int_{-pi}^{pi} frac{2 x}{1+cos ^{2} x} d x= | vedclass

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(B) Let $f(x) = \frac{2x}{1+\cos^2 x}$.Check if the function is odd or even by evaluating $f(-x)$:$f(-x) = \frac{2(-x)}{1+\cos^2(-x)} = \frac{-2x}{1+\

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(B) Let $f(x) = \frac{2x}{1+\cos^2 x}$.Check if the function is odd or even by evaluating $f(-x)$:$f(-x) = \frac{2(-x)}{1+\cos^2(-x)} = \frac{-2x}{1+\cos^2 x} = -f(x)$.Since $f(-x) = -f(x)$,the function $f(x)$ is an odd function.According to the property of definite integrals,if $f(x)$ is an odd function,then $\int_{-a}^{a} f(x) dx = 0$.Therefore,$\int_{-\pi}^{\pi} \frac{2x}{1+\cos^2 x} dx = 0$.

$\int_{-\pi}^{\pi} \frac{2 x}{1+\cos ^{2} x} d x=$

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