May not be substituted?

the given integral is

$\large{\displaystyle \int^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \ln(49 \cos^2 x+9 \sin^2 x) dx}$

Now

$\large{I(a)=\displaystyle \int^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \ln(a^2 \cos^2 x+b^2 \sin^2 x)dx}$

differentiating wrt $a$

$\large{I^{\prime}(a)=\displaystyle \int^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \frac{2a \cos^2 x}{a^2 \cos^2 x+b^2 \sin^2 x}}$

taking $\tan x=t$

$\large{I^{\prime}(a)=\frac{2a}{b^2-a^2} \left ( \displaystyle \int^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\frac{b^2}{a^2+b^2 t^2}dt-\displaystyle \int^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\frac{1}{1+t^2}dt \right)}$

$\large{I^{\prime}(a)=\frac{2 \pi}{a+b}}$

$\large{I(a)=2 \pi \ln (a+b)+c}$

for $a=b=1$ we get $c=-2 \pi \ln 2$

$\large{I(a)=2 \pi \ln \left( \frac{a+b}{2} \right)}$