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I am having a problem solving an integral. I am stuck in an infinite loop. Integral is:

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

I have separated it in dv and u on this way:

$u = \frac{dx}{\sqrt{1-x^2}}$ $dv = \frac{1}{\arcsin{x}}$

And the using:

$u v - \int{v \, du}$

I get again:

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

I dont know, but probably, I am doing something wrong. I am new at solving Integrals so I am learning :) According to my book the result should be:

$\ln({\arcsin{x}})-C$

And it will be true if I didn't had $\sqrt{1-x^2}$ but on this way I have no idea.

  • 1
    You set $dv = 1/arcsin(x)$? How did you even integrate that?2012-06-24

4 Answers 4

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Do not use Integration by Parts. Use $u$-substitution. Let $u=\sin^{-1}(x)$. Then $du=dx/\sqrt{1-x^2}$

So now your integral is $\int \frac{du}{u}$

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Note: Your comment at the end is incorrect. It is not true that $\int\frac{dx}{\arcsin x} = \ln(\arcsin x)+ C\tag{Wrong!}$

This is, unfortunately, a common mistake. Don't fall for it again!

While $\int\frac{dx}{x} = \ln|x|+C$ is true, in general, $\int\frac{dx}{f(x)}$ is not equal to $\ln|f(x)|+C$. If you differentiate $\ln|f(x)|$, you'll notice that you get $\frac{f'(x)}{f(x)}$. It is precisely because you have a $\frac{1}{\sqrt{1-x^2}}$ that you do get the natural logarithm of an arcsine.

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    Thank you for the advice :)2012-06-24
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"Another" method, which in fact is like substitution but, with some practice and care, perhaps is a little faster. Since $\,\displaystyle{(\arcsin x)'=\frac{1}{\sqrt{1-x^2}}}\,$ , and assuming we know $\,\displaystyle{\int\frac{f'}{f}\,dx=\log|f|+C}\,$, we can write

$\int\frac{dx}{\sqrt{1-x^2}\arcsin x}\,dx=\int \frac{1}{\arcsin x}\,\frac{1}{\sqrt{1-x^2}}\,dx=$$=\int \frac{1}{\arcsin x}\,d(\arcsin x)=\log|\arcsin x|+C$

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$\arcsin x = u$ then $\frac{1}{\sqrt{1-x^2}}dx = du $. In the integral it will then be: $ \int u^{-1}du $ which is $\ln \vert u \vert +c$ or simply $\ln \vert \arcsin x\vert+c$.