Micromass' big integral challenge

[micromass]

I'll post the solution to 7 tomorrow, unless somebody really wants to find this himself.

 

[StudentOfScience]

I'll post the solution to 7 tomorrow, unless somebody really wants to find this himself.

If possible, I would prefer if you wait to post the solution until a bit longer. Others may disagree with me, but right now, I have exams to worry about as well. Thus, I may need a couple of days to find time to return to the problem. Thank you for your consideration.

 

[Samy_A]

Half of my challenge problems are already solved. I did not expect this to happen so soon! The people who found these solutions are truly integral masters!

Anyway, I wish to make a little adertisement now for the book containing these problems. I got these problems (except 6 which is from Apostol) from the beautiful book "Inside interesting integrals" from Paul J. Nahin. So if you want to learn tricky solutions or are up for a challenge. This is the book for you!

https://www.amazon.com/dp/1493912763/?tag=pfamazon01-20

Judging by the integrals in this thread, this must be a great book.

Ordered it!

 

[Ssnow]

I agree with @Samy_A, I am curious on the techniques in this book, if this challenge was a way to publicize this book it works very well!

 

[Samy_A]

Just received the book.
Of course I first looked for the solution of integral 7.
Indeed very difficult.

A consolation for those how tried it: it was apparently solved by none other than the great G.H. Hardy!

 

[fresh_42]

Just received the book.
Of course I first looked for the solution of integral 7.
Indeed very difficult.

A consolation for those how tried it: it was apparently solved by none other than the great G.H. Hardy!

Now that's a motivation. It's like saying "Ramanujan solved this prime problem" or "Wiles could solve this problem on modular forms" ...

 

[micromass]

Now that's a motivation. It's like saying "Ramanujan solved this prime problem" or "Wiles could solve this problem on modular forms" ...

Gauss found the answer to the sum $1+2+3+4+5+...+1000$.

 

[fresh_42]

Gauss found the answer to the sum $1+2+3+4+5+...+1000$.

... at the age of about 6 ...

 

[micromass]

... at the age of about 6 ...

Details...

 

[fresh_42]

Details...

https://nrich.maths.org/2478

 

[Ssnow]

I ordered the book last week but I am far (I need at least two weeks...). So in this time I can drink tea at five in order to reproduce all Hardy-physical conditions that will permit me to solve the integral ...

 

[fresh_42]

Is anyone interested in the anecdote of Hardy's very special life insurance?

 

[Samy_A]

Of course!

 

[fresh_42]

It is not quite clear whether Hardy believed in God or was just superstitious. However, in any case he believed God will do everything to make his life tough and complicated. One day he was on a journey back home. (I've heard it with Harald Bohr and Copenhagen, but also found Norway on the internet.) Anyway. He had to take a ship and the boat he got didn't look very trustful. Typically, he thought, why me?
So he sent a postcard before boarding - say to Bohr - claiming he has found the proof of Riemann's hypothesis (conjecture).
When afterwards asked why he replied: Well, if the ship sank the proof would have been lost but I would have become the most famous mathematician of my generation. God won't allow this to happen. This way I only had to write Bohr another postcard in which I stated to have made a mistake.

 

[member 587159]

Can someone post the solution of the remaining integral? I have waited long enough ;)

 

[micromass]

OK, I'll post it soon. I just thought some people were still thinking about it.

 

[fresh_42]

I admit I gave up after many attempts on something like $\int q(x)^{-1} \arctan{x} dx$ with $q(x) = (3x^2+1)\sqrt{2x^2-1}$ or similar. (Don't count on it, it's just out of memory.)

 

[micromass]

Alright, here is the first part of the solution to $I = \int_0^{\pi/2} \cos^{-1}\left(\frac{\cos(x)}{1+2\cos(x)}\right)dx$. I will give people the chance to complete the solution. I will post the second half on monday.

Fist, notice that the double-angle formula from trigonometry say that for any $\theta$, we have $\cos(2\theta) = 2\cos^2(\theta)-1$. Call this $(*)$.

If we write $u=\cos(\theta)$ - and so $\theta = \cos^{-1}(u)$ - then $(*)$ says that $\cos(2\theta) = 2u^2 - 1$, from which immediately follows that
$$\cos^{-1}(2 u ^2 - 1) = \cos^{-1}(\cos(2\theta)) = 2\theta = 2\cos^{-1}(u)$$

So, since $u$ is simply an arbitrary variable, we can write

$$\cos^{-1}(2\theta^2 - 1) = 2\cos^{-1}(\theta)$$

Next, we write $\alpha = 2\theta^2 - 1$, which means that $\theta = \sqrt{\frac{1+\alpha}{2}}$. It follows that

$$\cos^{-1}(\alpha) = 2\cos^{-1}\left(\sqrt{\frac{1+\alpha}{2}}\right)$$

We write $\alpha = \frac{\cos(x)}{1+2\cos(x)}$, to get

$$\cos^{-1}\left(\frac{\cos(x)}{1 + 2\cos(x)}\right) = 2\cos^{-1}\left(\sqrt{ \frac{1 + \frac{\cos(x)}{1 + 2\cos(x)}}{2}}\right) = 2\cos^{-1}\left(\sqrt{\frac{1 + 3\cos(x)}{2 + 4\cos(x)}}\right).$$

Applying the Pythagorean theorem to a right triangle with acute angle whose cosine is $\sqrt{\frac{1 + 3\cos(x)}{2 + 4\cos(x)}}$, yu'll see that the tangent of that same angle is $\sqrt{\frac{1 + \cos(x)}{1 + 3\cos(x)}}$. Thus

$$\cos^{-1}\left(\frac{\cos(x)}{1 + 2\cos(x)}\right) = 2\tan^{-1}\left(\sqrt{\frac{1+\cos(x)}{1 + 3\cos(x)}}\right)$$

And so we have

$$I = 2\int_0^{\pi/2} \tan^{-1}\left(\sqrt{\frac{1 + \cos(x)}{1 + 3\cos(x)}}\right)dx$$

Make the chance of variable $x = 2y$, so $dx = 2dy$, to get

$$I = 4\int_0^{\pi/4} \tan^{-1}\left(\sqrt{\frac{1 + \cos(2y)}{1 + 3 \cos(2y)}}\right)dy$$

Using $(*)$ again, we can find

$$\sqrt{\frac{1 + \cos(2y)}{1 + 3\cos(2y)}} = \frac{\cos(y)}{\sqrt{2 - 3\sin^2(y)}}$$

And so

$$I = 4\int_0^{\pi/4} \tan^{-1}\left(\frac{\cos(y)}{\sqrt{2 - 3\sin^2(y)}} \right)dy$$

Notice that

$$\int_0^1 \frac{1}{1 + \left(\frac{\cos^2(y)}{2 - 3\sin^2(y)}\right)t^2}dt$$

is of the form

$$\int_0^1 \frac{1}{1 + b^2t^2}dt = \frac{1}{b^2}\int_0^1 \frac{1}{\frac{1}{b^2} + t^2}dt = \frac{1}{b^2} [b\tan^{-1}(bt)]_0^1 = \frac{1}{b} \tan^{-1}(b),$$
where $b = \frac{\cos(y)}{\sqrt{2 - 3\sin^2(y)}}$.
Thus

$$\int_0^1 \frac{1}{1 + \left(\frac{\cos^2(y)}{2 - 3\sin^2(y)}\right)t^2 } dt = \frac{\sqrt{2 - 3\sin^2(y)}}{\cos(y)} = \tan^{-1}\left(\frac{\cos(y)}{\sqrt{2 - 3\sin^2(y)}}\right)$$

That is,

$$\begin{eqnarray*} I & = & 4\int_0^{\pi/4} \frac{\cos(y)}{\sqrt{ 2 - 3\sin^2(y)}} \left(\int_0^1 \frac{1}{1 + \left(\frac{\cos^2(y)}{2 - 3\sin^2(y)}\right)t^2}\right)dy\\ & = & \int_0^{\pi/4} \int_0^1 \frac{4\cos(y)(2 - 3\sin^2(y))}{\sqrt{2 - 3\sin^2(y)}(2 - 3\sin^2(y) + t^2\cos^2(y))}dt dy\\ & = & \int_0^{\pi/4} \int_0^1 \frac{4\cos(y)\sqrt{ 2 - 3\sin^2(y)}}{2 - 3\sin^2(y) + t^2 - t^2\sin^2(y)}dt dy\\ & = & \int_0^{\pi/4} \int_0^1 \frac{4\cos(y) \sqrt{2 - 3\sin^2(y)} }{(t^2 + 2) - (t^2 + 3)\sin^2(y)}dt dy \end{eqnarray*}$$

Next, we change the variables $\sin(y) = \sqrt{\frac{2}{3}} \sin(w)$, and so $dy = \sqrt{\frac{2}{3}} \frac{\cos(w)}{\cos(y)}dw$. We have $w=0$ when $y=0$, and when $y = \pi/4$, we have $\sin(\pi/4) = \frac{1}{\sqrt{2}}$. And so $\sin(w) = \sqrt{\frac{3}{2}}\frac{1}{\sqrt{2}} = \frac{\sqrt{3}}{2}$, which says $w = \pi/3$. So

$$\begin{eqnarray*} I & = & \int_0^{\pi/3} \int_0^1 \frac{4\cos(y) \sqrt{2 - 3\frac{2}{3}\sin^2(w)}}{(t^2 + 2) - (t^2 + 3) \frac{2}{3} \sin^2(w)}dt \frac{\cos(w)}{\cos(y)} dw \sqrt{\frac{2}{3}}\\ & = & \int_0^{\pi/3} \int_0^1 \frac{4 \sqrt{2 - 2( 1 - \cos^2(w))}}{(t^2 + 2) - (t^2 + 3) \frac{2}{3} ( 1 - \cos^2(w))} dt \cos(w) dw \sqrt{\frac{2}{3}}\\ & = & \int_0^{\pi/3} \int_0^1\frac{4 \sqrt{2} \cos(w) \sqrt{2} \cos(w) }{(t^2 + 2) - (t^2 + 3) \frac{2}{3} ( 1 - \cos^2(w)} dt \frac{1}{\sqrt{3}} dw\\ & = & \int_0^{\pi/3} \int_0^1 \frac{8\sqrt{3}\cos^2(w)}{t^2 + (2t^2 + 6)cos^2(w)} dt dw \end{eqnarray*}$$

Our next step is another change of variable to $s = \tan(w)$. Thus as $\tan(w) = \frac{\sin(w)}{\cos(w)}$, we have

$$\frac{ds}{dw} = \frac{1}{\cos^2(w)}$$

and so $dw = \cos^2(w) ds$. Since

$$1 + s^2 = 1 + \tan^2(w) = \frac{1}{\cos^2(w)}$$

we have

$$\frac{1}{1 + s^2} = \cos^2(w)$$

and so

$$dw = \frac{ds}{ 1+ s^2}$$

Therefore, since $s=0$ when $w=0$ and $s= \sqrt{3}$, when $w = \pi/3$, we have

$$\begin{eqnarray*} I & = & \int_0^{\sqrt{3}} \int_0^1 \frac{8\sqrt{3} \frac{1}{1 + s^2}}{t^2 + (2t^2 + 6) \frac{1}{1 + s^2}} dt \frac{ds}{1 + s^2}\\ & = & \int_0^{\sqrt{3}} \int_0^1 \frac{8\sqrt{3}}{t^2(1 + s^2)^2 + (2t^2 + 6)(1 + s^2)} dt ds\\ & = & \int_0^{\sqrt{3}} \int_0^1 \frac{8\sqrt{3}}{(1 + s^2)(t^2s^2 + 3t^2 + 6)} dt ds \end{eqnarray*}$$

Let's make a partial fraction expansion:

$$\frac{1}{(1+ s^2)(t^2 s^2 + 3t^2 + 6)} = \frac{A}{1+s^2} + \frac{B}{t^2 s^2 + 3t^2 + 6}$$
where it is easy to confirm that
$$A = \frac{1}{2t^2 + 6}~~\text{and}~~ B = - \frac{t^2}{2t^2 + 6}$$

and so

$$\begin{eqnarray*} I & = & \int_0^{\sqrt{3}} \int_0^1 8\sqrt{3} \left(\frac{\frac{1}{2t^2 + 6}}{1 + s^2} - \frac{\frac{t^2}{2t^2 + 6}}{t^2 + 3t^2 + 6}\right)dt ds\\ & = & \int_0^1 \frac{4\sqrt{3}}{t^2 + 3}\left(\int_0^{\sqrt{3}} \frac{ds}{1+s^2} - \int_0^{\sqrt{3}} \frac{ds}{s^2 + 3 + \frac{6}{t^2}}\right) dt \end{eqnarray*}$$

The first inner intergral on the right is easy:

$$[\tan^{-1}(s)]_0^{\sqrt{3}} = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}$$
The second inner integral is almost as easy and equals

$$\begin{eqnarray*} \int_0^{\sqrt{3}} \frac{ds}{s^2 + \left(\sqrt{3 + \frac{6}{t^2}}\right)^2} & = & \frac{1}{\sqrt{3 + \frac{6}{t^2}}} \left. \left(\tan^{-1}\left(\frac{s}{\sqrt{3 + \frac{6}{t^2}}}\right)\right)\right|_0^{\sqrt{3}}\\ & = & \frac{t}{\sqrt{3}\sqrt{t^2 + 2}} \tan^{-1}\left.\left(\frac{st}{\sqrt{3}\sqrt{t^2 + 2}}\right)\right|_0^{\sqrt{3}}\\ & = & \frac{t}{\sqrt{3}\sqrt{t^2 + 2}} \tan^{-1}\left(\frac{t}{\sqrt{t^2 + 2}}\right) \end{eqnarray*}$$

Thus

$$\begin{eqnarray*} I & = & \int_0^1 \frac{4\sqrt{3}}{t^2 + 3}\left(\frac{\pi}{3} - \frac{t}{\sqrt{3}\sqrt{t^2 + 2}} \tan^{-1}\left(\frac{t}{\sqrt{t^2 + 2}}\right)\right)dt\\ & = & \frac{4\sqrt{3}\pi}{3}\int_0^1 \frac{dt}{t^2 + (\sqrt{3})^2} - 4\int_0^1 \frac{t}{(t^2 + 3)\sqrt{t^2 + 2}} \tan^{-1}\left(\frac{t}{\sqrt{t^2 + 2}}\right) dt\\ & = & \frac{4\sqrt{3}\pi}{3}\left.\left(\frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{t}{\sqrt{3}}\right)\right)\right|_0^1 - 4\int_0^1 \frac{t}{(t^2 + 3)\sqrt{t^2 + 2}} \tan^{-1}\left(\frac{t}{\sqrt{t^2 + 2}}\right) dt \end{eqnarray*}$$

Since

$$\frac{4\sqrt{3}\pi}{3}\left.\left(\frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{t}{\sqrt{3}}\right)\right)\right|_0^1 = \frac{4\pi}{3}\tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{2\pi^2}{9}$$

we have

$$I = \frac{2\pi^2}{9} - 4\int_0^1 \frac{t}{(t^2 + 3)\sqrt{t^2 + 2}} \tan^{-1}\left(\frac{t}{\sqrt{t^2 + 2}}\right) dt$$

We can do this by parts, let

$$u = \tan^{-1}\left(\frac{t}{\sqrt{t^2 + 2}}\right)$$

and

$$dv = \frac{t}{(t^2 + 3)\sqrt{t^2 + 2}} dt$$

Then

$$\frac{du}{dt} = \frac{1}{(t^2 + 1)\sqrt{t^2 + 2}}$$

and you can verify that

$$v = \tan^{-1}(\sqrt{t^2 + 2})$$

by simply sifferentiating this $v$ and observing we get the above $dv$ back. So, plugging all this into the integration by parts formula, we have

$$\begin{eqnarray*} I & = & \frac{2\pi^2}{9} - 4\left(\left.\left(\tan^{-1}\left(\frac{1}{\sqrt{t^2 + 2}}\right)\tan^{-1}(\sqrt{t^2 + 2})\right)\right|_0^1 - \int_0^1 \frac{\tan^{-1}(\sqrt{t^2 + 2})}{(t^2 + 1)\sqrt{t^2 + 2}}dt\right)\\ & = & \frac{2\pi^2}{9} - 4\left(\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)\tan^{-1}(\sqrt{3}) - \int_0^1 \frac{\tan^{-1}(\sqrt{t^2 + 2})}{(t^2 + 1)\sqrt{t^2 + 2}}dt\right)\\ & = & \frac{2\pi^2}{9} - 4\left(\frac{\pi}{6}\frac{\pi}{3} - \int_0^1 \frac{\tan^{-1}(\sqrt{t^2 + 2})}{(t^2 + 1)\sqrt{t^2 + 2}}dt\right)\\ & = & \frac{2\pi^2}{9} - \frac{2\pi^2}{9} + 4\int_0^1 \frac{\tan^{-1}(\sqrt{t^2 + 2})}{(t^2 + 1)\sqrt{t^2 + 2}}dt \end{eqnarray*}$$

And so

$$I = 4\int_0^1 \frac{\tan^{-1}(\sqrt{t^2 + 2})}{(t^2 + 1)\sqrt{t^2 + 2}}dt$$

This is the first part. I leave the second part up to you. I will post the solution monday unless somebody doesn't want me to

 

[member 587159]

How was anyone supposed to find that? :P

 

[micromass]

Alright, here is the second part. We let

$$I(u) = \int_0^1 \frac{\tan^{-1}\left(u\sqrt{2 +x^2}\right)}{(1+x^2)\sqrt{2 + x^2}}dx$$

Note that we wish to find $I(1)$. Notice that if $u\rightarrow +\infty$, then the argument for the inverse tangent also goes to $+\infty$ for all $x>0$. So since $\tan^{-1}(+\infty) = \frac{\pi}{2}$, we have

$$I(+\infty) = \frac{\pi}{2} \int_0^1 \frac{dx}{(1+x^2)\sqrt{2+x^2}}$$

This integral is easy to do if you remember the following standard formula:

$$\frac{d}{dx} \tan^{-1}(f(x)) = \frac{1}{1+f^2(x)}\frac{df}{dx}$$

We can use this formula to calculate

$$\frac{d}{dx} \tan^{-1}\left(\frac{x}{\sqrt{2+x^2}}\right) = \frac{1}{(1+x^2)\sqrt{2 + x^2}}$$

Thus

$$\begin{eqnarray*} I(+\infty) & = & \frac{\pi}{2}\int_0^1 \frac{d}{dx} \left(\frac{x}{\sqrt{2+x^2}}\right)dx\\ & = & \frac{\pi}{2} \left.\left(\tan^{-1}\left(\frac{x}{\sqrt{2+x^2}}\right)\right)\right|_0^1\\ & = & \frac{\pi}{2}\left(\tan^{-1}\left(\frac{1}{\sqrt{3}}\right) - \tan^{-1}(0)\right)\\ & = & \frac{\pi^2}{12} \end{eqnarray*}$$

Now we differentiate $I(u)$ with respect to $u$. We get

$$\frac{dI}{du} = \int_0^1\frac{dx}{(1+x^2)(1 + 2u^2 + u^2 x^2)}$$

With a partial fraction expansion, this becomes

$$\begin{eqnarray*} \frac{dI}{du} & = & \int_0^1 \frac{1}{1+u^2} \left(\frac{1}{1+x^2} - \frac{u^2}{1 + 2u^2 + u^2 x^2}\right)dx\\ & = & \frac{1}{1+u^2}\left(\int_0^1 \frac{dx}{1+x^2} - \int_0^1 \frac{dx}{\frac{1 + 2u^2}{u^2} + x^2}\right) \end{eqnarray*}$$

These last two integrals are of the form

$$\int \frac{dx}{a^2 + x^2} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right)$$

And so, doing the integrals, we have

$$\begin{eqnarray*} \frac{dI}{du} & = & \frac{1}{1+u^2} \left. \left( \tan^{-1}(x) - \frac{u}{\sqrt{1 + 2u^2}} \tan^{-1}\left(\frac{xu}{\sqrt{1 + 2u^2}}\right) \right) \right|_0^1\\ & = & \frac{1}{1+u^2}\left(\frac{\pi}{4} - \frac{u}{\sqrt{1 + 2u^2}}\tan^{-1} \left(\frac{u}{\sqrt{1 + 2u^2}}\right)\right) \end{eqnarray*}$$

Now we integrate both sides from $1$ to $+\infty$ with respect to $u$. On the left, we get:

$$\int_1^{+\infty} \frac{dI}{du} du = \int_1^{+\infty} dI = I(+\infty) - I(1)$$

On the right, we get

$$\frac{\pi}{4}\int_1^{+\infty} \frac{du}{1+u^2} - \int_1^{+\infty} \frac{u}{(1+u^2)\sqrt{1 + 2u^2}} \tan^{-1}\left(\frac{u}{\sqrt{1 + 2u^2}}\right)du$$

The first integral is easy:

$$\frac{\pi}{4} \int_1^{+\infty} \frac{du}{1+u^2} = \frac{\pi}{4} \left(\tan^{-1}(+\infty) - \tan^{-1}(1)\right) = \frac{\pi}{4}\left(\frac{\pi}{2} - \frac{\pi}{4}\right) = \frac{\pi^2}{16}$$

Thus

$$I(+\infty) - I(1) = \frac{\pi^2}{16} - \int_1^{+\infty} \frac{u}{(1+u^2)\sqrt{1 + 2u^2}}\tan^{-1} \left(\frac{u}{\sqrt{1 + 2u^2}}\right)du$$

In this final integral, we change variable $t = \frac{1}{u}$, and so $du = - \frac{1}{t^2} dt$ as follows:

$$\begin{eqnarray*} & & \int_1^{+\infty} \frac{u}{(1+u^2)\sqrt{1 + 2u^2}} \tan^{-1}\left(\frac{u}{\sqrt{1 + 2u^2}}\right) du\\ & = & \int_1^0 \frac{\frac{1}{t}}{\left(1 + \frac{1}{t^2}\right)\sqrt{1 + \frac{2}{t^2}}} \tan^{-1}\left(\frac{\frac{1}{t}}{\sqrt{1 + \frac{2}{t^2}}}\right) \left(-\frac{1}{t^2} dt\right)\\ & = & \int_0^1 \frac{\frac{1}{t}}{(t^2 + 1)\frac{\sqrt{t^2 + 2}}{t}} \tan^{-1} \left(\frac{\frac{1}{t}}{\frac{\sqrt{t^2 + 2}}{t}}\right)dt\\ & = & \int_0^1 \frac{1}{(t^2 + 1)\sqrt{t^2 + 2}} \tan^{-1}\left(\frac{1}{\sqrt{t^2 + 2}}\right)dt \end{eqnarray*}$$

Now recall the identity $\tan^{-1}(s) + \tan^{-1}\left(\frac{1}{s}\right) = \frac{\pi}{2}$, which becomes instantly obvious if you draw a right triangle with perpendicular sides of lengths $1$ and $s$ and remember that the two acute angles add to $\pi/2$. This says

$$\tan^{-1}\left(\frac{1}{\sqrt{t^2 + 2}}\right) = \frac{\pi}{2} - \tan^{-1}\left(\sqrt{t^2 + 2}\right)$$

And so we can write

$$\begin{eqnarray*} & & \int_0^1\frac{1}{(t^2 + 1)\sqrt{t^2 + 2}}\tan^{-1}\left(\frac{1}{\sqrt{t^2 + 2}}\right)dt\\ & = & \frac{\pi}{2}\int_0^1 \frac{dt}{(t^2 + 1)\sqrt{t^2 + 2}} -\int_0^1 \frac{\tan^{-1}\left(\sqrt{t^2 + 2}\right)}{(t^2 + 1)\sqrt{t^2 + 2}}dt \end{eqnarray*}$$

That is, we have

$$I(+\infty) - I(1) = \frac{\pi^2}{16} - \frac{\pi}{2}\int_0^1\frac{dt}{(t^2 + 1)\sqrt{t^2 + 2}} + \int_0^1 \frac{\tan^{-1}\left(\sqrt{t^2 + 2}\right)}{(t^2 + 1)\sqrt{t^2 + 2}}dt$$

The first integral is just $I(+\infty)$ and the last integral is just $I(1)$. That is, we have

$$I(+\infty) - I(1) = \frac{\pi^2}{16} - I(+\infty) + I(1)$$

and so

$$2 I (+\infty) - \frac{\pi^2}{16} = 2I(1)$$

and at last

$$I(1) = I(+\infty) - \frac{\pi^2}{32} = \frac{\pi^2}{12} - \frac{\pi^2}{32} = \frac{5\pi^2}{96}$$

Thus our original integral now becomes

$$\int_0^{\pi/2} \cos^{-1}\left(\frac{\cos(x)}{1 + 2\cos(x)}\right)dx = 4\int_0^1 \frac{\tan^{-1}\left(\sqrt{t^2 + 2}\right)}{(t^2 + 1)\sqrt{t^2 + 2}}dt = \frac{5\pi^2}{24}$$

This is very close to $\frac{\pi^2}{4}$, which two people got as answer. This is very interesting!

 

[micromass]

Thank you everybody who participated in this thread! I hope you found these integrals interesting. If you want more, get the book "Inside Interesting Integrals" by Paul J. Nahin.

https://www.amazon.com/dp/1493912763/?tag=pfamazon01-20

 

[ShayanJ]

That was the longest integration I've ever seen! To Hardy:

 

[StudentOfScience]

I would've never thought of that solution. Thank you for posting the solution before I went crazy trying to figure out the integral with elementary substitution techniques. If all the integrals in that book are this involved, I'm very interested in the book.