Can someone help me with this following integral? $$\left[2 ∫_0^{\frac 1{\sqrt 2}} \dfrac 1x~\arcsin(x)~~dx\right]-  \left[∫_0^1 \dfrac 1x~\arctan(x)~~dx\right]$$

I have tried putting $x=\sin\theta$ but in vain. In the tan inverse integral, how do I proceed? I know this involves integration by parts. When I tried to solve the sine inverse part independently, I got the integration of, $\theta\cot\theta$ (where $x=\sin\theta$) from $0$ to $\pi/4$. Also, if I try to solve them independently , I will have to take $x=\tan\varphi$ in the second part, and that would make the problem difficult. My sixth sense tells me that these are not meant to be solved separately i.e., each term independently, rather they are to be solved together. There is a link, mostly some terms cancel out on integrating by parts, but I haven't seen anything as such so far. Can somebody help?

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    $\begingroup$ I updated the math formatting in your question, please review to see some basic Latex formatting. Also, the link doesn't work - requires a password. In general, please type the equation in Latex rather than linking to an image. $\endgroup$
    – pshmath0
    Commented Mar 16, 2020 at 6:48
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    $\begingroup$ Please, use MathJax. Currently, in order to see the formula, one has to go to a link that doesn't even provide direct access to the file. So please type out the expressions with MathJax - it won't take longer than a few seconds. $\endgroup$
    – Matti P.
    Commented Mar 16, 2020 at 6:55
  • $\begingroup$ Although I couldn't do the MathJax, since I don't know how to use it. Maybe we need to write a code? Don't know. Need to learn more about it. For the time being, I have added the integral in the text. $\endgroup$ Commented Mar 16, 2020 at 8:34
  • $\begingroup$ @user. Yes, it must be there for the simplification. $\endgroup$ Commented Mar 16, 2020 at 9:03

1 Answer 1


Substituting $x=\sin t$ and $x=\tan t$ into the first and the second integrals, respectively, one obtains:

$$ I=2\int_0^{\pi/4}\frac{t}{\sin t}\cos t\,dt -\int_0^{\pi/4}\frac{t}{\tan t}\frac{dt}{\cos^2 t}\\ =\int_0^{\pi/4}2t\cot 2t\, dt=\frac12\int_0^{\pi/2} u\cot u\, du. $$ The last integral can be transformed by parts into: $$ \int_0^{\pi/2} u\cot u\, du=\int_0^{\pi/2} u d(\log (\sin u))=-\int_0^{\pi/2} \log(\sin u)\, du =\frac{\pi}2\log2. $$

Several proofs of the last equality can be found here.

  • $\begingroup$ Nicely done and +1 for sure. $\endgroup$ Commented Mar 16, 2020 at 10:36
  • $\begingroup$ @user, please clarify one thing, how can x be sin t and tan t at the same time? $\endgroup$ Commented Mar 17, 2020 at 15:58
  • $\begingroup$ @PrarabdhShukla We have two integrals. The integration variable in one of them has no relation to the other one. They have initially also different integration domains. $\endgroup$
    – user
    Commented Mar 17, 2020 at 17:54
  • $\begingroup$ @user, since we have two integrals, how can we add/subtract i.e. , combine them?(which you did to obtain 2tcot(2t).) To add/subtract them would mean that both the Ts are the same, won't it? $\endgroup$ Commented Mar 19, 2020 at 2:41
  • $\begingroup$ @PrarabdhShukla Here the very well-known linearity property of integral was used. $\endgroup$
    – user
    Commented Mar 19, 2020 at 6:57

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