Solve it !!!

$Let\quad { I }_{ n }=\int _{ 0 }^{ \pi /2 }{ \log { \frac { 4+3\sin { x } }{ 4+3\cos { x } } } } dx\\ then\quad by\quad \int _{ b }^{ a }{ f\left( x \right) dx } =\int _{ b }^{ a }{ f\left( a+b-x \right) dx } \\ we\quad get\quad { I }_{ n }=\int _{ 0 }^{ \pi /2 }{ \log { \frac { 4+3\sin { \pi /2-x } }{ 4+3\cos { \pi /2-x } } } } dx\\ and\quad can\quad be\quad written\quad as\quad { I }_{ n }=\int _{ 0 }^{ \pi /2 }{ \log { \frac { 4+3\cos { x } }{ 4+3\sin { x } } } } dx\\ and\quad 2{ I }_{ n }={ I }_{ n }+{ I }_{ n }\\ \quad \quad \quad \quad \quad \quad \quad =\int _{ 0 }^{ \pi /2 }{ \log { \frac { 4+3\cos { x } }{ 4+3\sin { x } } } } dx+\int _{ 0 }^{ \pi /2 }{ \log { \frac { 4+3\sin { x } }{ 4+3\cos { x } } } } dx\\ \quad \quad \quad \quad \quad \quad \quad \quad =\int _{ 0 }^{ \pi /2 }{ (\log { \frac { 4+3\cos { x } }{ 4+3\sin { x } } } -\log { \frac { 4+3\cos { x } }{ 4+3\sin { x } } ) } dx } \\ \quad \quad \quad \quad \quad \quad \quad \quad =\int _{ 0 }^{ \pi /2 }{ 0 } dx\\ \quad \quad \quad \quad \quad \quad \quad \quad =0$

Consider the symmetry: The integrand may be written as $I(x) = \log(4 + 3\sin x) - \log(4 + 3\cos x).$ This function is symmetric under mirroring $x' =\pi/2 - x$ : $I(x') = I(\pi/2 - x) = \log(4 + 3\cos x) - \log(4 + 3\sin x) = -I(x).$ Thus we may split the integral as $\int_0^{\pi/4} I(x)\ dx + \int_{\pi/4}^{\pi/2} I(x)\ dx \\ = \int_0^{\pi/4} I(x)\ dx - \int_0^{\pi/4} I(x')\ dx' \\ = \int_0^{\pi/4} (I(x) - I(x))\ dx = \boxed{0}.$