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Just playing with graph plotter I found some formulas for definite integrals of logarithmic functions such as $$ \int \limits_0^{\pi/2} \ln\left(a^2\cos^2 \theta + b^2\sin^2 \theta\right)\, \mathrm{d}{\theta} = \pi \ln \left(\frac{a+b}{2}\right), $$ $$ \int \limits_0^{\pi/2} \ln\left(4a^4\sin^4 \theta + b^4\cos^4\theta\right)\, \mathrm{d}{\theta}= \pi\ln\left(\frac{2a^2+b^2+2ab}{4}\right) $$ or even $$\int \limits_{0}^{1} \ln\left(a^2t^2 + b^2 (1-t)^2 \right) \, \mathrm{d}{t} = \frac{2a^2\ln a + \pi ab + 2b^2 \ln b}{a^2+b^2} - 2, $$ I hope you got the idea.

The formulas look relatively simple, like it is not impossible to prove, but I have no idea what method can be used. Not being a good mathematician, but rather the one who solves integrals by guessing in graph plotter, I asked a friend and he came up with an elegant way to prove the first one (partial derivatives along $a+b=\mathrm{const}$ are zero), but it does not seem to work well for the others as the contour lines have more complicated shapes.

Do you have any helpful thought about that? At least, where might some similar integrals have appeared before, what book/paper to search for?

Thank you.

Vlad the magnificent
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You can use differentiation over a parameter to get a simpler integral, for example

\begin{align} \frac{\partial}{\partial a}\int_0^{\pi/2} \ln (a^2\cos^2\theta +b^2\sin^2\theta)d\theta = \int_0^{\pi/2} \frac{2a\cos^2\theta}{a^2\cos^2\theta +b^2\sin^2\theta}d\theta \end{align}

Yu can calculate such integral with subsitution $t=\tan \theta$ or using contour integration. Then you integrate the result back (for the first integral you hjave an easy initial condition $I(a=b) = 0$): $$ \int_0^{\pi/2} \ln (a^2\cos^2\theta +b^2\sin^2\theta)d\theta = \int_b^a \Big(\int_0^{\pi/2} \frac{2a'\cos^2\theta}{a'^2\cos^2\theta +b^2\sin^2\theta}d\theta \Big) da' $$

The other integrals are more complicated, but I suspect similar method will work.

Adam Latosiński
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    Yeah the same method works for the at least the last one as well. Differentiate wrt $a$, complete the square in the denominator, make a substitution of the form $z = \alpha t - \beta$, where $\alpha$ and $\beta$ result from completing the square and then you will be left with some standard integrals that result in $\arctan(x)$ and $\log(x^2 + k^2)$. I admit it gets a little tedious, but I believe that's there is no neater way out of it. – sudeep5221 May 16 '19 at 01:38