Primitive of Arctangent of x over a
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Theorem
- $\ds \int \arctan \frac x a \rd x = x \arctan \frac x a - \frac a 2 \map \ln {x^2 + a^2} + C$
Proof 1
Let:
\(\ds u\) | \(=\) | \(\ds \arctan \frac x a\) | ||||||||||||
\(\text {(1)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds \tan u\) | \(=\) | \(\ds \frac x a\) | Definition of Real Arctangent | |||||||||
\(\text {(2)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds \sec u\) | \(=\) | \(\ds \sqrt {1 + \frac {x^2} {a^2} }\) | Difference of Squares of Secant and Tangent |
Then:
\(\ds \int \arctan \frac x a \rd x\) | \(=\) | \(\ds a \int u \sec^2 u \rd u\) | Primitive of Function of Arctangent | |||||||||||
\(\ds \) | \(=\) | \(\ds a \paren {u \tan u + \ln \size {\cos u} } + C\) | Primitive of $x \sec^2 a x$ with $a = 1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds a u \tan u - a \ln \size {\sec u} + C\) | Logarithm of Reciprocal and Secant is Reciprocal of Cosine | |||||||||||
\(\ds \) | \(=\) | \(\ds a u \frac x a - a \ln \size {\sec u} + C\) | Substitution for $\tan u$ from $\paren 1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds a u \frac x a - a \ln \size {\sqrt {1 + \frac {x^2} {a^2} } } + C\) | Substitution for $\sec u$ from $\paren 2$ | |||||||||||
\(\ds \) | \(=\) | \(\ds x \arctan \frac x a - a \ln \size {\sqrt {1 + \frac {x^2} {a^2} } } + C\) | Substitution for $u$ | |||||||||||
\(\ds \) | \(=\) | \(\ds x \arctan \frac x a - \frac a 2 \ln \size {\frac {x^2 + a^2 } {a^2} } + C\) | Logarithm of Power and simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds x \arctan \frac x a - \frac a 2 \ln \size {x^2 + a^2} - \ln \size {a^2} + C\) | Difference of Logarithms | |||||||||||
\(\ds \) | \(=\) | \(\ds x \arctan \frac x a - \frac a 2 \ln \size {x^2 + a^2} + C\) | subsuming $\ln \size {a^2}$ into arbitrary constant | |||||||||||
\(\ds \) | \(=\) | \(\ds x \arctan \frac x a - \frac a 2 \map \ln {x^2 + a^2} + C\) | $x^2 + a^2$ always positive |
$\blacksquare$
Proof 2
With a view to expressing the primitive in the form:
- $\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
\(\ds u\) | \(=\) | \(\ds \arctan \frac x a\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\d u} {\d x}\) | \(=\) | \(\ds \frac a {x^2 + a^2}\) | Derivative of $\arctan \dfrac x a$ |
and let:
\(\ds \frac {\d v} {\d x}\) | \(=\) | \(\ds 1\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds v\) | \(=\) | \(\ds x\) | Primitive of Constant |
Then:
\(\ds \int \arctan \frac x a \rd x\) | \(=\) | \(\ds x \arctan \frac x a - \int x \paren {\frac a {x^2 + a^2} } \rd x + C\) | Integration by Parts | |||||||||||
\(\ds \) | \(=\) | \(\ds x \arctan \frac x a - a \int \frac {x \rd x} {x^2 + a^2} + C\) | Primitive of Constant Multiple of Function | |||||||||||
\(\ds \) | \(=\) | \(\ds x \arctan \frac x a - a \paren {\frac 1 2 \ln \paren {x^2 + a^2} } + C\) | Primitive of $\dfrac x {x^2 + a^2}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds x \arctan \frac x a - \frac a 2 \ln \paren {x^2 + a^2} + C\) | simplifying |
$\blacksquare$
Also see
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving Inverse Trigonometric Functions: $14.483$
- 1989: Ephraim J. Borowski and Jonathan M. Borwein: Dictionary of Mathematics ... (previous) ... (next): Appendix $2$: Table of derivatives and integrals of common functions: Inverse trigonometric functions