Primitive of Function of Arctangent
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Theorem
- $\ds \int \map F {\arctan \frac x a} \rd x = a \int \map F u \sec^2 u \rd u$
where $u = \arctan \dfrac x a$.
Proof
First note that:
| \(\ds u\) | \(=\) | \(\ds \arctan \frac x a\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds a \tan u\) | Definition of Real Arctangent |
Then:
| \(\ds u\) | \(=\) | \(\ds \arctan \frac x a\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \frac {\d u} {\d x}\) | \(=\) | \(\ds \frac a {a^2 + x^2}\) | Derivative of Arctangent Function: Corollary | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \int \map F {\arctan \frac x a} \rd x\) | \(=\) | \(\ds \int \map F u \frac {a^2 + x^2} a \rd u\) | Primitive of Composite Function | ||||||||||
| \(\ds \) | \(=\) | \(\ds \int \map F u \frac {a^2 + a^2 \tan^2 u} a \rd u\) | Definition of $x$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \int \map F u a^2 \frac {1 + \tan^2 u} a \rd u\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \int \map F u a \sec^2 u \rd u\) | Difference of Squares of Secant and Tangent | |||||||||||
| \(\ds \) | \(=\) | \(\ds a \int \map F u \sec^2 u \rd u\) | Primitive of Constant Multiple of Function |
$\blacksquare$