The value of the integral $$\int_{-\pi/2}^{\pi/2} \frac{dx}{(1+e^x)(\sin^6 x + \cos^6 x)}$$ is equal to

We need to evaluate $$I = \int_{-\pi/2}^{\pi/2} \frac{dx}{(1+e^x)(\sin^6 x + \cos^6 x)}$$.

Using the property: for any integrable function $$g(x)$$, $$\int_{-a}^{a} \frac{g(x)}{1+e^x}\,dx = \int_0^a g(x)\,dx$$ when $$g(x) = g(-x)$$.

This follows because $$\frac{1}{1+e^x} + \frac{1}{1+e^{-x}} = 1$$, and adding $$I$$ with the substitution $$x \to -x$$ gives $$2I = \int_{-\pi/2}^{\pi/2} \frac{dx}{\sin^6 x + \cos^6 x}$$.

Since $$\sin^6 x + \cos^6 x$$ is an even function:

$$I = \int_0^{\pi/2} \frac{dx}{\sin^6 x + \cos^6 x}$$

Using the identity $$a^3 + b^3 = (a+b)^3 - 3ab(a+b)$$ with $$a = \sin^2 x$$, $$b = \cos^2 x$$:

$$\sin^6 x + \cos^6 x = (\sin^2 x + \cos^2 x)^3 - 3\sin^2 x \cos^2 x(\sin^2 x + \cos^2 x) = 1 - \frac{3}{4}\sin^2 2x$$

So: $$I = \int_0^{\pi/2} \frac{dx}{1 - \frac{3}{4}\sin^2 2x}$$

Let $$t = 2x$$, so $$dx = dt/2$$:

$$I = \frac{1}{2}\int_0^{\pi} \frac{dt}{1 - \frac{3}{4}\sin^2 t}$$

Since the integrand has period $$\pi$$, and is symmetric about $$t = \pi/2$$:

$$I = \int_0^{\pi/2} \frac{dt}{1 - \frac{3}{4}\sin^2 t}$$

Dividing numerator and denominator by $$\cos^2 t$$:

$$I = \int_0^{\pi/2} \frac{\sec^2 t\,dt}{\sec^2 t - \frac{3}{4}\tan^2 t} = \int_0^{\pi/2} \frac{\sec^2 t\,dt}{1 + \frac{1}{4}\tan^2 t}$$

Let $$u = \tan t$$, $$du = \sec^2 t\,dt$$:

$$I = \int_0^{\infty} \frac{du}{1 + \frac{u^2}{4}} = \int_0^{\infty} \frac{4\,du}{4 + u^2}$$

$$= \left[2\tan^{-1}\left(\frac{u}{2}\right)\right]_0^{\infty} = 2 \cdot \frac{\pi}{2} - 0 = \pi$$

The correct answer is Option C.