# Primitive of Function of Arctangent

## Theorem

$\displaystyle \int F \left({\arctan \frac x a}\right) \ \mathrm d x = a \int F \left({u}\right) \sec^2 u \ \mathrm d u$

where $u = \arctan \dfrac x a$.

## Proof

First note that:

 $\displaystyle u$ $=$ $\displaystyle \arctan \frac x a$ $\displaystyle \implies \ \$ $\displaystyle x$ $=$ $\displaystyle a \tan u$ Definition of Arctangent

Then:

 $\displaystyle u$ $=$ $\displaystyle \arctan \frac x a$ $\displaystyle \implies \ \$ $\displaystyle \frac{\mathrm d u} {\mathrm d x}$ $=$ $\displaystyle \frac a {a^2 + x^2}$ Derivative of Arctangent Function: Corollary $\displaystyle \implies \ \$ $\displaystyle \int F \left({\arctan \frac x a}\right) \ \mathrm d x$ $=$ $\displaystyle \int F \left({u}\right) \ \frac {a^2 + x^2} a \ \mathrm d u$ Primitive of Composite Function $\displaystyle$ $=$ $\displaystyle \int F \left({u}\right) \ \frac {a^2 + a^2 \tan^2 u} a \ \mathrm d u$ Definition of $x$ $\displaystyle$ $=$ $\displaystyle \int F \left({u}\right) \ a^2 \frac {1 + \tan^2 u} a \ \mathrm d u$ $\displaystyle$ $=$ $\displaystyle \int F \left({u}\right) \ a \sec^2 u \ \mathrm d u$ Difference of Squares of Secant and Tangent $\displaystyle$ $=$ $\displaystyle a \int F \left({u}\right) \ \sec^2 u \ \mathrm d u$ Primitive of Constant Multiple of Function

$\blacksquare$