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I was trying to review calculus integration techniques before my differential equations class. I came across $\int \frac{1}{\sqrt{1-2x^2}}\,\mathrm{d}x$. I can't exactly figure out a good way to solve an integral of this form with ease. I was trying to get it in the form $\int \frac{1}{\sqrt{1-x^2}}\,\mathrm{d}x$ which is just $\sin^{-1}(x)$, but the 2 in-front of the $x^2$ is throwing me off.

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    $\begingroup$ Try $\sqrt{2}\,x=u$. $\endgroup$ – André Nicolas Aug 26 '15 at 2:05
  • $\begingroup$ Quick question, how did you know to set $u=\sqrt{2}x$ $\endgroup$ – JuliusDariusBelosarius Aug 26 '15 at 2:07
  • $\begingroup$ We want $2x^2=u^2$. $\endgroup$ – André Nicolas Aug 26 '15 at 2:07
  • $\begingroup$ Ok that makes sense, but lets say you had an integral like $\int \frac{1}{\sqrt{3-4x^2}}\,\mathrm{d}x$. Would you attempt to remove the 4 from the $x^2$ via division and remove the $\frac{3}{4}$ with reciprocal multiplication? $\endgroup$ – JuliusDariusBelosarius Aug 26 '15 at 2:28
  • $\begingroup$ Quick is $4x^2=3u^2$ or equivalently $2x=\sqrt{3}\,u$. $\endgroup$ – André Nicolas Aug 26 '15 at 2:34
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Notice, we have $$\int\frac{1}{\sqrt {1-2x^2}}dx=\int\frac{1}{\sqrt {1-(x\sqrt 2)^2}}dx$$ Let $x\sqrt 2 =t \implies \sqrt2 dx=dt$ or $dx=\frac{dt}{\sqrt 2}$ $$\int\frac{1}{\sqrt {1-t^2}}\frac{dt}{\sqrt 2}$$

$$=\frac{1}{\sqrt 2}\int\frac{1}{\sqrt {1-t^2}}dt$$ $$=\frac{1}{\sqrt 2}\sin^{-1}(t)+C$$ Setting $t=x\sqrt 2$ $$=\frac{1}{\sqrt 2}\sin^{-1}(x\sqrt 2)+C$$ Hence, we get $$\bbox[5pt, border:2.5pt solid #FF0000]{\color{blue}{\int\frac{1}{\sqrt {1-2x^2}}dx=\frac{1}{\sqrt 2}\sin^{-1}(x\sqrt 2)+C}}$$

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Because we need the integrand in the form $\int\! \frac{1}{\sqrt{1-u^2}} \ \mathrm{d}u$, we need $2x^2$ to be equal to $u$. So let us make the substitution $u = \sqrt2 \ x$. Now we have $\mathrm{d}u = \sqrt2 \ \mathrm{d}x$. So $\mathrm{d}x = \dfrac{1}{\sqrt{2}}\mathrm{d}u$. So our integral simplifies to \begin{align*} \int \! \frac{1}{\sqrt{1-2x^2}}\ \mathrm{d}x &= \int \! \frac{\frac{1}{\sqrt{2}}}{\sqrt{1-u^2}}\ \mathrm{d}u \\ &= \frac{1}{\sqrt2}\int\! \frac{1}{\sqrt{1-u^2}}\ \mathrm{d}u \\ &=\frac{1}{\sqrt{2}}\sin^{-1}(u)+C \\ &= \frac{1}{\sqrt{2}}\sin^{-1}(\sqrt{2}\ x)+C \end{align*}

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The same "trick", verily.

We have $$ \int_{x} \frac{1}{\sqrt{1 - 2x^{2}}} = \int_{u := \sqrt{2}x} \frac{1}{\sqrt{1-u^{2}}}\frac{1}{\sqrt{2}} \simeq \frac{1}{\sqrt{2}}\sin^{-1}u = \frac{1}{\sqrt{2}}\sin^{-1}(\sqrt{2}x), $$ where $\simeq$ denotes the equality "up to" a constant.

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$$\int\frac{1}{\sqrt{1-2x^2}} dx$$ Using trigonometric substitution, we have $$x=\frac1{\sqrt 2}\sin\phi \Rightarrow dx=\frac1{\sqrt 2}\cos\phi\ d\phi$$ So now $$\frac1{\sqrt 2}\int\frac{\cos\phi}{\sqrt{1-\sin^2\phi}} d\phi=\frac1{\sqrt 2}\int\frac{\cos\phi}{\sqrt{\cos^2\phi}} d\phi$$ $$=\frac1{\sqrt 2}\int\frac{\cos\phi}{\cos\phi} d\phi=\frac1{\sqrt 2}\int\ d\phi$$ $$=\frac{1}{\sqrt 2}\phi+C=\frac{1}{\sqrt 2}\arcsin\left(\sqrt 2x\right)+C$$

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