0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Let $8x=\sin(u)$. $\endgroup$ – Donald Splutterwit Sep 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. $\endgroup$ – Jean-Claude Arbaut Sep 20 at 17:13
5
$\begingroup$

Use the substitution $t=8x,\quad\mathrm d t=8\,\mathrm dx$. The integral becomes $$\int \frac{3}{\sqrt{1-64x^2}}\,\mathrm dx =\int \frac{3}{\sqrt{1-t^2}}\,\frac{\mathrm dt}8=\frac38\int \frac{\mathrm dt}{\sqrt{1-t^2}}.$$

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Ponder

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

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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

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

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.