Mathematics

The value of $$\int _{ 0 }^{ \dfrac { \pi  }{ 2 }  }{ \log { \left( \dfrac { 4+3\sin { x }  }{ 4+3\cos { x }  }  \right) dx }  }$$ is


ANSWER

$$0$$


SOLUTION
As,
$$\begin{array}{l}\int_0^a {f\left( x \right)} dx = \int_0^a {f\left( {x - a} \right)} dx\\So,\\I = \int_0^{\frac{\pi }{2}} {\log \left( {\frac{{4 + 3\sin x}}{{4 + 3\cos x}}} \right)} \\ = \int_0^{\frac{\pi }{2}} {\log \left( {\frac{{4 + 3\sin \left( {\frac{\pi }{2} - x} \right)}}{{4 + 3\cos \left( {\frac{\pi }{2} - x} \right)}}} \right)} \\ = \int_0^{\frac{\pi }{2}} {\log \left( {\frac{{4 + 3\cos x}}{{4 + 3\sin x}}} \right)} \end{array}$$
Adding the above two integral,
$$\begin{array}{l}I + I = \int_0^{\frac{\pi }{2}} {\log \left( {\frac{{4 + 3\sin x}}{{4 + 3\cos x}}} \right)} dx + \int_0^{\frac{\pi }{2}} {\log \left( {\frac{{4 + 3\cos x}}{{4 + 3\sin x}}} \right)} dx\\2I = \int_0^{\frac{\pi }{2}} {\left[ {\log \left( {\frac{{4 + 3\sin x}}{{4 + 3\cos x}}} \right) + \log \left( {\frac{{4 + 3\cos x}}{{4 + 3\sin x}}} \right)} \right]dx} \\As,\\\log a + \log b = \log ab\\So,\\2I = \int_0^{\frac{\pi }{2}} {\log \left[ {\left( {\frac{{4 + 3\sin x}}{{4 + 3\cos x}}} \right) \times \left( {\frac{{4 + 3\cos x}}{{4 + 3\sin x}}} \right)} \right]dx} \\ = \int_0^{\frac{\pi }{2}} {\log \left( 1 \right)dx} \\ = \int_0^{\frac{\pi }{2}} {0dx} \\2I = 0\\I = 0\end{array}$$

Thus Option C


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