Primitive of Arccotangent Function/Proof 1

Theorem

$\ds \int \arccot x \rd x = x \arccot x + \frac {\map \ln {x^2 + 1} } 2 + C$


Proof

Let:

\(\ds u\) \(=\) \(\ds \arccot x\)
\(\text {(1)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \cot u\) \(=\) \(\ds x\) Definition of Arccotangent
\(\text {(2)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \csc u\) \(=\) \(\ds \sqrt {1 + x^2}\) Difference of Squares of Cosecant and Cotangent


Then:

\(\ds \int \arccot x \rd x\) \(=\) \(\ds -\int u \csc^2 u \rd u\) Primitive of Function of Arccotangent
\(\ds \) \(=\) \(\ds -\paren {-u \cot u + \ln \size {\sin u} } + C\) Primitive of $x \csc^2 a x$ with $a := 1$
\(\ds \) \(=\) \(\ds u \cot u - \ln \size {\sin u} + C\) simplifying
\(\ds \) \(=\) \(\ds u \cot u + \ln \size {\csc u} + C\) Logarithm of Reciprocal and Cosecant is Reciprocal of Sine
\(\ds \) \(=\) \(\ds u x + a \ln \size {\csc u} + C\) Substitution for $\cot u$ from $(1)$
\(\ds \) \(=\) \(\ds x u + a \ln \size {\sqrt {1 + x^2} } + C\) Substitution for $\csc u$ from $(2)$
\(\ds \) \(=\) \(\ds x \arccot x + \ln \size {\sqrt {1 + x^2} } + C\) Substitution for $u$
\(\ds \) \(=\) \(\ds x \arccot x + \frac 1 2 \ln \size {x^2 + 1} + C\) Logarithm of Power and simplifying
\(\ds \) \(=\) \(\ds x \arccot x + \frac {\map \ln {x^2 + 1} } 2 + C\) $x^2 + 1$ always positive

$\blacksquare$