The integral $\displaystyle\int_{0}^{\pi}\frac{8x}{4\cos^2x + \sin^2x}\,dx$ is equal to:

Step 1: Apply the property of definite integrals

We use the property of definite integrals which states that for an integral from $0$ to $a$, we can replace $x$ with $a-x$ without changing the value of the integral. In this case, $a = \pi$. Let the given integral be $I$.

Step 2: Substitute $x = \pi - x$ into the integral

Substitute $x = \pi - x$ into the original integral. Recall that $\cos(\pi-x) = -\cos x$ and $\sin(\pi-x) = \sin x$. Therefore, $\cos^2(\pi-x) = (-\cos x)^2 = \cos^2 x$ and $\sin^2(\pi-x) = (\sin x)^2 = \sin^2 x$.

Step 3: Simplify the integral

After substituting and simplifying the trigonometric terms, the denominator remains unchanged. We can split this integral into two parts.

Step 4: Add the original and modified integrals

Adding the original integral $I$ and the modified integral $I$ (from the previous step) gives $2I$. The denominators are the same, so we can combine the numerators.

Step 5: Combine and simplify the numerators

The $8x$ terms in the numerator cancel out, leaving $8\pi$. Now we have a simpler integral to evaluate.

Step 6: Evaluate the simplified integral

Divide the numerator and denominator by $\cos^2x$. This transforms the integral into a form involving $\sec^2x$ and $\tan^2x$, which can be solved using a substitution. Recall that $\sec^2x = 1 + \tan^2x$.

Step 7: Perform substitution and integrate

Let $t = \tan x$. Then $dt = \sec^2x\,dx$. When $x=0$, $t=\tan 0 = 0$. When $x=\pi$, $t=\tan \pi = 0$. However, the function $\tan x$ is discontinuous at $x=\pi/2$. We need to split the integral into two parts: from $0$ to $\pi/2$ and from $\pi/2$ to $\pi$. Due to symmetry, $\int_{0}^{\pi}\frac{\sec^2x}{4 + \tan^2x}\,dx = 2 \int_{0}^{\pi/2}\frac{\sec^2x}{4 + \tan^2x}\,dx$. For the integral from $0$ to $\pi/2$, when $x=\pi/2$, $t \to \infty$.

Step 8: Complete the integration

The integral $\int \frac{dt}{a^2+t^2} = \frac{1}{a}\tan^{-1}\left(\frac{t}{a}\right)$. Here $a=2$. So, $\int_{0}^{\pi/2}\frac{\sec^2x}{4 + \tan^2x}\,dx = \left[ \frac{1}{2}\tan^{-1}\left(\frac{\tan x}{2}\right) \right]_{0}^{\pi/2} = \frac{1}{2}\tan^{-1}(\infty) - \frac{1}{2}\tan^{-1}(0) = \frac{1}{2} \cdot \frac{\pi}{2} - 0 = \frac{\pi}{4}$. Therefore, $2I = 8\pi \cdot 2 \cdot \frac{\pi}{4} = 4\pi^2$.

Step 9: Solve for I

Divide by 2 to find the value of $I$.

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