Primitive of Arccotangent of x over a/Proof 1

Theorem

$\ds \int \arccot \frac x a \rd x = x \arccot \frac x a + \frac a 2 \map \ln {x^2 + a^2} + C$

Proof

Let:

\(\ds u\)

\(=\)

\(\ds \arccot \frac x a\)

\(\text {(1)}: \quad\)

\(\ds \leadsto \ \ \)

\(\ds \cot u\)

\(=\)

\(\ds \frac x a\)

Definition of Arccotangent

\(\text {(2)}: \quad\)

\(\ds \leadsto \ \ \)

\(\ds \csc u\)

\(=\)

\(\ds \sqrt {1 + \frac {x^2} {a^2} }\)

Difference of Squares of Cosecant and Cotangent

Then:

\(\ds \int \arccot \frac x a \rd x\)

\(=\)

\(\ds -a \int u \csc^2 u \rd u\)

Primitive of Function of Arccotangent

\(\ds \)

\(=\)

\(\ds -a \paren {-u \cot u + \ln \size {\sin u} } + C\)

Primitive of $x \csc^2 a x$ with $a = 1$

\(\ds \)

\(=\)

\(\ds a u \cot u - a \ln \size {\sin u} + C\)

simplifying

\(\ds \)

\(=\)

\(\ds a u \cot u + a \ln \size {\csc u} + C\)

Logarithm of Reciprocal and Cosecant is Reciprocal of Sine

\(\ds \)

\(=\)

\(\ds a u \frac x a + a \ln \size {\csc u} + C\)

Substitution for $\cot u$ from $(1)$

\(\ds \)

\(=\)

\(\ds x u + a \ln \size {\sqrt {1 + \frac {x^2} {a^2} } } + C\)

Substitution for $\csc u$ from $(2)$

\(\ds \)

\(=\)

\(\ds x \arccot \frac x a + a \ln \size {\sqrt {1 + \frac {x^2} {a^2} } } + C\)

Substitution for $u$

\(\ds \)

\(=\)

\(\ds x \arccot \frac x a + \frac a 2 \ln \size {\frac {x^2 + a^2 } {a^2} } + C\)

Logarithm of Power and simplifying

\(\ds \)

\(=\)

\(\ds x \arccot \frac x a + \frac a 2 \ln \size {x^2 + a^2} - \ln \size {a^2} + C\)

Difference of Logarithms

\(\ds \)

\(=\)

\(\ds x \arccot \frac x a + \frac a 2 \ln \size {x^2 + a^2} + C\)

subsuming $\ln \size {a^2}$ into arbitrary constant

\(\ds \)

\(=\)

\(\ds x \arccot \frac x a + \frac a 2 \, \map \ln {x^2 + a^2} + C\)

$x^2 + a^2$ always positive

$\blacksquare$