Primitive of Reciprocal of x squared plus a squared/Arctangent Form/Proof 1
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Theorem
- $\ds \int \frac {\d x} {x^2 + a^2} = \frac 1 a \arctan \frac x a + C$
Proof
Let:
- $a \tan \theta = x$
for $\theta \in \openint {-\dfrac \pi 2} {\dfrac \pi 2}$.
From Shape of Tangent Function, this substitution is valid for all real $x$.
Then:
\(\ds x\) | \(=\) | \(\ds a \tan \theta\) | from above | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\d x} {\d \theta}\) | \(=\) | \(\ds a \sec^2 \theta\) | Derivative of Tangent Function | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \int \frac 1 {x^2 + a^2} \rd x\) | \(=\) | \(\ds \int \frac {a \ \sec^2 \theta} {a^2 \tan^2 \theta + a^2} \rd \theta\) | Integration by Substitution | ||||||||||
\(\ds \) | \(=\) | \(\ds \frac a {a^2} \int \frac {\sec^2 \theta} {\tan^2 \theta + 1} rd \theta\) | Primitive of Constant Multiple of Function | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 a \int \frac {\sec^2 \theta} {\sec^2 \theta} \rd \theta\) | Difference of Squares of Secant and Tangent | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 a \int \rd \theta\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 a \theta + C\) | Integral of Constant |
As $\theta$ was stipulated to be in the open interval $\openint {-\dfrac \pi 2} {\dfrac \pi 2}$:
- $\tan \theta = \dfrac x a \iff \theta = \arctan \dfrac x a$
Thus:
- $\ds \int \frac 1 {x^2 + a^2} \rd x = \frac 1 a \arctan \frac x a + C$
$\Box$
When $a = 0$, both $\dfrac x a$ and $\dfrac 1 a$ are undefined.
However, consider the limit of the above primitive as $a \to 0$:
\(\ds \lim_{a \mathop \to 0} \frac 1 a \arctan {\frac x a}\) | \(=\) | \(\ds \lim_{a \mathop \to 0} \frac {\arctan {\frac x a} } a\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \) | \(=\) | \(\ds \lim_{a \mathop \to 0} \frac {-x a^{-2} } {1 + \frac {x^2} {a^2} }\) | L'Hôpital's Rule, Derivative of Arctangent Function | ||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{a \mathop \to 0} \frac {a^{-2} } {a^{-2} } \frac {-x} {x^2 + a^2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -\frac 1 x\) |
This corresponds with the result:
- $\ds \int \frac 1 {x^2} \rd x = \frac {-1} x + C$
which follows from Primitive of Power.
$\blacksquare$
Sources
- For a video presentation of the contents of this page, visit the Khan Academy.