Definite integral of a trigonometric function.

Question

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?

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.