Surprisingly concise answer

I did a long solution, not sure if it's the smartest one. Still:

$\large \displaystyle I = \int_0^{\frac{\pi}{2}} \ln(49 \sin^2 \theta + 81 \cos^2 \theta) d \theta$

$\large \displaystyle I = \int_0^{\frac{\pi}{2}} \ln(49 \sin^2 \theta + 49 \cos^2 \theta + 32 \cos^2 \theta) d \theta$

$\large \displaystyle I = \int_0^{\frac{\pi}{2}} \ln(49 + 32 \cos^2 \theta) d \theta$

$\large \displaystyle I = \int_0^{\frac{\pi}{2}} \ln \left [ (49) \left (1 + \frac{32}{49} \cos^2 \theta \right ) \right ] d \theta$

$\large \displaystyle I = \int_0^{\frac{\pi}{2}} \ln (49) d\theta + \int_0^{\frac{\pi}{2}} \ln \left (1 + \frac{32}{49} \cos^2 \theta \right ) d \theta$

From the Maclaurin Series of $\ln(1+x), |x| \leq 1, x \neq -1$ :

$\large \displaystyle I = \frac{\pi}{2} \ln(7^2) + \int_0^{\frac{\pi}{2}} \sum_{k=1}^{\infty} \frac{(-1)^{k+1} \left( \frac{32 \cos^2 \theta}{49} \right )^k }{k} d \theta$

$\large \displaystyle I = \frac{\pi}{2} \ln(7^2) - \int_0^{\frac{\pi}{2}} \sum_{k=1}^{\infty} \frac{ \left( - \frac{32 \cos^2 \theta}{49} \right )^k }{k} d \theta$

$\large \displaystyle I = \pi \ln(7) - \sum_{k=1}^{\infty} \frac{ \left(- \frac{32}{49} \right )^k }{k} \int_0^{\frac{\pi}{2}} \cos^{2k} \theta d\theta$

$\large \displaystyle I = \pi \ln(7) - \sum_{k=1}^{\infty} \frac{ \left(- \frac{32}{49} \right )^k }{k} \cdot \frac12 \cdot B \left ( \frac12, k+\frac12 \right )$

Where $B(x,y)$ is the Beta Function , which is equal to:

$\large \displaystyle I = \pi \ln(7) - \frac12 \sum_{k=1}^{\infty} \frac{ \left(- \frac{32}{49} \right )^k }{k} \cdot \frac{ \Gamma \left( \frac12 \right ) \Gamma \left( k + \frac12 \right ) }{\Gamma (k+1) }$

Where $\Gamma(x)$ is the Gamma Function . We have $\Gamma(k+1) = k!$ , $\Gamma(\frac12) = \sqrt{\pi}$ and $\Gamma(x+1) = x\Gamma(x)$ , which leads to the following property of $\Gamma(k+\frac12)$ :

$\large \displaystyle I = \pi \ln(7) - \frac12 \sum_{k=1}^{\infty} \frac{ \left( - \frac{32}{49} \right )^k }{k} \cdot \frac{\sqrt{\pi}}{k!} \cdot \frac{1 \times 3 \times 5 \times ... \times (2k-1)}{2^k} \sqrt{\pi}$

$\large \displaystyle I = \pi \ln(7) - \frac{\pi}{2} \sum_{k=1}^{\infty} \frac{ \left( - \frac{32}{49} \right )^k }{k} \cdot \frac{1}{k!} \cdot \frac{1 \times 3 \times 5 \times ... \times (2k-1)}{2^k} \cdot \frac{2 \times 4 \times 6 \times ... \times (2k)}{2 \times 4 \times 6 \times ... \times (2k)}$

$\large \displaystyle I = \pi \ln(7) - \frac{\pi}{2} \sum_{k=1}^{\infty} \frac{ \left( - \frac{32}{49} \right )^k }{k} \cdot \frac{1}{k!} \cdot \frac{(2k)!}{4^k k!}$

$\large \displaystyle I = \pi \ln(7) - \frac{\pi}{2} \sum_{k=1}^{\infty} \frac{ \left( - \frac{8}{49} \right )^k }{k} {2k \choose k}$

Now recall that:

$\large \displaystyle \sum_{k=1}^{\infty} {2k \choose k} x^k = \frac{1}{\sqrt{1-4x}} -1, |x| < \frac14$

Dividing by $x$ and then Integrating with respect to $x$ from $0$ to $x$ on both sides (See NOTE below for further details):

$\large \displaystyle \sum_{k=1}^{\infty} {2k \choose k} \frac{x^k}{k} = 2 \left [ \ln(2) - \ln \left( \sqrt{1-4x} + 1 \right ) \right ], |x| < \frac14$

So:

$\large \displaystyle I = \pi \ln(7) - \frac{\pi}{2} \cdot 2 \cdot \left [ \ln(2) - \ln \left( \sqrt{1-4 \left ( -\frac{8}{49} \right ) } + 1 \right ) \right ]$

$\large \displaystyle I = \pi \ln(7) - \pi \left [ \ln(2) - \ln \left( \frac{16}{7} \right ) \right ]$

$\large \displaystyle I = \pi \ln(7) - \pi\ln(2) + \pi \ln(16) - \pi \ln(7)$

$\large \displaystyle I = \pi \ln(8)$

$\color{#20A900} \boxed{\large \displaystyle I = 3 \pi \ln(2)}$

So:

$\color{#3D99F6} \large \displaystyle a = 3, b = 2, \boxed{\large \displaystyle a+b=5}$

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NOTE: The integration steps (for $|x| < \frac14$ ):

From Maclaurin Series :

$\large \displaystyle (1-4x)^{-\frac12} = \frac{1}{0!}(1-4x)^{-\frac12} |_{x=0} +\frac{1}{1!} \left ( -\frac12 \cdot -4 \right ) (1-4x) ^{-\frac32} |_{x=0} \cdot x + \frac{1}{2!}\left ( -\frac12 \cdot - \frac32 \cdot (-4)^2 \right ) (1-4x) ^{-\frac52} |_{x=0} \cdot x^2 + ...$

$\large \displaystyle (1-4x)^{-\frac12} = \sum_{k=0}^{\infty} \frac{4^k}{2^k \cdot k!} 1 \times 3 \times 5... \times (2k-1) \cdot x^k$

$\large \displaystyle (1-4x)^{-\frac12} = \sum_{k=0}^{\infty} \frac{2^k}{k!} 1 \times 3 \times 5... \times (2k-1) \cdot \frac{ 2 \times 4 \times 6 \times ... \times 2k}{ 2 \times 4 \times 6 \times ... \times 2k} \cdot x^k$

$\large \displaystyle (1-4x)^{-\frac12} = \sum_{k=0}^{\infty} \frac{2^k}{k!} \frac{(2k)!}{2^k k!} x^k$

$\large \displaystyle (1-4x)^{-\frac12} = \sum_{k=0}^{\infty} \frac{(2k)!}{k! k!} x^k$

$\large \displaystyle (1-4x)^{-\frac12} = \sum_{k=0}^{\infty}{ 2k \choose k} x^k$

$\large \displaystyle (1-4x)^{-\frac12} - 1 = \sum_{k=1}^{\infty}{ 2k \choose k} x^k$

Now dividing by $x$ on both sides:

$\large \displaystyle \sum_{k=1}^{\infty}{ 2k \choose k} x^{k-1} = \frac{1}{x \sqrt{1-4x}} - \frac{1}{x}$

Integrating on both sides:

$\large \displaystyle \int \sum_{k=1}^{\infty}{ 2k \choose k} x^{k-1} = \int \frac{1}{x \sqrt{1-4x}} dx - \int \frac{1}{x} dx$

On the first integral, make $u = 1-4x$ , then $x = \frac{1-u}{4}$ and $dx = - \frac{du}{4}$ . ( $C_i$ is an arbitrary constant:)

$\large \displaystyle \sum_{k=1}^{\infty}{ 2k \choose k} \frac{x^k}{k} + C_1 = -\int \frac{1}{(1-u) \sqrt{u}} du - \ln(x) + C_2$

Now make $t = \sqrt{u}$ , $u = t^2$ and $\frac{du}{\sqrt{u}} = 2dt$ :

$\large \displaystyle \sum_{k=1}^{\infty}{ 2k \choose k} \frac{x^k}{k} + C_1 = -\left ( \int \frac{2}{1-t^2} dt + \ln(x) \right ) + C_2$

Expanding in partial fractions:

$\large \displaystyle \sum_{k=1}^{\infty}{ 2k \choose k} \frac{x^k}{k} + C_1 = -\left [ \int \left ( \frac{1}{1+t} + \frac{1}{1-t} \right ) dt + \ln(x) \right ] + C_2$

$\large \displaystyle \sum_{k=1}^{\infty}{ 2k \choose k} \frac{x^k}{k} + C_1 = -\left [ \ln(1+t) -\ln(1-t) + \ln(x) \right ] + C_2$

Since $t = \sqrt{u} = \sqrt{1-4x}$ :

$\large \displaystyle \sum_{k=1}^{\infty}{ 2k \choose k} \frac{x^k}{k} + C_1 = - \ln \left ( \frac{x(1 + \sqrt{1-4x})}{1 - \sqrt{1-4x}} \right ) + C_2$

$\large \displaystyle \sum_{k=1}^{\infty}{ 2k \choose k} \frac{x^k}{k} + C_1 = - \ln \left ( \frac{x(1 + \sqrt{1-4x})(1 + \sqrt{1-4x})}{(1 - \sqrt{1-4x})(1 + \sqrt{1-4x})} \right ) + C_2$

$\large \displaystyle \sum_{k=1}^{\infty}{ 2k \choose k} \frac{x^k}{k} + C_1 = - \ln \left ( \frac{x \left (1 + \sqrt{1-4x} \right )^2}{4x} \right ) + C_2$

$\large \displaystyle \sum_{k=1}^{\infty}{ 2k \choose k} \frac{x^k}{k} + C_1 = - \ln \left ( \frac{ \left (1 + \sqrt{1-4x} \right )^2}{4} \right ) + C_2$

$\large \displaystyle \sum_{k=1}^{\infty}{ 2k \choose k} \frac{x^k}{k} + C_1 = - \ln \left ( \left (1 + \sqrt{1-4x} \right )^2\right ) + \ln(4) + C_2$

$\large \displaystyle \sum_{k=1}^{\infty}{ 2k \choose k} \frac{x^k}{k} + C_1 = -2 \ln \left (1 + \sqrt{1-4x}\right ) + C_3$

Making the limits of integration from $0$ to $x$ :

$\color{#20A900} \boxed{\large \displaystyle \sum_{k=1}^{\infty}{ 2k \choose k} \frac{x^k}{k} = 2 \left [ \ln(2) - \ln \left (1 + \sqrt{1-4x}\right ) \right ] }$