The value of ∫^(π)/2_0log≤ft((5+4sinx)/(5+4cosx))dx is :

The correct answer is Option (1) → 0

$I=\int\limits^{\frac{\pi}{2}}_{0}\log\left(\frac{5+4\sin x}{5+4\cos x}\right)dx$  ...(1)

$I=\int\limits^{\frac{\pi}{2}}_{0}\log\left(\frac{5+4\sin(\frac{\pi}{2}-x)}{5+4\cos(\frac{\pi}{2}-x)}\right)dx=\int\limits^{\frac{\pi}{2}}_{0}\log\left(\frac{5+4\cos x}{5+4\sin x}\right)dx$  ...(2)

Adding (1) and (2)

$2I=\int\limits^{\frac{\pi}{2}}_{0}\log\left(\frac{5+4\sin x}{5+4\cos x}×\frac{5+4\cos x}{5+4\sin x}\right)dx$

$2I=\int\limits^{\frac{\pi}{2}}_{0}\log 1dx=\int\limits^{\frac{\pi}{2}}_{0}0dx=0⇒I=0$