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

Question

Aicharya · Accepted Answer

Let $f(x)=\frac{2 x}{1+\cos ^{2} x} \Rightarrow f(-x)=\frac{-2 x}{1+\cos ^{2} x}$ Thus $\mathrm{f}(-\mathrm{x})=-\mathrm{f}(\mathrm{x}) \Rightarrow \mathrm{I}=0$

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

Solution