Primitive of Arctangent of x over a/Proof 1

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

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

$\blacksquare$