2}(1/2-u/2)^{1/2}}$

$\int\frac{dx}{\sqrt{1-2x^{2}}}=\frac{-1}{4}\int\frac{du}{(u/2-u^{2}/2)^{1/2}}$

$\int\frac{dx}{\sqrt{1-2x^{2}}}=\frac{-\sqrt{2}}{4}\int\frac{du}{(u-u^{2})^{1/2}}$

I don't know where to go from there. Any help is greatly appreciated.

**Plato** · August 26th 2009, 11:56 AM

Go to this website Wolfram|Alpha
In the input window, type in this exact expression: integrate 1/sqrt[1-2x^2]dx (you can copy & paste)
Click the equals bar at the right-hand end of the input window.
Be sure you click ‘show steps’ to see the solution.

**sinewave85** · August 26th 2009, 12:06 PM

Originally Posted by Plato

Go to this website Wolfram|Alpha
In the input window, type in this exact expression: integrate 1/sqrt[1-2x^2]dx (you can copy & paste)
Click the equals bar at the right-hand end of the input window.
Be sure you click ‘show steps’ to see the solution.

Thanks very much for the link. The problem is that it shows the answer, but even the "show steps" window does not really show how to get there. Could you explain it a little bit more?

I realize that this is a basic integral:

$\int\frac{dx}{\sqrt{1-x^{2}}}=\sin^{-1}(x) + c$

but is that enough to get from this problem to the answer they give without any intermediate steps?

$\int\frac{dx}{\sqrt{1-2x^{2}}}=\frac{sin^{-1}(x\sqrt{2})}{\sqrt{2}}+c$

**Plato** · August 26th 2009, 12:15 PM

$u=\sqrt{2}x$

**sinewave85** · August 26th 2009, 12:22 PM

Originally Posted by Plato

$u=\sqrt{2}x$

I see (now that you have positioned me directly in front of the solution, that is)! Thanks very much.

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