Mathematics

The value of $$\displaystyle \int_0^{\cfrac {\pi}{2}}\log \left (\frac {4+3 \sin x}{4+3 \cos x}\right )dx$$ is


ANSWER

$$0$$


SOLUTION
Let $$\displaystyle I=\int_0^{\frac {\pi}{2}}\log \left (\frac {4+3 \sin x}{4+3 \cos x}\right )dx$$ .............. (1)
$$\Rightarrow\displaystyle I=\int_0^{\frac {\pi}{2}}log \left [\frac {4+3\sin \left (\frac {\pi}{2}-x\right )}{4+3 \cos \left (\frac {\pi}{2}-x\right )}\right ]dx,  \; (\because \int_0^a f(x) dx=\int_0^af(a-x)dx)$$
$$\Rightarrow\displaystyle I=\int_0^{\frac {\pi}{2}}\log \left (\frac {4+3 \cos x}{4+3 \sin x}\right )dx$$ ........... (2)
Adding (1) and (2), we obtain
$$2I\displaystyle =\int_0^{\frac {\pi}{2}}\left \{\log \left (\frac {4+3 \sin x}{4+3 \cos x}\right )+\log \left (\frac {4+3 \cos x}{4+3 \sin x}\right )\right \}dx$$
$$\Rightarrow\displaystyle 2I=\int_0^{\frac {\pi}{2}}\log \left (\frac {4+3 \sin x}{4+3 \cos x}\times \frac {4+3 \cos x}{4+3 \sin x}\right )dx$$
$$\Rightarrow 2I\displaystyle=\int_0^{\frac {\pi}{2}}\log 1 dx=\int_0^{\frac {\pi}{2}}0 dx$$
$$\Rightarrow I=0$$
Hence, the correct Answer is C.
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Single Correct Medium Published on 17th 09, 2020
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