Calculate Value of Expression Involving Definite Integral

To find the value of $27 I^2$, we begin by analyzing the given integral:

$I = \frac{2}{\pi} \int_{-\pi/4}^{\pi/4} \frac{dx}{(1 + e^{\sin x})(2 - \cos 2x)}$

Let $J = \int_{-\pi/4}^{\pi/4} \frac{dx}{(1 + e^{\sin x})(2 - \cos 2x)}$, so that $I = \frac{2}{\pi} J$.

Using the property of definite integrals $\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx$, with $a = -\pi/4$ and $b = \pi/4$, we replace $x$ with $-x$:

$J = \int_{-\pi/4}^{\pi/4} \frac{dx}{(1 + e^{-\sin x})(2 - \cos(-2x))}$

Since $\cos(-2x) = \cos(2x)$, we rewrite $1 + e^{-\sin x}$ as $1 + \frac{1}{e^{\sin x}} = \frac{e^{\sin x} + 1}{e^{\sin x}}$:

$J = \int_{-\pi/4}^{\pi/4} \frac{e^{\sin x} \, dx}{(1 + e^{\sin x})(2 - \cos 2x)}$

Adding the two expressions for $J$:

$2J = \int_{-\pi/4}^{\pi/4} \frac{1 + e^{\sin x}}{(1 + e^{\sin x})(2 - \cos 2x)} \, dx$

$2J = \int_{-\pi/4}^{\pi/4} \frac{dx}{2 - \cos 2x}$

Since the integrand $\frac{1}{2 - \cos 2x}$ is an even function of $x$, we can simplify the integral bounds:

$2J = 2 \int_{0}^{\pi/4} \frac{dx}{2 - \cos 2x} \implies J = \int_{0}^{\pi/4} \frac{dx}{2 - \cos 2x}$

Using the trigonometric identity $\cos 2x = \frac{1 - \tan^2 x}{1 + \tan^2 x}$:

$2 - \cos 2x = 2 - \frac{1 - \tan^2 x}{1 + \tan^2 x} = \frac{2(1 + \tan^2 x) - (1 - \tan^2 x)}{1 + \tan^2 x} = \frac{1 + 3\tan^2 x}{1 + \tan^2 x}$

Since $1 + \tan^2 x = \sec^2 x$, the integral becomes:

$J = \int_{0}^{\pi/4} \frac{\sec^2 x \, dx}{1 + 3\tan^2 x}$

Let $t = \tan x$, so $dt = \sec^2 x \, dx$.

• When $x = 0$, $t = 0$.
• When $x = \pi/4$, $t = 1$.

Substituting $t$ into the integral:

$J = \int_{0}^{1} \frac{dt}{1 + (\sqrt{3}t)^2} = \left[ \frac{1}{\sqrt{3}} \tan^{-1}(\sqrt{3}t) \right]_{0}^{1}$

$J = \frac{1}{\sqrt{3}} \left( \tan^{-1}(\sqrt{3}) - \tan^{-1}(0) \right) = \frac{1}{\sqrt{3}} \left( \frac{\pi}{3} - 0 \right) = \frac{\pi}{3\sqrt{3}}$

Now, substituting $J$ back to find $I$:

$I = \frac{2}{\pi} J = \frac{2}{\pi} \cdot \frac{\pi}{3\sqrt{3}} = \frac{2}{3\sqrt{3}}$

Squaring $I$:

$I^2 = \left( \frac{2}{3\sqrt{3}} \right)^2 = \frac{4}{27}$

Finally, we calculate the required value:

$27 I^2 = 27 \times \frac{4}{27} = 4$