Having trouble of finding this integral in the standard integral list I am facing difficult problem to deal with this integral 
$$\int_{0}^{\pi/2}\ln(9-4\cos^2\theta)\,\mathrm{d}\theta \tag*{(1)}$$
note $(3)^2-(2\cos{\theta})^2=(3-2\cos{\theta})(3+2\cos{\theta}) \tag*{(2)}$
also note that, $$\log(AB)=\log A+\log B \tag*{(3)}$$ using $(2)$ and $(3)$ we have,
$$\int_{0}^{\pi/2}\ln(3-2\cos\theta)\,d\theta+\int_{0}^{\pi/2}\ln(3+2\cos\theta)\,\mathrm{d}\theta$$
Looking through my table of standard integrals I couldn't find the result for this integral$\int \ln(a+b\cos{\theta})$
Standard table of integral
Integral (44) (in the table) has the closest of resemblance
$$\int \ln(a\theta+b)\,d\theta=(\theta+{b\over a})\ln(a\theta+b)-\theta$$
In the original problem is $\cos{\theta}$ and not $\theta$
I am facing quite a problem here, I need some help. 
Very much appreciated if you can help.
 A: Here a slightly different approach will be considered. We will make use of the generating function for the central binomial coefficients of
\begin{align*}
\sum_{n=0}^{\infty}\binom{2n}{n}x^n=\frac{1}{\sqrt{1-4x}}, \qquad |x|<\frac{1}{4}, \tag1
\end{align*}
a result that follows directly from the binomial series.

Let
$$I = \int^{\pi/2}_0 \ln (9 - 4 \cos^2 \theta) \, d\theta.$$
After rearranging the argument for the log function we can write
\begin{align*}
I &= \int_0^{\pi/2} \ln \left [9 \left (1 - \frac{4}{9} \cos^2 \theta \right ) \right ] \, d\theta\\
&= \pi \ln (3) + \int_0^{\pi/2} \ln \left (1 - \frac{4}{9} \cos^2 \theta \right ) \, d\theta\\
&= \pi \ln (3) + I_\alpha.
\end{align*}
Making use of the following Maclaurin series expansion for $\ln (1 - x)$, namely
$$\ln (1 - x) = -\sum_{n = 1}^\infty \frac{x^n}{n}, \qquad |x| < 1,$$
we have, after interchanging the infinite sum with the integration
$$I_\alpha =  -\sum_{n = 1}^\infty \frac{4^n}{n 9^n} \int^{\pi/2}_0 \cos^{2n} \theta \, d\theta.$$
The integral that appears here can be evaluated using a Beta function. Here
\begin{align*}
\int_0^{\pi/2} \cos^{2n} \theta \, d\theta &= \frac{1}{2} \cdot 2 \int_0^{\pi/2} \cos^{2(n + \frac{1}{2}) - 1}\theta \sin^{2(\frac{1}{2}) - 1} \theta \, d\theta\\
&= \frac{1}{2} \text{B} \left (n + \frac{1}{2}, \frac{1}{2} \right )\\
&= \frac{1}{2} \frac{\Gamma (n + 1/2) \Gamma (1/2)}{\Gamma (n + 1)}.
\end{align*}
Since $n \in \mathbb{N}$, we have
$$\Gamma (n + 1) = n! \quad {\rm and} \quad \Gamma \left (n + \frac{1}{2} \right ) = \frac{\sqrt{\pi}}{2^{2n}} \frac{(2n)!}{n!},$$
giving
$$\int_0^{\pi/2} \cos^{2n} \theta \, d\theta = \frac{\pi}{2} \cdot \frac{1}{2^{2n}} \frac{(2n)!}{(n!)^2},$$
Thus
$$I_\alpha = - \frac{\pi}{2} \sum_{n = 1}^\infty \binom{2n}{n} \frac{1}{n 9^n},$$
and yields
$$I = \pi \ln (3) - \frac{\pi}{2} \sum_{n = 1}^\infty \binom{2n}{n} \frac{1}{n 9^n}. \tag2$$

To find the sum in (2) we write the generating function for the central binomial coefficients given by (1) as
$$1 + \sum_{n = 1}^\infty \binom{2n}{n} x^n = \frac{1}{\sqrt{1 - 4x}}.$$
Dividing the above expression by $x$ before integrating it from zero to $x$ we have
\begin{align*}
\sum_{n = 1}^\infty \binom{2n}{n} \int_0^x t^{n - 1} \, dt &= 
 \int_0^x \frac{1}{t} \left (\frac{1}{\sqrt{1 - 4t}} - 1 \right ) \, dt\\
\sum_{n = 1}^\infty \binom{2n}{n} \left [\frac{t^n}{n} \right ]^x_0 &= \int_0^x \frac{1 - \sqrt{1 - 4t}}{t \sqrt{1 - 4t}} \, dt\\
\sum_{n = 1}^\infty \binom{2n}{n} \frac{x^n}{n} &= \int_0^x \frac{1 - \sqrt{1 - 4t}}{t \sqrt{1 - 4t}} \, dt.
\end{align*}
The remaining integral can be found by enforcing a substitution of $x \mapsto \frac{1}{4} \cos^2 x$. When this is done we find
$$\sum_{n = 1}^\infty \binom{2n}{n} \frac{x^n}{n} = 2 \ln \left (\frac{2}{1 + \sqrt{1 - 4x}} \right ), \quad |x| < \frac{1}{4}.$$
If we set $x = 1/9$ in the above generating function we find
$$\sum_{n = 1}^\infty \binom{2n}{n} \frac{1}{n 9^n} = 2 \ln \left (\frac{6}{3 + \sqrt{5}} \right ),$$
and is the required sum we were after.

Returning to (2) we find
$$I = \pi \ln (3) - \pi \ln \left (\frac{6}{3 + \sqrt{5}} \right ),$$
or
$$\int_0^{\pi/2} \ln (9 - 4 \cos^2 \theta) \, d\theta = \pi \ln \left (\frac{3 + \sqrt{5}}{2} \right ).$$
A: Do you know differentiation under the integral sign, also known as Feynman's trick? It is pretty useful here. For any $a\in(-1,1)$, let
$$ F(a) = \int_{0}^{\pi/2}\log(1-a\cos^2\theta)\,d\theta\stackrel{\theta\mapsto\arctan u}{=} \int_{0}^{+\infty}\frac{\log\left(1-\frac{a}{1+u^2}\right)}{1+u^2}\,du. $$
We have
$$ F'(a) = -\int_{0}^{+\infty}\frac{du}{(1-a+u^2)(1+u^2)}\stackrel{\text{PFD}}{=}-\frac{\pi}{2}\cdot\frac{1}{(1-a)+\sqrt{1-a}} $$
and since $F(0)=0$, 
$$ F(a)=-\frac{\pi}{2}\int_{0}^{a}\frac{dv}{(1-v)+\sqrt{1-v}}=\color{red}{\pi\log\left(\tfrac{1+\sqrt{1-a}}{2}\right)}.$$
Now just plug in $a=\frac{4}{9}$. You'll get that the wanted integral equals $\color{red}{2\pi\log\varphi}$ where $\varphi=\frac{1+\sqrt{5}}{2}$ is the golden ratio.

There also is an approach through trigonometric identities and Riemann sums, have a look at page 48 of my notes.
