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As part of going through a set of definite integrals that are solvable using the Feynman Trick, I am now solving the following:

$$ \int_{0}^{\frac{\pi}{2}} \ln\left|2 + \tan^2(x) \right| \:dx $$

I'm seeking methods using the Feynman Trick (or any method for that matter) that can be used to solve this definite integral.

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If one wishes to use "Feynman's Trick," then begin by defining a function $I(a)$, $a>1$ as given by

$$I(a)=\int_0^{\pi/2}\log(a+\tan^2(x))\,dx \tag1$$

Differentiation of $(1)$ reveals

$$\begin{align} I'(a)&=\int_0^{\pi/2} \frac{1}{a+\tan^2(x)}\,dx\\\\ &=\frac{\pi/2}{a-1}-\frac{\pi/2}{\sqrt a (a-1)}\tag2 \end{align}$$

Integration of $(2)$ yields

$$\begin{align} I(a)&=\frac\pi2\left(\log(a-1)+\log\left(\frac{\sqrt a+1}{\sqrt{a}-1}\right) \right)\\\\ &=\pi \log(\sqrt a+1)\tag3 \end{align}$$

Finally, setting $a=2$ in $(3)$, we obtain the coveted result

$$\int_0^{\pi/2}\log(2+\tan^2(x))\,dx=\pi \log(\sqrt 2+1)$$

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  • $\begingroup$ How did you resolve the constant of Integration in Step (3)? $\endgroup$ – user150203 Nov 21 '18 at 4:38
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    $\begingroup$ @davidg Apology for the sign error. Note $I(0)=0$. $\endgroup$ – Mark Viola Nov 21 '18 at 5:08
  • $\begingroup$ Hi Mark ! Long time no speak. Nice solution $\to +1$. Cheers. $\endgroup$ – Claude Leibovici Nov 21 '18 at 6:22
  • $\begingroup$ @MarkViola do you know of any other ‘tricks that would work? $\endgroup$ – user150203 Nov 21 '18 at 6:51
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    $\begingroup$ Hi David. You could try writing the logarithm as $\log(1+\cos^2(x))-\log(\sin^2(x))$ and expanding the first term as $\sum_{n=1}^\infty \frac{(-1)^{n-1}\cos^{2n}(x)}{n}$ and proceeding. I haven't tried this, but it might be worth pursuing. $\endgroup$ – Mark Viola Nov 21 '18 at 15:47
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My approach

Let

\begin{align} \int_{0}^{\frac{\pi}{2}} \ln\left|2 + \tan^2(x) \right| \:dx &= \int_{0}^{\frac{\pi}{2}} \ln\left|1 + \left(1 + \tan^2(x)\right) \right| \:dx \\ &= \int_{0}^{\frac{\pi}{2}} \ln\left|1 + \sec^2(x) \right| \:dx \\ &= \int_{0}^{\frac{\pi}{2}} \ln\left|\frac{\cos^2(x) + 1}{\cos^2(x)} \right| \:dx \\ &= \int_{0}^{\frac{\pi}{2}} \left[ \ln\left|\cos^2(x) + 1 \right| - \ln\left|\cos^2(x)\right| \right]\:dx \\ &= \int_{0}^{\frac{\pi}{2}} \ln\left|\cos^2(x) + 1 \right|\:dx - \int_{0}^{\frac{\pi}{2}} \ln\left|\cos^2(x)\right|\:dx \end{align}

Now

$$ \int_{0}^{\frac{\pi}{2}} \ln\left|\cos^2(x)\right|\:dx = 2\int_{0}^{\frac{\pi}{2}} \ln\left|\cos(x)\right|\:dx = 2\cdot-\frac{\pi}{2}\ln(2) = -\pi \ln(2)$$

For detail on this definite integral, see guidance here

We now need to solve

$$ \int_{0}^{\frac{\pi}{2}} \ln\left|\cos^2(x) + 1 \right|\:dx $$

Here, Let

$$ I(t) = \int_{0}^{\frac{\pi}{2}} \ln\left|\cos^2(x) + t \right|\:dx $$

Thus,

$$ \frac{dI}{dt} = \int_{0}^{\frac{\pi}{2}} \frac{1}{\cos^2(x) + t}\:dx = \int_{0}^{\frac{\pi}{2}} \frac{1}{\frac{\cos(2x) + 1}{2} + t}\:dx = 2\int_{0}^{\frac{\pi}{2}} \frac{1}{\cos(2x) + 2t + 1}\:dx $$

Employ a change of variable $u = 2x$:

$$\frac{dI}{dt} = \int_{0}^{\pi} \frac{1}{\cos(u) + 2t + 1}\:du $$

Employ the Weierstrass substitution $\omega = \tan\left(\frac{u}{2} \right)$:

\begin{align} \frac{dI}{dt} &= \int_{0}^{\infty} \frac{1}{\frac{1 - \omega^2}{1 + \omega^2} + 2t + 1}\:\frac{2}{1 + \omega^2}\cdot d\omega \\ &= \int_{0}^{\infty} \frac{1}{t\omega^2 + t + 1} \:d\omega \\ &= \frac{1}{t}\int_{0}^{\infty} \frac{1}{\omega^2 + \frac{t + 1}{t}} \:d\omega \\ &= \frac{1}{t}\left[\frac{1}{\sqrt{\frac{t+1}{t}}}\arctan\left( \frac{\omega}{\sqrt{\frac{t+1}{t}}}\right)\right]_{0}^{\infty} \\ &= \frac{1}{t}\frac{1}{\sqrt{\frac{t+1}{t}}}\frac{\pi}{2} \\ &= \frac{1}{\sqrt{t}\sqrt{t + 1}}\frac{\pi}{2} \end{align}

And so,

$$I(t) = \int \frac{1}{\sqrt{t}\sqrt{t + 1}}\frac{\pi}{2}\:dt = \pi\ln\left| \sqrt{t} + \sqrt{t + 1}\right| + C$$

Now

$$I(0) = \int_{0}^{\frac{\pi}{2}} \ln\left|\cos^2(x) + 0 \right|\:dx = -\pi \ln(2) = \pi\ln\left|\sqrt{0} + \sqrt{0 + 1} \right| + C \rightarrow C = -\pi \ln(2)$$

And so,

$$I(t) = \pi\ln\left| \sqrt{t} + \sqrt{t + 1}\right| -\pi \ln(2) = \pi\ln\left|\frac{\sqrt{t} + \sqrt{t + 1}}{2} \right|$$

Thus,

$$I = I(1) = \int_{0}^{\frac{\pi}{2}} \ln\left|\cos^2(x) + 1 \right|\:dx = \pi\ln\left|\frac{\sqrt{1} + \sqrt{1 + 1}}{2} \right| = \pi\ln\left|\frac{1 + \sqrt{2}}{2} \right| $$

And Finally

\begin{align} \int_{0}^{\frac{\pi}{2}} \ln\left|2 + \tan^2(x) \right| \:dx &= \int_{0}^{\frac{\pi}{2}} \ln\left|\cos^2(x) + 1 \right|\:dx - \int_{0}^{\frac{\pi}{2}} \ln\left|\cos^2(x)\right|\:dx \\ &= \pi\ln\left|\frac{1 + \sqrt{2}}{2} \right| - \left(-\pi \ln(2)\right) \\ &= \pi\ln\left|1 + \sqrt{2} \right| \end{align}

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  • $\begingroup$ Your question and your answer have a gap of just 1 minute. If you did solve the question earlier then you must've included your approach in the question itself. What is the need to post it as an answer? $\endgroup$ – Rohan Shinde Nov 21 '18 at 3:56
  • $\begingroup$ @Digamma - Sorry, although I've been a member of Math StackExchange for 4 years, I've only become regularly active recently. When I asked a friend who is a member as to what is the best way of presenting a solution whilst asking for other solutions I was told this is the approach taken within the community. If this violate any of the rules, please advise and I will reposition accordingly. Also, I never assign my solution as 'the solution' to any post of this nature. $\endgroup$ – user150203 Nov 21 '18 at 4:00
  • $\begingroup$ @Digamma The site encourages people to share the questions with answers. $\endgroup$ – Tianlalu Nov 21 '18 at 4:03
  • $\begingroup$ @Tianlalu - Sorry, just to be clear - Have I employed the correct process here? $\endgroup$ – user150203 Nov 21 '18 at 4:10
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    $\begingroup$ Thanks @mick - much appreciated :-) $\endgroup$ – user150203 Dec 3 '18 at 3:42

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