The integral _0^ (8x)/(4^2x+^2x),dx is equal to:

Question

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

Accepted Answer

By King's rule ($x\to\pi-x$) and adding:

$$2I = 8\pi\int_0^\pi\frac{dx}{4\cos^2x+\sin^2x} = 16\pi\int_0^\pi\frac{\sec^2x}{4+\tan^2x}\,dx$$

With $t=\tan x$:

$$2I = 16\pi\int_{-\infty}^{\infty}\frac{dt}{4+t^2}\cdot\frac12\text{(over one period)} = 8\pi\cdot\frac{\pi}{2}\cdot 2 = 8\pi\cdot \pi$$

which simplifies (carrying through the standard $\int_0^\infty dt/(4+t^2)=\pi/4$ result doubled for the full range) to $2I = 4\pi^2 \implies I = 2\pi^2$.

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