The value of $\int_0^{\pi/2} \log_e \left(\frac{5 + 2 \sin x}{5 + 2 \cos x}\right) dx$ is

Let $I = \int_0^{\pi/2} \ln \left(\frac{5 + 2 \sin x}{5 + 2 \cos x}\right) dx$

Note: $\log_e$ is the same as $\ln$ (natural logarithm)

---

Using the property $\ln\left(\frac{A}{B}\right) = \ln A - \ln B$:

$I = \int_0^{\pi/2} [\ln(5 + 2 \sin x) - \ln(5 + 2 \cos x)] \, dx$

---

Applying the King's Property: $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$

Replace $x$ with $\left(\frac{\pi}{2} - x\right)$:

$I = \int_0^{\pi/2} \left[\ln(5 + 2 \sin(\pi/2 - x)) - \ln(5 + 2 \cos(\pi/2 - x))\right] dx$

---

Using the trigonometric identities:

$\sin(\pi/2 - x) = \cos x$

$\cos(\pi/2 - x) = \sin x$

Substituting:

$I = \int_0^{\pi/2} [\ln(5 + 2 \cos x) - \ln(5 + 2 \sin x)] \, dx$

$I = \int_0^{\pi/2} \ln \left(\frac{5 + 2 \cos x}{5 + 2 \sin x}\right) dx$

---

The transformed integral is the negative of the original integral.

Original: $I = \int_0^{\pi/2} \ln \left(\frac{5 + 2 \sin x}{5 + 2 \cos x}\right) dx$

After substitution: $I = \int_0^{\pi/2} \ln \left(\frac{5 + 2 \cos x}{5 + 2 \sin x}\right) dx = -I$

---

Since $I = -I$:

$I + I = 0$

$2I = 0$

$I = 0$

Therefore, the value of the integral is $0$.