Can you figure this out (4)

Lets consider the complex notation of $\cos x = \dfrac {e^{ix} + e^{-ix}}{2}$

Now our integral transforms to $I = \displaystyle \int_0^{\frac π2} \ln(e^{ix} + e^{-ix})dx = \displaystyle \int_0^{\frac π2} \ln(e^{ix}(1 + e^{-2ix}) dx = {\color{#20A900}{\displaystyle \int_0^{\frac π2} \ln(e^{ix})dx}} + {\color{#D61F06}{\displaystyle \int_0^{\frac π2} \ln(1+ e^{-2ix}) dx}} = {\color{#20A900}{I_1}} + {\color{#D61F06}{I_2}}$

Now, ${\color{#20A900}{I_1 = \displaystyle \int_0^{\frac π2} (ix) dx = [\dfrac {ix^2}{2}]_0^{\frac π2} = \dfrac {iπ^2}{8} = I_1}}$

Consider Taylor Series for $\ln(1 + x) = x - \dfrac {x^2}{2} + \dfrac {x^3}{3} - \dfrac {x^4}{4} + ........= \displaystyle \sum_{n=1}^\infty \dfrac {(-1)^{n+1} x^n}{n}$

Now, $\ln(1+ e^{-2ix}) = e^{-2ix} - \dfrac {e^{(-2ix)2}}{2} + \dfrac {e^{(-2ix)3}}{3} -........$

Now, ${\color{#D61F06}{\displaystyle \int_0^{\frac π2} \ln( 1 + e^{-2ix})dx = \dfrac {1}{-2i} [e^{-2ix} - \dfrac {e^{(-2ix)2}}{2^2} + \dfrac {e^{(-2ix)3}}{3^2} - .......]_0^{\frac π2} = [S]_0^{\frac π2}}}$

Observe, $S_{x={\frac π2}} = \dfrac {1}{-2i} [ e^{-iπ} - \dfrac {e^{-2iπ}}{2^2} + \dfrac {e^{-3iπ}}{3^2} - ........= \dfrac {1}{-2i} [ -1 - \dfrac {1}{2^2} + \dfrac {-1}{3^2} - \dfrac {1}{4^2} + .........]$

Now $S_{x=0} = \dfrac {1}{-2i}[ 1 - \dfrac {1}{2^2} + \dfrac {1}{3^2} - \dfrac {1}{4^2} + .........]$

We know that, ${\color{#D61F06}{S = S_{\frac π2} - S_0 = \dfrac {-i}{1} [ \dfrac {1}{1^2} + \dfrac {1}{3^2} + \dfrac {1}{5^2} +.......] = \dfrac {-iπ^2}{8} = I_2}}$

$I = I_1 + I_2 = \dfrac {(0)iπ^2}{8} = 0$

Hence, $a= \boxed{0}$