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Hello we all know that $$\sqrt {a^2}=|a|$$ so when we have $$a^2=5$$ that is $$|a|=\sqrt5$$ and $$a=\pm\sqrt5$$ but i very often see that when solving integrals only the positive value is usually considered, for eg.

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Here in this example since $$a^2=5$$ that means$$a=\pm\sqrt5$$ but the book considers only the positive value why?

Arnaldo
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Hfdssdjns
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  • No there is no $$a$$ inside the square root there is only $$a^2$$ which is always positive – Hfdssdjns Feb 23 '23 at 12:41
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    I am sorry, I deleted my comment when I saw my mistake. – student91 Feb 23 '23 at 12:42
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    In this particular case, it is because $$\frac{\mathrm{d}}{\mathrm{d}u}\sin^{-1}(u/a)=\frac1{a\sqrt{1-(\frac ua)^2}}=\frac1{\frac{a}{|a|}\sqrt{a^2-u^2}}$$ is only equal to $\frac1{\sqrt{a^2-u^2}}$ when $a> 0$ – student91 Feb 23 '23 at 12:46
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    They want a number that squares to $5$ in order to make the substitution to a term of the form $a^2 ,(=5)$. It is not obligatory that they choose $a=+\sqrt{5}$, but neither is there any reason to choose $-\sqrt{5}$, so they use positive as it is the simpler option. Often in math, we opt for things because they are simpler and more natural. – Jam Feb 23 '23 at 12:46
  • Jarm so you are saying that both choices are correct? Even though the resulting integral will be different by a multiple of -1 – Hfdssdjns Feb 23 '23 at 12:48
  • @Hfdssdjns Check again whether the solved integral would be any different with $-a$ instead of $a$. – Jam Feb 23 '23 at 12:51
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    Shouldn't the section "Forms involving $\sqrt{a^2-u^2}$" insist that $a>0$? Otherwise the formula will be incorrect. Basically, you probably need to take $a>0$ because that's the region where the formula given in the book works. If you take $a<0$ then it fails. All you need is that $a^2=5$, but if you choose $a = -\sqrt{5}$ I'm sure you can't use the formula in that section so you need to choose $a = \sqrt{5}$. – Sarvesh Ravichandran Iyer Feb 23 '23 at 12:51
  • @SarveshRavichandranIyer The integral is even. It is well defined for both negative and positive $a$ (provided they are sufficiently small). – Jam Feb 23 '23 at 12:53
  • Actually, excuse my earlier comments. We do need $|a|$ rather than $a$ and the solved form is incorrect when $a$ is negative. – Jam Feb 23 '23 at 12:57
  • @Jam Indeed, if we directly replaced $a$ by $-a$ then the argument of $\sin^{-1}$ flips sign, and hence that term flips sign since $\sin(-x) = -\sin(x)$ implies that $\sin^{-1}(y) = -\sin^{-1}(-y)$. – Sarvesh Ravichandran Iyer Feb 23 '23 at 13:28

2 Answers2

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The "correct" formula is

$$\int\frac{u^2}{\sqrt{a^2-u^2}}\,du=-\frac u2\sqrt{a^2-u^2}+\frac{a^2}2\sin^{-1}\left(\frac u{\color{red}|a\color{red}|}\right)+C$$

and it does not matter which sign you give $a$.

For this reason the tables of integrals usually assume $a>0$, implicitly or explicitly, as discussing the negatives is of no practical interest.

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If we use the pre-packaged solution with $a^2=\frac54$: $$=-\frac x4\sqrt{5/4-x^2}+\frac5{16}\sin^{-1}\frac{2x}{\sqrt5}+K$$ and then use $a=-\frac{\sqrt5}2$, the arcsine term would be flipped in sign, but then differentiating $$-\frac x4\sqrt{5/4-x^2}-\frac5{16}\sin^{-1}\frac{2x}{\sqrt5}+K$$ does not give the original integrand but $-\frac14\sqrt{5 - 4 x^2}$ instead.

To be fully correct the integral should have an absolute value sign where the single $a$ appears. There is a reason only the plus sign is accounted for though – the principal square root of a positive number remains positive, so it is the more "natural" choice of sign.

Parcly Taxel
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    This does not tell why the case of $-\sqrt5$ can be ignored. (On the opposite !) –  Feb 23 '23 at 12:58
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    Just the opposite. Plugging $-\sqrt 5$ in the formula seems to give another antiderivative. Why should it be ignored ? –  Feb 23 '23 at 13:00
  • Because that's precisely the point of the question. Curse the question, not me. –  Feb 23 '23 at 13:01
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    As the integrand is sign agnostic, the book is wrong in its expression of the antiderivative. I don't consider this a valid motivation for ignoring the negative case. –  Feb 23 '23 at 13:07
  • @YvesDaoust I edited my answer. Does this please you now? – Parcly Taxel Feb 23 '23 at 13:49
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    The question is not about correctness of the formula but about why only the plus sign is considered. Of course, explaining the reason with a partially wrong formula is more challenging. –  Feb 23 '23 at 14:09