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I noticed that people solved the following integral using Lobachevsky trick. $$ \mbox{However, notice that in}\quad \int_{0}^{\infty}\frac{\sin^{2}\left(x\right)}{x^{2}} \frac{\ln\left(\cos^{2}\left(x\right)\right)}{\sin^{2}\left(x\right)}{\rm d}x $$ $$ \mbox{the function}\quad\frac{\ln\left(\cos^{2}\left(x\right)\right)}{\sin^{2}\left(x\right)} \quad\mbox{is not even continuous at}\quad \left(0,\infty\right) $$

So how is Lobachevsky trick a valid step for this integral ???.

Or is the continuous function condition actually unnecessary for Lobachevsky trick ???.

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    $\begingroup$ A paper on the L-trick the only required condition is Reimann integrability which is much less restraining than continuity $\endgroup$
    – RandomGuy
    Commented May 21 at 1:22

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This is not an answer to the question.

@RandomGuy provided the link to a paper which very clearly explains the conditions.

For the fun of using once more my favored $1,400^+$ years old approximations of the sine and cosine functions, $$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad \text{for} \qquad 0\leq x\leq\pi$$ $$\cos(x) \simeq\frac{\pi ^2-4x^2}{\pi ^2+x^2}\qquad \text{for} \qquad -\frac \pi 2 \leq x\leq\frac \pi 2$$ $$\int_{0}^{\pi}\frac{\log(\cos^2(x))}{x^2}\,dx\sim 2\int_0^{\frac \pi 2}\frac 1 {x^2}\log \left(\frac{\pi ^2-4 x^2}{\pi ^2+x^2}\right)\,dx+$$ $$2\int_{\frac \pi 2}^\pi \frac 1{\left(x+\frac{\pi }{2}\right)^2 } \log \left( \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x} \right)\,dx$$

The antiderivatives do not present any difficulty $$ 2\int_0^{\frac \pi 2}\frac 1 {x^2}\log \left(\frac{\pi ^2-4 x^2}{\pi ^2+x^2}\right)\,dx=-\frac 4 \pi \left(\log \left(\frac{16}{5}\right)+\cot ^{-1}(2)\right) =-2.07130$$ to be compared to the $-2.06589$ obtained by numerical integration.

$$2\int_{\frac \pi 2}^\pi \frac 1{\left(x+\frac{\pi }{2}\right)^2 } \log \left( \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x} \right)\,dx=-1 +\frac 1{3\pi}\left(\log \left(\frac{327680}{59049}\right)+6 \tan^{-1}(2)\right)$$ which is $-0.113341$ to be compared to the $-0.113556$ obtained by numerical integration.

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    $\begingroup$ you have immediately solved tremendous a mount of integrals. Could you please tell what kind of handy tool you used? $\endgroup$
    – MathArt
    Commented May 21 at 7:51
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    $\begingroup$ @MathArt. As many people, I use Mathematica plus a mountain of old software of mine accumulated over the last 65 years of professinal experience. Just to give you an idea, I used ISABELLE written in $1979$ at University of Bordeaux (my alma mater) which became PARI /GP in $1985$. $\endgroup$ Commented May 21 at 8:35

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