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I'm given the following problem:

$\int \frac{3}{\sqrt{1-64x^2}}dx$

As it seems, the solution to this integral is:

$\frac{3}{8}\sin^{-1}{(8x)}$

How does this follow from the trig "rules" of integration? I guess I'm not seeing it. I'm in an accelerated Calc II class, and I want to be sure I really get this stuff down.

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    $\begingroup$ Let $8x=\sin(u)$. $\endgroup$ Sep 20 '20 at 17:08
  • $\begingroup$ It may help to know that the derivative of $\arcsin(x)$ is $\dfrac{1}{\sqrt{1-x^2}}$. If you don't know this yet, prove it using the chain rule. $\endgroup$ Sep 20 '20 at 17:13
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Use the substitution $t=8x,\quad\mathrm d t=8\,\mathrm dx$. The integral becomes $$\int \frac{3}{\sqrt{1-64x^2}}\,\mathrm dx =\int \frac{3}{\sqrt{1-t^2}}\,\frac{\mathrm dt}8=\frac38\int \frac{\mathrm dt}{\sqrt{1-t^2}}.$$

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Ponder

$$(\arcsin(x))'=\frac1{\sqrt{1-x^2}}.$$

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With $(\sin^{-1}u)’= \frac1{\sqrt{1-u^2}}$

$$\int \frac{3}{\sqrt{1-64x^2}}dx = \frac38\int \frac{d(8x)}{\sqrt{1-(8x)^2}}dx =\frac38 \sin^{-1}8x+C $$

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Alternatively, let $x=\frac{\sin(u)}{8}$ then $dx=\frac{1}{8}\cos(u)du$ and substituting gives $$\int \frac{3}{\sqrt{1-64x^2}}dx=\frac{1}{8}\int\frac{3\cos(u)}{\sqrt{1-64(\frac{\sin(u)}{8})^{2}}}du$$ $$=\frac{3}{8}\int \frac{\cos(u)}{\sqrt{1-\sin^2(u)}}du=\frac{3}{8}\int1du$$ $$=\frac{3}{8}u+C=\frac{3}{8}\sin^{-1}(8x)+C.$$

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Any integral of the form $\sqrt {a^2-x^2}$ follows the substitution $x=a\sin\theta$ so $\sqrt {a^2-x^2}$= $\sqrt {a^2-{a^2\sin^2\theta}}=a\sqrt {1-{\sin^2\theta}} = a\cos\theta$

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