The integral $$\displaystyle\int_0^{\pi} \dfrac{8x \, dx}{4\cos^2 x + \sin^2 x}$$ is equal to

Let the given definite integral be $$I$$:

$$I = \int_0^{\pi} \frac{8x}{4\cos^2 x + \sin^2 x} \, dx$$

Applying the definite integral property $$\int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx$$:

$$I = \int_0^{\pi} \frac{8(\pi - x)}{4\cos^2(\pi - x) + \sin^2(\pi - x)} \, dx$$

Since $$\cos(\pi - x) = -\cos x \implies \cos^2(\pi - x) = \cos^2 x$$ and $$\sin(\pi - x) = \sin x \implies \sin^2(\pi - x) = \sin^2 x$$:

$$I = \int_0^{\pi} \frac{8\pi - 8x}{4\cos^2 x + \sin^2 x} \, dx$$

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Adding these two integral expressions together:

$$2I = \int_0^{\pi} \frac{8\pi}{4\cos^2 x + \sin^2 x} \, dx$$

$$I = 4\pi \int_0^{\pi} \frac{1}{4\cos^2 x + \sin^2 x} \, dx$$

Using the property $$\int_0^{2a} f(x) \, dx = 2\int_0^a f(x) \, dx$$ because the integrand is symmetric about $$\frac{\pi}{2}$$:

$$I = 8\pi \int_0^{\frac{\pi}{2}} \frac{1}{4\cos^2 x + \sin^2 x} \, dx$$

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To evaluate this integral, divide the numerator and the denominator by $$\cos^2 x$$:

$$I = 8\pi \int_0^{\frac{\pi}{2}} \frac{\sec^2 x}{4 + \tan^2 x} \, dx$$

Let $$\tan x = t$$. Differentiating both sides gives $$\sec^2 x \, dx = dt$$.

The limits of integration change from $$x = 0 \to t = 0$$ and $$x = \frac{\pi}{2} \to t = \infty$$.

Substituting these into the integral:

$$I = 8\pi \int_0^{\infty} \frac{1}{4 + t^2} \, dt$$

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Using the standard integration formula $$\int \frac{1}{a^2 + t^2} \, dt = \frac{1}{a}\tan^{-1}\left(\frac{t}{a}\right)$$:

$$I = 8\pi \left[ \frac{1}{2}\tan^{-1}\left(\frac{t}{2}\right) \right]_0^{\infty}$$

$$I = 4\pi \left[ \tan^{-1}(\infty) - \tan^{-1}(0) \right]$$

$$I = 4\pi \left[ \frac{\pi}{2} - 0 \right] = 2\pi^2$$

Therefore, the final value of the integral is equal to $$2\pi^2$$.