The value of $\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{4+3\sin x}{4+3\cos x} \right)dx}$ is
$2$
$\dfrac{3}{4}$
$0$
$-2$
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Ans: Let $I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{4+3\sin x}{4+3\cos x} \right)dx}$ …..(1)
Since,
$\int\limits_{0}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}$
Therefore,
$\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{4+3\sin \left( \dfrac{\pi }{2}-x \right)}{4+3\cos \left( \dfrac{\pi }{2}-x \right)} \right)dx}$
$\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{4+3\cos x}{4+3\sin x} \right)dx}$ …….(2)
On adding equation 1 and 2-
$2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\log 1dx}$
$\Rightarrow 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{0dx}$
$\Rightarrow I=0$
Hence, the correct option is C.