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Within this collocation of definite integrals number $30.$ is given by

$$\int_0^{\frac{\pi}2}4\cos^2(x)\log^2(\cos x)\mathrm dx~=~-\pi\log 2+\pi\log^2 2-\frac{\pi}2+\frac{\pi^3}{12}\tag1$$

I figured out a way to evaluate the integral using a substitution and further the values of the derivatives of the Beta Function. My approach can be found on the bottom of the post. Another solution can be found here within the original post of the $(1)$.

I was just curious whether there is an easier attempt. First of all I thought about applying a Weierstrass Substitution to get rid of the trigonometry and instead solving an algebraic integral. Anyway it did not worked out quite well since the substitution made the new integrals even more complicated. Hence the logarithm is squared I guess series expansions $-$ although the linked answer invoked the Fourier Series Expansion of $\log(\cos x)$ $-$ or IBP are not the right way to approach to this integral either.

Therefore I am asking for interesting or elegant ways to evaluate $(1)$. Please provide a different attempt than the two I suggested within this post if you are aware of one.

Thanks in advance!


Own evaluation

First of all rewrite the integral sligthly $$\small\int_0^{\frac{\pi}2}4\cos^2(x)\log^2(\cos x)\mathrm dx=\int_0^{\frac{\pi}2}\cos^2(x)\log^2(\cos^2 x)\mathrm dx$$ To notice that the subsititution $\cos^2 x=y$ is suitable in this case. Computing $\mathrm dy$ and changing the borders of integration yields to $$\begin{align} \small\int_0^{\frac{\pi}2}\cos^2(x)\log^2(\cos^2 x)\mathrm dx&=\small\int_1^0 (y)\log^2(y)~\left(\frac12\frac{-\mathrm dy}{\sqrt{y(1-y)}}\right)\\ &=\small\int_0^1\frac{\sqrt{y}\log^2(y)}{\sqrt{1-y}}\mathrm dy \end{align}$$ One may note the familiar structure similiar to one given by the second derivative of the Beta Function. Therefore this integral can be written as $$\small\int_0^1\frac{\sqrt{y}\log^2(y)}{\sqrt{1-y}}\mathrm dy=\left.B_{xx}\left(x,\frac32\right)\right|_{x=\frac12}$$ The second derivative of the Beta Function is given in terms of the Beta Function connected with the Digamma and Trigamma Function as $$\small B_{xx}(x,y)=B(x,y)[(\psi^{(0)}(x)+\psi^{(0)}(x+y))^2+\psi^{(1)}(x)-\psi^{(1)}(x+y)]$$ Plugging in the values $x=\frac12$ and $y=\frac32$ yields to $$\begin{align} \small\left.B_{xx}\left(x,\frac32\right)\right|_{x=\frac12}&=\small B\left(\frac12,\frac32\right)\left[\left(\psi^{(0)}\left(\frac12\right)+\psi^{(0)}\left(\frac32+\frac12\right)\right)^2+\psi^{(1)}\left(\frac12\right)+\psi^{(1)}\left(\frac12+\frac32\right)\right]\\ &=\small\frac14\pi\left[\left(\left(\psi^{(0)}\left(\frac12\right)+\psi^{(0)}\left(2\right)\right)^2\right)+\psi^{(1)}\left(\frac12\right)+\psi^{(1)}\left(2\right)\right]\\ &=\small\frac14\pi\left[(-\gamma-2\log 2-1+\gamma)^2+\frac{\pi^2}2-\frac{\pi^2}6+1\right]\\ &=\small\frac14\pi\left[4\log^2 2+4\log 2+\frac{\pi^2}3+2\right]\\ &=\small -\pi\log 2+\pi\log^2 2-\frac{\pi}2+\frac{\pi^3}{12} \end{align}$$ Form hereon the desired equality follows $$\small\therefore~\int_0^{\frac{\pi}2}4\cos^2(x)\log^2(\cos x)\mathrm dx~=~-\pi\log 2+\pi\log^2 2-\frac{\pi}2+\frac{\pi^3}{12}$$

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    $\begingroup$ Why not solve it directly: $$I(n) =\int_0^\frac{\pi}{2} \cos^n x dx$$ Use beta function, $I(n) =\frac12 B\left(\frac{n+1}{2}, \frac12\right) $derivate with respect to $n$ how many log powers you need then set $n$ at your wish. $\endgroup$ – Zacky Oct 19 '18 at 14:29
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    $\begingroup$ @Dahaka Well, this is a little bit easier than my own way. I did not thought about this simple way. Thank you for the hint! $\endgroup$ – mrtaurho Oct 19 '18 at 14:34
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Let’s consider the general case.$$\mathfrak{I}=\int\limits_0^{\pi/2}\mathrm dx\,\cos^mx\sin^nx$$Make the substitution $x\mapsto\sin x$ to get that$$\mathfrak{I}=\int\limits_0^1\mathrm dx\, \left(1-x^2\right)^{(m+1)/2}x^n$$And now make the transformation $x\mapsto x^2$ so that$$\mathfrak{I}=\frac 12\int\limits_0^1\mathrm dx\,\left(1-x\right)^{(m-1)/2}x^{(n-1)/2}=\frac 12\operatorname{B}\left(\frac {m+1}2,\frac {n+1}2\right)$$Now substitute $n=0$ and differentiate with respect to $m$.

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  • $\begingroup$ Basically it is the same approach as mine but generalized to a class of integrals. Furthermore Dahaka mentioned a similiar attempt within the comment section. Anyway thank you for your response! To be honest I totally forgot to include the Beta Function while solving the integral which is in this case I would say slightly obvious. My bad. $\endgroup$ – mrtaurho Oct 19 '18 at 16:51
  • $\begingroup$ @mrtaurho Nothing wrong with generalizing it even further, right? I just provided a way to directly get towards the answer. But yes your way is basically the same. $\endgroup$ – Frank W. Oct 20 '18 at 0:58
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You can first substitute $x=\arctan u$: \begin{align} 4\int_0^{\pi/2} \cos^2 (x) \log^2 \cos x\,dx&=\int_0^\infty \frac{\log^2(1+u^2)}{(1+u^2)^2} \,du \\ &=\frac{\partial^2}{\partial^2 \beta}\Biggr|_{\beta=0} \int_0^\infty \frac{1}{(1+u^2)^{2-\beta}}\,du \\ &=\frac{\partial^2}{\partial^2 \beta}\Biggr|_{\beta=0} \frac{1}{\Gamma(2-\beta)}\int_0^\infty \int_0^\infty \nu^{1-\beta} e^{-\nu(1+u^2)}\,d\nu\, du \\ &=\frac{\partial^2}{\partial^2 \beta}\Biggr|_{\beta=0} \frac{1}{\Gamma(2-\beta)}\int_0^\infty e^{-\nu}\nu^{1-\beta} \int_0^\infty e^{-\nu u^2}\,du\, d\nu \\ &= \frac{\partial^2}{\partial^2 \beta}\Biggr|_{\beta=0} \frac{\sqrt{\pi}}{\Gamma(2-\beta)}\int_0^\infty e^{-\nu}\nu^{1/2 -\beta} d\nu \\ &=\frac{\partial^2}{\partial^2 \beta}\Biggr|_{\beta=0} \frac{\sqrt{\pi}\,\Gamma\left(3/2-\beta\right)}{\Gamma(2-\beta)} \end{align}

After doing the computations, we can find that this is equal to

\begin{align} \frac{\partial^2}{\partial^2 \beta}\Biggr|_{\beta=0} \frac{\sqrt{\pi}\,\Gamma\left(3/2-\beta\right)}{\Gamma(2-\beta)} &= \sqrt{\pi}\frac{\Gamma(3/2)}{\Gamma(2)}\left(\psi^{(0)}(3/2)^2-2\psi^{(0)}(2)\psi^{(0)}(3/2)+\psi^{(1)}(3/2)+\psi^{(0)}(2)^2-\psi^{(1)}(2)\right) \\ &=\frac{\pi}{4}\left(\frac{\pi^2}{3}+4\log^2-4\log 2 -2\right) \\ &=\frac{\pi^3}{12}+\pi\log^2 2 -\pi\log 2 -\frac{\pi}{2} \end{align}

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