1 Differentiation

Chapter 1Differentiation

${\int _{0}^{\pi /2}\frac{\cos x}{1+\sin ^{2}x}dx}{=\int _{0}^{1}\frac{du}{1+u^{2}}}{=\arctan u\bigg ]_{u=0}^{u=1}}{=\arctan 1-\arctan 0=}\frac{\pi }{4}-0=\frac{\pi }{4}=\int _{0}^{\pi /2}\frac{\cos x}{1+\sin ^{2}x}dx=\int _{0}^{1}\frac{du}{1+u^{2}}=\arctan u\bigg ]_{u=0}^{u=1}=\arctan 1-\arctan 0=\frac{\pi }{4}-0=\frac{\pi }{4}.$

Definition 1 This is the definition of a differentiable function $$f$$