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I am trying to solve this derivation:

$$\int \frac{1}{\sqrt{(1 - x^2)}\arcsin^3x} dx $$

using substitution. My problem is that I can not find the right way to use substitution in this case.

The result should be:

$$ C- \frac{1}{2(\arcsin^2x)}$$

note: I guess that I should use this formula:

$$ \int \frac{1}{\sqrt{1-x^2}}dx = \arcsin x + C $$

Thank you.

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    Your formula for the antiderivative of $\frac{1}{1-x^2}$ is not correct. If you fix that (double check the derivative of arcsin x), you should see easily how to do the problem. – 2017-01-09
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    Ohh, I know... I have just wrote it wrong – 2017-01-09

1 Answers 1

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Notice that the derivative of $\arcsin(x)$ appears in this integrand. In general, this tells us that $u$-substitution would be a good thing to try; in our case, if we let $u = \arcsin(x)$, we get $\displaystyle du = \frac{dx}{\sqrt{1 - x^2}}$. Thus our integral becomes:

$$\int \frac{1}{\sqrt{1 - x^2} \ \arcsin^3(x)} \ dx \ =\ \int \frac{1}{u^3} \ du$$

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    So easy... :) Thanks. – 2017-01-09
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    Yep! Just takes practice doing a ton of these sorts of problems – 2017-01-09
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    I know! Sometimes I just do not see it (need more practice). :) – 2017-01-09