The value of $$\int_0^{\pi} \frac{e^{\cos x} \sin x}{1 + \cos^2 x \cdot e^{\cos x} + e^{-\cos x}} dx$$ is equal to

We need to evaluate $$I = \int_0^{\pi} \frac{e^{\cos x} \sin x}{(1 + \cos^2 x)(e^{\cos x} + e^{-\cos x})} \, dx$$.

Step 1: Substitution $$t = \cos x$$.

Let $$t = \cos x$$, so $$dt = -\sin x \, dx$$, which gives $$\sin x \, dx = -dt$$.

When $$x = 0$$: $$t = 1$$. When $$x = \pi$$: $$t = -1$$.

$$I = \int_{1}^{-1} \frac{e^t}{(1 + t^2)(e^t + e^{-t})} \cdot (-dt) = \int_{-1}^{1} \frac{e^t}{(1 + t^2)(e^t + e^{-t})} \, dt$$ $$-(1)$$

Step 2: Apply the substitution $$t \to -t$$.

Replacing $$t$$ by $$-t$$ in $$(1)$$:

$$I = \int_{-1}^{1} \frac{e^{-t}}{(1 + t^2)(e^{-t} + e^{t})} \, dt$$ $$-(2)$$

(We used the facts that $$(-t)^2 = t^2$$ and the limits remain $$-1$$ to $$1$$.)

Step 3: Add equations $$(1)$$ and $$(2)$$.

$$2I = \int_{-1}^{1} \frac{e^t + e^{-t}}{(1 + t^2)(e^t + e^{-t})} \, dt$$

The factor $$(e^t + e^{-t})$$ cancels in the numerator and denominator:

$$2I = \int_{-1}^{1} \frac{1}{1 + t^2} \, dt$$

Step 4: Evaluate the standard integral.

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

$$= \frac{\pi}{4} - \left(-\frac{\pi}{4}\right) = \frac{\pi}{2}$$

Step 5: Solve for $$I$$.

$$2I = \frac{\pi}{2}$$

$$I = \frac{\pi}{4}$$

The answer is $$\frac{\pi}{4}$$, which matches Option B.