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I´m trying to enhance my integral skills, as I´m planning to tutor some students next semester, but I´m stuck with this integrals. I`m searching for a method or an idea that would work for both of them. I tried integrating by parts but that failed, the usual substitutions to get rid of the square roots did not look very promising either. So can someone please help me to solve the following 2 integrals, and not just solve, but providing a method that works for both of them and, if possible, explaining why they are equal and what was the idea behind it?

$$\int_0^{\frac{\sqrt2}{2}} \frac{\log(1-x^2)}{2(1-x)^{\frac{3}{2}}}dx$$ and $$\int_1^{\frac{\sqrt2}{2}} \frac{\log (\frac{1}{x})}{x^2\sqrt{1-x^2}}dx$$

Thanks!

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  • $\begingroup$ $x = 1/t$ works. $\endgroup$ – user8277998 Nov 12 '17 at 20:02
  • $\begingroup$ $(1-x^2) = (1+x)(1-x).$ $\log(1-x^2) = \log(1+x) + \log(1-x).$ $\log(\frac 1x) = -\log(x)$ $\endgroup$ – Namaste Nov 12 '17 at 20:04
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    $\begingroup$ Could the upvoter explain explicitly as to why this post received an upvote? (Unless, of course, it was expressing pity on someone getting a downvote :-( (boo-hoo.) It does not show research or effort, nor is it terribly useful. Hence, my conclusion. $\endgroup$ – Namaste Nov 12 '17 at 20:11
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    $\begingroup$ Verbe Not adding any context whatsoever (not explaining where you got stuck in using each method on each question, which is best shown by including your work, or explaining the source and motivation for the question. So yes, describing your approaches, where you get stuck, etc. is very useful, and not doing so, makes your question far less worthwhile to remain on the site. $\endgroup$ – Namaste Nov 12 '17 at 20:42
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    $\begingroup$ What is virtually worthless is merely transcribing a problem statement pretending to be a question, and claiming "I tried". Such questions can justifiably be closed. $\endgroup$ – Namaste Nov 12 '17 at 20:44
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In the second integral, take $x=\frac{1}{z}$. Then you have

$$\int_{\sqrt{2}}^{1}\frac{zlogz}{\sqrt{z^{2}-1}}dz$$.

If we wrote $z=sec\ t$, then we get $$\int_{\frac{\pi}{4}}^{0}(sec^{2}t)\ (log(sec \ t))dt$$. Now apply by parts integration, and you would get $$(log(sec t))(tan\ t)\big|_{\frac{\pi}{4}}^{0}-\int_{\frac{\pi}{4}}^{0}tan^{2}t\ dt$$.

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First of all, many thanks for the answers, especially the answer of Aritro Pathak gave me some ideas to solve it by myself.

For the first one, using the substitution $x=sin(t)$ gave me $$\int_0^{\frac{\pi}{4}} \frac{\log(cos(t)^2)}{cos(t)^2}dt$$ which is solvable by using integration by parts and gave me $\frac{1}{4}(4-\pi-\log(4)) $

The second one can be solved in the same spirit, but starting with $x=cos(t)$ and using the hint of amWhy, $\log(\frac{1}{x})=-\log(x)$.

Thanks to all, it`s always more fun if you can inspire me to solve it by myself!

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For the first one, integrate by parts to get rid of the logarithm and then, in the remaining integral, set $t=\sqrt{1-x}$.

For the second one, it might be a good idea to integrate by parts, first noting that $$ \int\frac{1}{x^2\sqrt{1-x^2}}\,dx = \int\frac{1}{x^3\sqrt{1/x^2-1}}\,dx =-\sqrt{1/x^2-1} =-\frac{\sqrt{1-x^2}}{x}. $$ Then, another integration by parts will be useful.

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