Primitive of Function of Arctangent

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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

$\blacksquare$


Also see