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The integral is

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

I tried solving it by parts, but that didn't work out. I couldn't integrate the result of substituting $t=1-x^2$ either.

The answer is

$$\ln\left|\dfrac{1-\sqrt{1-x^2}}{x}\right|$$

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    $\begingroup$ Try $x=\operatorname{sech} t$ $\endgroup$ Jun 15, 2021 at 10:07
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    $\begingroup$ @NinadMunshi I don't want to do it by trignometric substitution. The final answer contains a trigonometric term as a result, but in the answer given in the original question, there's only a log term. $\endgroup$
    – csmathhc
    Jun 15, 2021 at 10:09
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    $\begingroup$ If you look closely, I have not suggested a trig substitution. You will get your log from $\operatorname{sech}t = \frac{2}{e^t+e^{-t}}$ $\endgroup$ Jun 15, 2021 at 10:11
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    $\begingroup$ Why do you not want to use any trigonometric substitution? $\endgroup$ Jun 15, 2021 at 10:27
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    $\begingroup$ @Joe most likely OP does not know what they want, but it's good practice understand the relationship between inverse hyperbolic and log, which is very different from inverse trig not being a real log $\endgroup$ Jun 15, 2021 at 11:01

3 Answers 3

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Let $\sqrt{-x^{2}+1}=xt+1$ to transform the integral into an easier rational function. Rearrange for $x$ and we have $$x = \frac{2t}{-1-t^{2}}$$

Find the derivative $\frac{dx}{dt}$, then substitute $x$ and $dx$ into the integral and work towards your answer.

Solution:

$$\frac{dx}{dt}=\frac{2t^{2}-2}{\left(-1-t^{2}\right)^{2}}$$

Our integral becomes thus $$\int\frac{\frac{2t^{2}-2}{\left(-1-t^{2}\right)^{2}}}{\frac{2t}{-1-t^{2}}\left(\frac{2t}{-1-t^{2}}+1\right)}dt$$ which simplifies nicely to $$\int\frac{1}{t}dt\ =\ \ln\left|t\right|\ + C$$ where $t=\frac{\sqrt{1-x^{2}}-1}{x}$

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  • $\begingroup$ To anyone who sees this answer: how do you make the integral sign only bigger? I am "new" to TeX and I don't know how to change it's size $\endgroup$
    – user71207
    Jun 15, 2021 at 10:40
  • $\begingroup$ I tried something I thought would work and then reverted it. $\endgroup$
    – Ian
    Jun 15, 2021 at 10:46
  • $\begingroup$ @user71207 There is a network website for such questions, though the solutions there may not be supported by the TeX interface here. Example. ;) $\endgroup$
    – Pedro
    Jun 15, 2021 at 10:52
  • $\begingroup$ @user71207, Use four '$' instead of two '$' (dolar sign), $\endgroup$
    – sirous
    Jun 15, 2021 at 18:31
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With the substitution $t=\sqrt{1-x^2}$, the integral can be integrated as follows $$\int\frac{1}{x\sqrt{1-x^2}}dx=-\int \frac1{1-t^2}dt= \frac12 \ln\frac{1-t}{1+t}= \frac12\ln \frac{1-\sqrt{1-x^2}}{1+\sqrt{1-x^2}}+C $$ which is the same as $\ln\frac{1-\sqrt{1-x^2}}{x}$ after rationalizing the denominator.

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  • $\begingroup$ Perfect! Just the kind of answer I was looking for. $\endgroup$
    – csmathhc
    Jun 15, 2021 at 14:37
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Edit: I just realised that you wanted to evaluate this integral without trigonometric substitution. I'm sorry if this answer is not of any use to you. I will keep it up for the benefit of other readers.


Because of the identity $\sin^2\theta+\cos^2\theta=1$, a good candidate for a substitution is $x=\sin\theta$ (with $-\pi/2\le\theta\le\pi/2$). If we rearrange $\cos^2\theta+\sin^2\theta=1$, we get $$ \sqrt{\cos^2\theta} = |\cos\theta|=\sqrt{1-\sin^2\theta} $$ but since $\cos\theta$ is nonnegative for $\theta\in[-\pi/2,\pi/2]$, this simplifies to $$ \cos\theta=\sqrt{1-\sin^2\theta} $$ Let us try using this substitution for the integral at hand. If $x=\sin\theta$, then $dx=\cos\theta \, d\theta$, and so \begin{align} \int \frac{dx}{x\sqrt{1-x^2}} &= \int \frac{\cos\theta \, d\theta}{\sin\theta\cos\theta} \\[5pt] &= \int \csc\theta \\[5pt] &= -\log|\csc\theta+\cot\theta| + C \label{*}\tag{*} \\[5pt] &= -\log\left|\frac{1+\cos\theta}{\sin\theta}\right| + C \\[5pt] &= -\log\left|\frac{1+\sqrt{1-x^2}}{x}\right| + C \end{align} Using the identity $-\log|y|=\log|y^{-1}|$, this becomes $$ \int \frac{dx}{x\sqrt{1-x^2}} = \log\left|\frac{1-\sqrt{1-x^2}}{x}\right| + C $$


If you are not familiar with how to integrate $\csc\theta$ as in $\eqref{*}$, then note that $$ \int \csc\theta \, d\theta = \int\frac{\sin\theta}{1-\cos^2\theta} \, d\theta $$ Try making the substitution $u=\cos\theta$ and be prepared to simplify your answer a lot!

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