The value of ∫limits_0^π/2log_e≤ft((5+2sin x)/(5+2cos x))dx is

The correct answer is Option (1) → 0

$I=\int_{0}^{\frac{\pi}{2}}\log_{e}\!\left(\frac{5+2\sin x}{5+2\cos x}\right)\,dx$

$I=\int_{0}^{\frac{\pi}{2}}\log(5+2\sin x)\,dx\;-\;\int_{0}^{\frac{\pi}{2}}\log(5+2\cos x)\,dx$

In the second integral use $x=\frac{\pi}{2}-t$:

$\cos x=\sin t,\;dx=-dt$

$\int_{0}^{\frac{\pi}{2}}\log(5+2\cos x)\,dx =\int_{\frac{\pi}{2}}^{0}\log(5+2\sin t)(-dt) =\int_{0}^{\frac{\pi}{2}}\log(5+2\sin t)\,dt$

Both integrals are equal, therefore their difference is $0$.

The value of the integral is $0$.