Where does the Integral went wrong .... $ \displaystyle \int _{0}^{\pi} \frac {\ln ( 1 + \sin \alpha \cos x )}{\cos x} \ \mathrm {d}x $ This is how I solved :
Let $$ \displaystyle y = \int_{0}^{\pi} \frac {\ln ( 1 + \sin \alpha  \cos x )}{\cos x} \ \mathrm {d}x 
 $$
Then I used expansion of $\displaystyle \ln (1+x) = \sum _{n=0}^{\infty} (-1)^n \frac {x^{n+1}}{n+1} $
Now, $\displaystyle y = \sum _{n=0}^{\infty} (-1)^n \frac {\sin ^{n+1}  \alpha }{n+1} \int_{0}^{\pi} \cos ^n x \ \mathrm {d}x \implies y = \sum _{n=0}^{\infty} [1+(-1)^n] \frac {\sin ^{n+1}  \alpha }{n+1} \int_{0}^{\frac {\pi}{2} } \cos ^n x \ \mathrm {d}x $
Replacing $n$ by $2k$ (as we will end up with $0$ for odd $n$) and using gamma function  :
$$ \displaystyle y =  2 \sum _{k=0}^{\infty} \frac {\sin ^{2k+1}  \alpha }{2k+1} \cdot \frac { \Gamma \left ( \frac {2k+1}{2} \right ) \sqrt {\pi} }{ \Gamma \left ( \frac {2k+2}{2} \right )} $$
Which can be simplified to $\displaystyle y =  2 \pi \sum _{k=0}^{\infty} \frac {\sin ^{2k+1} \alpha }{2k+1} \cdot \frac {(2k-1)!!}{k! \cdot 2^k}$
Modification :
$$ \displaystyle y = 2 \pi \sum _{k=0}^{\infty} \frac {\sin ^{2k+1} \alpha }{2k+1} \cdot \frac {(2k-1)!!}{k! \cdot 2^k} \cdot \frac {k! 2^k}{k! 2^k} \implies y = 2 \pi \sum _{k=0}^{\infty}  \frac {\sin ^{2k+1} \alpha }{2k+1} \binom {2k}{k} \frac {1}{4^k} $$
We get : $\displaystyle \frac {\mathrm {d}y }{\mathrm {d} ( \sin \alpha ) } = 2 \pi \sum _{k=0}^{\infty} \binom {2k}{k}  \left ( \frac {\sin ^{2} \alpha}{4} \right ) ^k$
As $\displaystyle (1-4x)^{-1/2} = \sum _{n=0}^{\infty}  \binom {2n}{n} x^n$
We will finally get : $\displaystyle y = 2 \pi \int \frac {\mathrm {d} (\sin \alpha )}{ \cos \alpha }   \implies  y = 2 \pi \alpha + C $
Now as $\alpha = 0 $ the summation is zero $\implies C = 0  $
Hence,
$$ \displaystyle \int_{0}^{\pi} \frac {\ln ( 1 + \sin \alpha  \cos x )}{\cos x} \ \mathrm {d}x = 2 \pi \alpha $$
But : 
This suggests : $ \displaystyle \int_{0}^{\pi} \frac {\ln ( 1 + \sin \alpha  \cos x )}{\cos x} \ \mathrm {d}x = \pi \alpha $

Your mistake is thinking $\int_0^{\pi/2}\cos^nx\mathrm{d}x$ is a Beta function; it's only half of one, viz.$$\operatorname{B}(a,\,b)=2\int_0^{\pi/2}\sin^{2a-1}x\cos^{2b-1}x\mathrm{d}x.$$
