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I saw this problem $$\int_0^{\frac{\pi}{2}} \frac{\ln(\cos(x))}{1+\sin^2(x)} \, dx$$ on my problem book but I have no idea how to evaluate it. I reduced the integral like this

by king's rule $$I_1= \int_0^{\frac{\pi}{2}} \frac{\ln(\cos(x))}{1+\sin^2(x)}dx =\int_0^{\frac{\pi}{2}} \frac{\ln(\sin(x))}{1+\cos^2(x)}dx $$ by using the double angle rule

$$I_1=\int_0^{\frac{\pi}{2}} \frac{\ln(2\sin(x/2) \cos(x/2))}{2\cos(x/2)^2}dx$$ we let $x/2=x$ then $dx=2dx$ $$I_1=\int_0^{\frac{\pi}{4}} \frac{\ln(2)}{\cos(x)^2}dx+\int_0^{\frac{\pi}{4}} \frac{\ln(\sin(x))}{\cos(x)^2}dx+\int_0^{\frac{\pi}{4}} \frac{\ln(\cos(x))}{\cos(x)^2}dx$$

now I will denote $\int_0^{\frac{\pi}{4}}\frac{\ln(2)}{\cos(x)^2}dx$ by $I_2$ , and I will denote $\int_0^{\frac{\pi}{4}}\frac{\ln(\sin(x))}{\cos(x)^2}dx$ by $I_3$ , and I will denote $\int_0^{\frac{\pi}{4}} \frac{\ln(\cos(x))}{\cos(x)^2}dx$ by $I_4$

$I_2$ is easy to do $$I_3=\frac{\ln(\tan(x))}{\cos(x)^2}dx +I_4$$ I will denote $\int_0^{\frac{\pi}{4}}\frac{\ln(\tan(x))}{\cos(x)^2}dx$ by $I_5$ which is easy to do.

the problem was to evaluate $2I_4$ which I couldn't do

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    $\begingroup$ How do you arrive at $1+\cos^2x = 2\cos^2\left(\frac{x}{2}\right)$ in your second line? That incorrect identity completely invalidates the rest of your work. You don't have to delete your work. The last version of your post showed effort and the current version would lead to a downvote and closing of your question. I have rolled back the edit for you. $\endgroup$ Jul 11, 2023 at 6:04
  • $\begingroup$ @Ninad Munshi oh you right I forgot about the power of 2 $\endgroup$
    – pie
    Jul 11, 2023 at 6:05
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    $\begingroup$ @pie integration by parts will solve $I_4$. $\endgroup$ Jul 11, 2023 at 6:10
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    $\begingroup$ @Ercanayan I got it thank you , I din't even try to use this as the original integral was non elementary so when I made a mistake I didn't notice so I expected there must be a non elementary part $\endgroup$
    – pie
    Jul 11, 2023 at 6:15
  • $\begingroup$ For your $I_4$ perform the change of variable $u=\tan x$ then perform integration by parts $\endgroup$
    – FDP
    Mar 3 at 7:38

7 Answers 7

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Alternatively

\begin{align} &\int_0^{\frac{\pi}{2}} \frac{\ln(\cos x)}{1+\sin^2x}dx\\ =& -\int_0^{\frac{\pi}{2}} \int_0^1 \frac{t\sin^2x}{(1+\sin^2x)(1-t^2 \sin^2 x)}dt\ dx\\ =& -\frac\pi{2\sqrt2}\int_0^1 \frac1{\sqrt2+t}dt = {-\frac{\pi}{2\sqrt{2}}\ln\bigg(1+\frac{1}{\sqrt{2}}\bigg)} \end{align}

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  • $\begingroup$ how do you know that would work , I mean why did you think that making that double integral would solve the problem ? $\endgroup$
    – pie
    Jul 11, 2023 at 13:20
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    $\begingroup$ @pie - I’ve learnt quite a number of not-so-obvious integrals are doubles in disguise. $\endgroup$
    – Quanto
    Jul 11, 2023 at 13:34
  • $\begingroup$ it is a very common trick to first differentiate and then integrate a function with constant limits. Very useful and very Overpowered $\endgroup$ Feb 21 at 10:31
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Your second form of the integral can be rewritten as

$$I = \int_0^{\frac{\pi}{2}}\frac{\sec^2x\cdot\log(\sin x)}{2+\tan^2x}dx = \frac{\log (\sin x)}{\sqrt{2}}\tan^{-1}\left(\frac{\tan x}{\sqrt{2}}\right)\Biggr|_0^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}}\frac{\cot x}{\sqrt{2}}\tan^{-1}\left(\frac{\tan x}{\sqrt{2}}\right)\:dx$$

$$= 0 - \int_0^{\frac{\pi}{2}}\frac{\cot x}{\sqrt{2}}\tan^{-1}\left(\frac{\tan x}{\sqrt{2}}\right)\:dx$$

To solve the remaining piece, denote

$$I[a] = \int_0^{\frac{\pi}{2}}\cot x \tan^{-1}(a\tan x)\:dx \implies I'[a] = \int_0^{\frac{\pi}{2}}\frac{dx}{1+a^2\tan^2x}$$

$$ = \int_0^{\frac{\pi}{2}}\frac{\sec^2x\:dx}{(1+a^2\tan^2x)(1+\tan^2x)} = \frac{1}{1-a^2}\int_0^{\frac{\pi}{2}}\left(\frac{1}{1+\tan^2x}-\frac{a^2}{1+a^2\tan^2x}\right)\sec^2x\:dx$$

In other words

$$ I'[a] = \frac{\pi}{2(1+a)} \implies I[a] = \frac{\pi}{2}\log(a+1)+C$$

Plugging in $I[0]=0$ gives $C=0$, thus the integral we want is

$$I = -\frac{1}{\sqrt{2}}I\left[\frac{1}{\sqrt{2}}\right] = \boxed{-\frac{\pi}{2\sqrt{2}}\log\left(1+\frac{1}{\sqrt{2}}\right)}$$

$$$$

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Map $x \mapsto \arctan{x}$ so that

$$ \begin{align} I &:= \int_{0}^{\frac{\pi}{2}}\frac{\ln\left(\cos x\right)}{1+\sin^{2}x}dx \\ &= \int_{0}^{\infty}\frac{\ln\left(\cos\left(\arctan x\right)\right)}{1+\sin^{2}\left(\arctan x\right)}d\left(\arctan x\right) \\ &= -\int_{0}^{\infty}\frac{\ln\left(1+x^{2}\right)}{4x^{2}+2}dx \\ &= -\frac{1}{4}\int_{-\infty}^{\infty}\frac{\ln\left(1+x^{2}\right)}{2x^{2}+1}dx. \\ \end{align} $$

Let $f(z)$ be the integrand as a function of $z$. We can write it as $$ \begin{align} f(z) &= \frac{\log\left(z+i\right)+\log\left(z-i\right)}{2z^{2}+1} \\ &= \frac{1}{2z^{2}+1}\cdot\left(\ln\left|z+i\right|+\ln\left|z-i\right|+i\arg\left(z+i\right)+i\arg\left(z-i\right)\right) \\ \end{align} $$ where we can justify splitting $\log\left(z^2+1\right)$ by defining $\arg(z-i) \in (\pi/2, 5\pi/2)$ and $\arg(z+i) \in (-5\pi/2, -\pi/2)$. We integrate over the following keyhole contour $C$ illustrated below.

Keyhole Contour C

This contour has a radius of $R$. The green x symbols are the poles $i/\sqrt{2}$ and $-i/\sqrt{2}$ and the branch cuts are highlighted in orange. And $\Gamma$ is the union of the two outer circular arcs.

Cauchy's Residue Theorem allows us to express $\displaystyle \oint_C f(z)dz$ as

$$ 2\pi i \mathop{\mathrm{Res}}_{z = i/\sqrt{2}} f(z) = \left(\int_{-R}^{R} + \int_{\Gamma} + \int_{\lambda_1} + \int_{\gamma} + \int_{\lambda_2}\right)f(z)dz $$

As $R \to \infty$, the integrals over $\Gamma$ decay to $0$. As the gap between the line segments shrinks, the radius of $\gamma$ also shrinks and forces the integral over $\gamma$ to decay to $0$ as well.

When $\lambda_1,\lambda_2 \to \Lambda$, we get

$$ \begin{align} \lim_{\lambda_1,\lambda_2 \to \Lambda} \left(\int_{\lambda_1} + \int_{\lambda_2}\right)f(z)dz =& \lim_{\lambda_1 \to \Lambda}\int_{\lambda_1}\frac{dz}{2z^{2}+1}\left(\ln\left|z+i\right|+\ln\left|z-i\right|+i\arg\left(z+i\right)+i\arg\left(z-i\right)\right) \\ & + \lim_{\lambda_2 \to \Lambda}\int_{\lambda_2}\frac{dz}{2z^{2}+1}\left(\ln\left|z+i\right|+\ln\left|z-i\right|+i\arg\left(z+i\right)+i\arg\left(z-i\right)\right) \\ =& -\int_{\Lambda} \frac{dz}{2z^{2}+1}\left(\ln\left|z+i\right|+\ln\left|z-i\right|+i\left(-\frac{3\pi}{2}\right)+i\left(\frac{5\pi}{2}\right)\right) \\ =& + \int_{\Lambda} \frac{dz}{2z^{2}+1}\left(\ln\left|z+i\right|+\ln\left|z-i\right|+i\left(-\frac{3\pi}{2}\right)+i\left(\frac{\pi}{2}\right)\right) \\ =& -2\pi i \int_{i}^{i \infty}\frac{dz}{2z^2+1} \\ =& -\sqrt{2} \pi \coth^{-1}{\sqrt{2}}. \\ \end{align} $$

Next, we evaluate the residue at the simple pole $i/\sqrt{2}$ of $f(z)dz$ here:

$$ \begin{align} 2\pi i \mathop{\mathrm{Res}}_{z = i/\sqrt{2}} f(z) &= 2\pi i \lim_{z \to i/\sqrt{2}}\frac{z-\frac{i}{\sqrt{2}}}{2z^{2}+1}\cdot\left(\ln\left|z+i\right|+\ln\left|z-i\right|+i\arg\left(z+i\right)+i\arg\left(z-i\right)\right) \\ &= 2\pi i\frac{-i}{2\sqrt{2}}\left(\ln\left|\frac{i}{\sqrt{2}}+i\right|+\ln\left|\frac{i}{\sqrt{2}}-i\right|+i\arg\left(\frac{i}{\sqrt{2}}+i\right)+i\arg\left(\frac{i}{\sqrt{2}}-i\right)\right) \\ &= 2\pi i\frac{-i}{2\sqrt{2}}\left(\ln\left|\frac{i}{\sqrt{2}}+i\right|+\ln\left|\frac{i}{\sqrt{2}}-i\right|+i\left(-\frac{3\pi}{2}\right)+i\left(\frac{3\pi}{2}\right)\right) \\ &= \frac{\pi}{\sqrt{2}}\left(\ln\left(1+\frac{1}{\sqrt{2}}\right)+\ln\left(1-\frac{1}{\sqrt{2}}\right)\right). \\ \end{align} $$

Finally, we get $$ \begin{align} I &= -\frac{1}{4}\left(\sqrt{2}\pi\coth^{-1}\left(\sqrt{2}\right)+\frac{\pi}{\sqrt{2}}\left(\ln\left(1+\frac{1}{\sqrt{2}}\right)+\ln\left(1-\frac{1}{\sqrt{2}}\right)\right)\right) \\ &= -\frac{\pi}{2\sqrt{2}}\ln\left(1+\frac{1}{\sqrt{2}}\right) \\ \end{align} $$

and we're done!

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  • $\begingroup$ I wish to know why you chose the set of arguments you did for $arg(z-i)$ and $arg(z+i)$, we could have also chosen $arg(z+i)\in (\frac {-\pi}{2},\frac{3\pi}{2}]$ but why go with that $(\frac {-5\pi}{2},\frac{-\pi}{2}]$? $\endgroup$ Feb 21 at 10:35
  • $\begingroup$ @MadhavAsthana In general, $\log\left(z^2+1\right)=\ln\left|z^2+1\right|+i\arg\left(z^2+1\right)$ where $\arg$ makes the log function single-valued. I wanted the $\arg$ to equal $0$ so that the integral stays a real number. Since $\log\left(z+i\right)+\log\left(z-i\right) = \ln\left|z^2+1\right|+i\left(\arg(z+i)+\arg(z-i)\right)$, setting it to equal $\ln\left|z^2+1\right|+i\cdot 0$ would make sense if we choose $\arg(z+i)$ to point down from the keyhole contour. In other words, letting $\arg(z+i) \in (-5\pi/2, -\pi/2)$ forces $\arg(z-i) \in (\pi/2, 5\pi/2)$. $\endgroup$ Feb 22 at 0:39
  • $\begingroup$ You can definitely let $\arg(z+i) \in (-\pi/2, 3\pi/2)$, but that forces $\arg(z-i) \in (-3\pi/2,\pi/2)$ if you want the integral to stay a real number. $\endgroup$ Feb 22 at 1:10
  • $\begingroup$ so you have to choose diametrically opposing sets of arguments for conjugated branch points to make that property of logarithms hold. I posted a question about choosing the right set of arguments for branch points so that logarithms can be split. Was my first statement correct? $\endgroup$ Feb 22 at 1:23
  • $\begingroup$ @MadhavAsthana I don't think it's possible to let $\arg(z+i) \in (-\pi/2, 3\pi/2)$ and $\arg(\pi/2,5\pi/2)$ simultaneously and recover the original integral. You see, my thought process wasn't just, "oh wait if I let $\arg(z+i) \in (-5\pi/2, -\pi/2)$ I can just switch around some numbers and negative signs and define the other $\arg$ that way." I'm more of a fan of rigorous mathematical statements (I majored in pure math a few years ago) where everything needs a proof, theorem, etc. You can prove those intervals for the arguments work by doing a little algebra I described in my first reply. $\endgroup$ Feb 22 at 1:29
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With a few substitutions and integration by parts, we find

$$\begin{align*} I &= \int_0^{\tfrac\pi2} \frac{\ln(\cos x)}{1 + \sin^2x} \, dx \\ &= \int_0^1 \frac{\ln y}{(2-y^2) \sqrt{1-y^2}} \, dy & x=\arcsin\sqrt{1-y} \\ &= - \int_0^1 \frac{dy}{(2-y^2) \sqrt{1-y^2}} \left[\int_y^1 \frac{dz}z\right] \\ &= - \int_0^1 \frac{dz}z \left[\int_0^z \frac{dy}{(2-y^2) \sqrt{1-y^2}}\right] \\ &= - \frac1{\sqrt2} \int_0^1 \arctan \left(\frac1{\sqrt2} \frac z{\sqrt{1-z^2}}\right) \, \frac{dz}z \\ &= -\frac1{2\sqrt2} \int_0^\infty \frac{\arctan w}{w\left(\frac12+w^2\right)} \, dw & z=\frac{\sqrt2\,w}{\sqrt{1+2w^2}} \end{align*}$$


Edit: On second thought, the earlier suggestion to consider the parameterized integral,

$$I(a,b) = \int_0^\infty \frac{\arctan x}{x^a \left(x^2+b^2\right)} \, dx, \qquad (a,b) \in (0,1)^2,$$

by following along with Svyatoslav's approach shown here, won't work because the integrals to either side of the branch cut due to $x^a$ would vanish as $a\to1$.


Instead, we can take advantage of the integrand's symmetry,

$$I = -\frac1{4\sqrt2} \int_{-\infty}^\infty \frac{\arctan w}{w \left(\frac12+w^2\right)} \, dw$$

and tackle $$\oint_C \frac{\arctan z}{z \left(\frac12+z^2\right)} \, dz = \oint_C \frac{i \ln \left\lvert\frac{1-iz}{1+iz}\right\rvert - \arg \frac{1-iz}{1+iz}}{z \left(1+2z^2\right)} \, dz$$ along a semicircular contour $C$ as shown below. Pink points are poles at $z=\pm \dfrac i{\sqrt2}$ and a removable singularity at $z=0$; blue points are branch points at $z=\pm i$, and the dashed lines indicate where we take the cuts.

plot of contour

At this point we basically reproduce Accelerator's solution with some slight differences peppered in.

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  • $\begingroup$ Out of curiosity, what program did you use to make that picture? $\endgroup$ Jul 11, 2023 at 23:32
  • $\begingroup$ @Accelerator Mathematica, this is a ComplexPlot of the integrand with some graphics primitives thrown in. $\endgroup$
    – user170231
    Jul 12, 2023 at 1:42
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Just for fun, let's see what happens if we use the approach that Lars Ahlfors uses to evaluate $\int_{0}^{\pi} \ln(\sin x) \, \mathrm dx$ in the textbook Complex Analysis: An Introduction to the Theory of Analytic Functions of One Complex Variable.

Let $\ln$ be the principal branch of the logarithm.

For $- \frac{\pi}{2} < x < \frac{\pi}{2}$, we have the identity $$\Re \ln(1+e^{2ix}) = \ln \left(\sqrt{\left(1+\cos(2x)\right)^{2}+ \sin^{2}(x)} \right) = \frac{1}{2} \, \ln \left(4 \cos^{2}(x) \right) = \ln(2 \cos x).$$

Using this identity, we have $$ \begin{align} \int_{0}^{\pi/2} \frac{\ln \left(\cos x \right)}{1+\sin^{2}(x)} \, \mathrm dx &= \frac{1}{2} \int_{-\pi/2}^{\pi/2} \frac{\ln \left(\cos x \right)} {1+\sin^{2}(x)} \, \mathrm dx \\ &= \frac{1}{2} \, \Re \int_{-\pi/2}^{\pi /2} \frac{\ln \left(1+e^{2ix} \right)}{1+ \sin^{2}(x)} \, \mathrm dx - \frac{1}{2}\int_{- \pi/2}^{\pi/2} \frac{\ln(2)}{1+\sin^{2}(x)} \, \mathrm dx \\ &= \frac{1}{2} \, \Re \int_{-\pi/2}^{\pi /2} \frac{\ln \left(1+e^{2ix} \right)}{1+ \sin^{2}(x)} \, \mathrm dx - \frac{1}{2}\int_{-\infty}^{\infty} \frac{\ln(2)}{1+2t^{2}} \, \mathrm dt \\ &=\frac{1}{2} \, \Re \int_{-\pi/2}^{\pi /2} \frac{\ln \left(1+e^{2ix} \right)}{1+ \sin^{2}(x)} \, \mathrm dx - \frac{\pi \ln(2)}{2 \sqrt{2}}. \end{align}$$

The function $1+e^{2iz}$ is real and nonpositive when $\Re(z) = \pi \, \frac{2n+1}{2}$ ($n \in \mathbb{Z}$) and $\Im(z) \le 0$.

If we omit these half-lines, the principal branch of $\ln(1+e^{2iz})$ is well defined and analytic.

So let's integrate $$f(z) = \frac{\ln \left(1+e^{2iz} \right)}{1+ \sin^{2}(z)} $$ around a tall rectangle contour with vertices at $z= - \pi/2$, $z= \pi/2$, $z= \pi/2 + i R$, $z= -\pi/2 + iR$.

The contour also includes small semicircles about the branch points at $z= \pm \pi/2$, but the integrals along the semicircles vanish as the radii of the semicircles go to zero since $\lim_{z \to \pm \pi/2} z f(z) =0$.

(Near $z= \pm \pi/2$, $\ln(1+e^{2iz}) \sim \ln\left(z \mp \frac{\pi}{2} \right)$.)

The integrals on the vertical sides of the contour are purely imaginary.

And the integral on the top of the contour vanishes as $R \to \infty$ since the magnitude of $f(z)$ decays exponentially to zero.

Since $ \sin(z) =\pm i$ when $z= n \pi \pm i \operatorname{arsinh}(1)$, the only singularity inside the contour is a simple pole at $z= i \operatorname{arsinh}(1) = i\ln (1+ \sqrt{2})$. (See appendix.)

So we have

$$ \begin{align} \Re \int_{-\pi/2}^{\pi/2} \frac{\ln \left(1+e^{2ix} \right)}{1+ \sin^{2}(x)} \, \mathrm dx &= \Re \, 2 \pi i \operatorname{Res} \left[f(z) , z=i \ln(1+ \sqrt{2}) \right] \\ &= \Re \, 2 \pi i \lim_{z \to i \ln(1+\sqrt{2})} \frac{\ln \left(1+e^{2iz} \right)}{\sin(2z)} \\ &= \Re \, 2 \pi i \, \frac{\ln \left(1+ \frac{1}{(1+\sqrt{2})^{2}} \right)}{\frac{1}{2i} \left(\frac{1}{(1+\sqrt{2})^{2}} - (1+\sqrt{2})^{2}\right)} \\&= \pi \, \ln \left(\frac{4+2 \sqrt{2}}{3+2 \sqrt{2}}\right) \frac{3+2 \sqrt{2}}{4+3 \sqrt{2}} \\ &= \frac{\pi}{\sqrt{2}} \, \ln (4-2\sqrt{2}). \end{align}$$

And therefore

$$ \begin{align} \int_{0}^{\pi/2} \frac{\ln \left(\cos x \right)}{1+\sin^{2}(x)} \, \mathrm dx &= \frac{1}{2} \left(\frac{\pi }{\sqrt{2}} \, \ln \left(4-2 \sqrt{2}\right) \right) - \frac{\pi \ln(2)}{2 \sqrt{2}} \\ &= \frac{\pi}{2 \sqrt{2}} \, \ln \left(2- \sqrt{2} \right) \\ &= \frac{\pi}{2 \sqrt{2}} \, \ln \left(\frac{2}{2+\sqrt{2}} \right) \\ &= - \frac{\pi}{2 \sqrt{2}} \, \ln \left(1+ \frac{1}{\sqrt{2}} \right). \end{align}$$


Appendix

$\sin(\pm iz) = \pm i \sinh(z) $

$\sin (n \pi +z) = (-1)^n \sin(z) $

$\sin \left(n \pi \pm i \operatorname{arsinh}(1)\right) = (-1)^n \sin\left( \pm i \operatorname{arsinh}(1) \right) = \pm i (-1)^{n} \sinh\left(\operatorname{arsinh}(1) \right) = \pm i (-1)^{n}$

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Let $t=\tan x$, then $$ \begin{aligned} \int_0^{\frac{\pi}{2}} \frac{\ln (\cos x)}{1+\sin ^2 x} d x = & \int_0^{\frac{\pi}{2}} \frac{\sec ^2 x \ln (\cos x)}{\sec ^2 x+\tan ^2 x} d x \\ = & -\frac{1}{2} \int_0^{\infty} \frac{\ln \left(1+t^2\right)}{1+2 t^2} d t \end{aligned} $$ Using contour integration along anti-clockwise direction of the path $$\gamma=\gamma_{1} \cup \gamma_{2} \textrm{ where } \gamma_{1}(t)=t+i 0(-R \leq t \leq R) \textrm{ and } \gamma_{2}(t)=R e^{i t} (0<t<\pi), $$ and the identity $\ln(a^2+b^2)=2\Re (\ln(a-bi))$, we have $$ \begin{aligned} \int_0^{\infty} \frac{\ln \left(1+t^2\right)}{1+2 t^2} d t&=\frac{1}{2} \int_{-\infty}^{\infty} \frac{\ln \left(1+t^2\right)}{1+2 t^2} d t \\ & =2 \Re \int_{-\infty}^{\infty} \frac{\ln (1-t i)}{1+2 t^2} d t \\ & =2 \Re \int_\gamma \frac{\ln (1-z i)}{1+2 z^2} d z \\ & =\Re\left[2 \pi i \cdot \lim _{z \rightarrow \frac{i}{\sqrt{2}}}\left(z-\frac{i}{\sqrt{2}}\right) \frac{\ln (1-z i)}{2\left(z^2+\frac{1}{2}\right)}\right] \\ & =\frac{\pi}{\sqrt{2}} \ln \left(1+\frac{1}{\sqrt{2}}\right) \\ & \end{aligned} $$

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Let $t=\tan x$, then $$ \begin{aligned} \int_0^{\frac{\pi}{2}} \frac{\ln (\cos x)}{1+\sin ^2 x} d x = & \int_0^{\frac{\pi}{2}} \frac{\sec ^2 x \ln (\cos x)}{\sec ^2 x+\tan ^2 x} d x \\ = & -\frac{1}{2} \int_0^{\infty} \frac{\ln \left(1+t^2\right)}{1+2 t^2} d t \end{aligned} $$ Considering the parametrised integral $$ I(a)=\int_0^{\infty} \frac{\ln \left(1+a t^2\right)}{1+2 t^2} d t $$ Differentiating $I(a)$ w.r.t $a$ yields $$ \begin{aligned} I^{\prime}(a) & =\int_0^{\infty} \frac{t^2}{\left(1+a t^2\right)\left(1+2 t^2\right)} d t \\ & =\frac{1}{2-a} \int_0^{\infty}\left(\frac{1}{1+a t^2}-\frac{1}{1+2 t^2}\right) d t \\ & =\frac{1}{2-a}\left[\frac{1}{\sqrt{a}} \tan ^{-1}(\sqrt{a} t)-\frac{1}{\sqrt{2}} \tan ^{-1} \sqrt{2} t\right]_0^{\infty}\\&= \frac{\pi}{2}\left[\frac{1}{(2-a) \sqrt{a}}-\frac{1}{(2-a) \sqrt{2}}\right]\\&= \frac{\pi}{2 \sqrt{2}} \cdot \frac{1}{(\sqrt{2}+\sqrt{a}) \sqrt{a}} \end{aligned} $$ Integrating back from $0$ to $1$ gives $$ \begin{aligned} I(1) & = \frac{\pi}{2 \sqrt{2}} \int_0^1 \frac{1}{(\sqrt{2}+\sqrt{a}) \sqrt{a}} d x \\ & = \frac{\pi}{2 \sqrt{2}}[2 \ln (\sqrt{x}+\sqrt{2})]_0^1 \\ & =\frac{\pi}{\sqrt{2}} \ln \left(1+\frac{1}{\sqrt{2}}\right) \end{aligned} $$ Hence $$ \int_0^{\infty} \frac{\ln \left(1+t^2\right)}{1+2 t^2} d t=-\frac{1}{2} I(1)=-\frac{\pi}{2 \sqrt{2}} \ln \left(1+\frac{1}{\sqrt{2}}\right) $$

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