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**Prove It** If you don't want to use a formula, then you can use trigonometric substitution.

Notice that

$\displaystyle \int{\frac{1}{\sqrt{1 - 9x^2}}\,dx} = \int{\frac{1}{\sqrt{9\left(\frac{1}{9} - x^2\right)}}\,dx}$

$\displaystyle = \int{\frac{1}{3\sqrt{\frac{1}{9} - x^2}}\,dx}$

$\displaystyle = \frac{1}{3}\int{\frac{1}{\sqrt{\frac{1}{9} - x^2}}\,dx}$.

Now make the substitution $\displaystyle x = \frac{1}{3}\sin{\theta}$ so that $\displaystyle dx = \frac{1}{3}\cos{\theta}\,d\theta$.

Also notice that $\displaystyle \sin{\theta} = 3x$ and so $\displaystyle \theta = \arcsin{3x}$.

Then the integral becomes

$\displaystyle = \frac{1}{3}\int{\frac{1}{\sqrt{\frac{1}{9} - \left(\frac{1}{3}\sin{\theta}\right)^2}}\,\frac{1} {3}\cos{\theta}\,d\theta}$

$\displaystyle = \frac{1}{9}\int{\frac{\cos{\theta}}{\sqrt{\frac{1} {9} - \frac{1}{9}\sin^2{\theta}}}\,d\theta}$

$\displaystyle = \frac{1}{9}\int{\frac{\cos{\theta}}{\sqrt{\frac{1} {9}(1 - \sin^2{\theta})}}\,d\theta}$

$\displaystyle = \frac{1}{9}\int{\frac{\cos{\theta}}{\frac{1}{3}\sq rt{\cos^2{\theta}}}\,d\theta}$

$\displaystyle = \frac{1}{3}\int{\frac{\cos{\theta}}{\cos{\theta}}\ ,d\theta}$

$\displaystyle = \frac{1}{3}\int{1\,d\theta}$

$\displaystyle = \frac{1}{3}\theta + C$

$\displaystyle = \frac{1}{3}\arcsin{3x} + C$.