Equality of integrals: $ \int_{0}^{\infty} \frac {1}{1+x^2} \, \mathrm{d}x = 2 \cdot \int_{0}^{1} \frac {1}{1+x^2} \, \mathrm{d}x $
Showing that $\int_{0}^{\infty} \frac{dx}{1 + x^2} = 2 \int_0^1 \frac{dx}{1 + x^2}$