Integration using trig substitution or substitution 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.
 A: 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
$$\bbox[5pt, border:2.5pt solid #FF0000]{\color{blue}{\int\frac{1}{\sqrt {1-2x^2}}dx=\frac{1}{\sqrt 2}\sin^{-1}(x\sqrt 2)+C}}$$
A: 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*}
A: 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.
A: $$\int\frac{1}{\sqrt{1-2x^2}} dx$$
Using trigonometric substitution, we have
$$x=\frac1{\sqrt 2}\sin\phi \Rightarrow dx=\frac1{\sqrt 2}\cos\phi\ d\phi$$
So now
$$\frac1{\sqrt 2}\int\frac{\cos\phi}{\sqrt{1-\sin^2\phi}} d\phi=\frac1{\sqrt 2}\int\frac{\cos\phi}{\sqrt{\cos^2\phi}} d\phi$$
$$=\frac1{\sqrt 2}\int\frac{\cos\phi}{\cos\phi} d\phi=\frac1{\sqrt 2}\int\ d\phi$$
$$=\frac{1}{\sqrt 2}\phi+C=\frac{1}{\sqrt 2}\arcsin\left(\sqrt 2x\right)+C$$
