Prove that $\int_0^\pi \ln(1+\alpha \cos x)\, dx= \pi \ln\left(\frac{1+\sqrt{1-\alpha^2}}{2}\right)$

Question

Prove that

$$\int_0^\pi \ln(1+\alpha \cos x)\, dx= \pi \ln\left(\frac{1+\sqrt{1-\alpha^2}}{2}\right).$$

This question is under the Leibnitz's Rule section of my book, but I'm not sure if you're supposed to use it here. Using it simply gives the derivative of $\ln(1+\alpha \cos x)$ with respect to $\alpha$, which isn't really useful. Can anyone help?

Edit: So I've taken some suggestions from the comments and worked out the question. I was able to get pretty close to the answer, but I'm unable to get the answer exactly. I've skipped a couple of steps here, but do let me know if I did something wrong.

$$I(a)=\int_0^\pi \ln(1+\alpha \cos x)\, dx$$

$$I'(a)=\int^\pi_0\frac{cosx}{1+acosx}dx$$

$$I'(a)=\frac{1}{a}\int^\pi_0\frac{acosx+1-1}{1+acosx}dx$$

$$I'(a)=\frac{1}{a}[\pi-\int^\pi_0\frac{1}{1+acosx}dx]$$

By the Tangent Half-Angle Substitution,

$$I'(a)=\frac{1}{a}[\pi-\int^\pi_0\frac{1}{1+acosx}dx]$$

$$I'(a)=\pi[\frac{1}{a}-\frac{1}{\sqrt {1-a^2}}]$$

$$I(a)=\int\pi[\frac{1}{a}-\frac{1}{\sqrt {1-a^2}}]da$$

Using normal integrating techniques,

$$I(a)=\pi[ln\sqrt{1-a^2}+1]$$

Note that I can't get the last term, $\pi ln2$. If I can get this last term, I can get the answer. Now, I'm wondering if it is because it is a constant term, and thus when I have differentiated by $a$, in the beginning, the information in that term is lost. Can someone help me continue or point out where I went wrong?

If $f(\alpha) = \int_0^{\pi} \log(1+\alpha\cos{x}),dx$, consider $\frac{d}{d\alpha} [f'(\alpha)/\alpha]$. — Jason, Aug 27 '18 at 02:45
$\partial_\alpha (\log(1+\alpha \cos(x)))$ is actually continuous on $[0,\pi]$ when $|\alpha|<1$, hence you could take the derivative. Don't forget the integral symbol. — xbh, Aug 27 '18 at 02:46
@Jason Hi, how's that supposed to work? I tried what you proposed but I got an even more complicated function instead. — koifish, Aug 27 '18 at 03:03
@xbh Sure, I tried that taking the derivative w.r.t. $a$, and I got $\int^0 _\pi \frac{cosx}{1+acosx}dx$ using Leibnitz rule, but it brings me no closer to solving the question. — koifish, Aug 27 '18 at 03:05
@YipJungHon Then you calculate this integral, which is totally solvable. — xbh, Aug 27 '18 at 03:06
@xbh How can you solve that integral? I tried substitution and breaking it up into parts, but to no avail. — koifish, Aug 27 '18 at 03:30
I can't seem to get it to work. A get a polynomial of t of degree 4 in a demoniator and polynomial of degree 2 in the numerator. — koifish, Aug 27 '18 at 03:47
Did you look at my answer for $\int \frac{\cos(x)}{1+a\cos(x)},dx$ — Claude Leibovici, Aug 27 '18 at 06:04
@xbh, I have tried to use your methods suggested to get the answer. However, I still fall short. I've edited my question to reflect my progress so far. Could you advise please. — koifish, Aug 27 '18 at 12:42
@YipJungHon $I(a) = I(0) + \int_0^a I'(t) \mathrm dt ; [a \in (-1,1)]$. Maybe check your calculation would find out the miss or the answer is wrong… — xbh, Aug 27 '18 at 12:48
Hmm...I seemed to have missed out the $I(0)$ part. I'll try it tomorrow and see if it works — koifish, Aug 27 '18 at 12:54
Possible duplicate of Verification the integral identity $\int_0^\pi \ln(1+\alpha\cos(x))dx=\pi\ln\left(\frac{1+\sqrt{1-\alpha^2}}{2}\right)$. — tarit goswami, Aug 27 '18 at 14:17
In my notes you can find a simple proof through Riemann sums. — Jack D'Aurizio, Aug 27 '18 at 15:01

Prove that $\int_0^\pi \ln(1+\alpha \cos x)\, dx= \pi \ln\left(\frac{1+\sqrt{1-\alpha^2}}{2}\right)$

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