Primitive of Root of x squared minus a squared/Logarithm Form

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Theorem

$\ds \int \sqrt {x^2 - a^2} \rd x = \frac {x \sqrt {x^2 - a^2} } 2 - \frac {a^2} 2 \ln \size {x + \sqrt {x^2 - a^2} } + C$

for $\size x \ge a$.


Proof

We have that $\sqrt {x^2 - a^2}$ is defined only when $x^2 \ge a^2$, that is, either:

$x \ge a$

or:

$x \le -a$

where it is assumed that $a > 0$.


First let $x \ge a$.

\(\ds x\) \(=\) \(\ds a \cosh u\)
\(\text {(1)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \frac {\d x} {\d u}\) \(=\) \(\ds a \sinh u\) Derivative of Hyperbolic Cosine


Also:

\(\ds x\) \(=\) \(\ds a \cosh u\)
\(\ds \leadsto \ \ \) \(\ds x^2 - a^2\) \(=\) \(\ds a^2 \cosh^2 u - a^2\)
\(\ds \) \(=\) \(\ds a^2 \paren {\cosh^2 u + 1}\)
\(\ds \) \(=\) \(\ds a^2 \sinh^2 u\) Difference of Squares of Hyperbolic Cosine and Sine
\(\text {(2)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \sqrt {x^2 - a^2}\) \(=\) \(\ds a \sinh u\)


and:

\(\ds x\) \(=\) \(\ds a \cosh u\)
\(\text {(3)}: \quad\) \(\ds \leadsto \ \ \) \(\ds u\) \(=\) \(\ds \arcosh \frac x a\) Definition of Real Area Hyperbolic Cosine


Thus:

\(\ds \int \sqrt {x^2 - a^2} \rd x\) \(=\) \(\ds \int \sqrt {x^2 - a^2} \, a \sinh u \rd u\) Integration by Substitution from $(1)$
\(\ds \) \(=\) \(\ds \int a^2 \sinh^2 u \rd u\) substituting for $\sqrt {x^2 - a^2}$ from $(2)$
\(\ds \) \(=\) \(\ds a^2 \int \sinh^2 u \rd u\) Primitive of Constant Multiple of Function
\(\ds \) \(=\) \(\ds a^2 \frac {\sinh u \cosh u - u} 2 + C\) Primitive of $\sinh^2 u$: Corollary
\(\ds \) \(=\) \(\ds \frac 1 2 a \sinh u a \cosh u - \frac {a^2 u} 2 + C\) rearranging
\(\ds \) \(=\) \(\ds \frac 1 2 x a \sinh u - \frac {a^2 u} 2 + C\) substituting $x = a \cosh u$
\(\ds \) \(=\) \(\ds \frac 1 2 x \sqrt {x^2 - a^2} - \frac {a^2 u} 2 + C\) substituting $\sqrt {x^2 - a^2} = a \sinh u$ from $(2)$
\(\ds \) \(=\) \(\ds \frac {x \sqrt {x^2 - a^2} } 2 - \frac {a^2} 2 \cosh^{-1} \frac x a + C\) substituting for $\theta$ from $(3)$
\(\ds \) \(=\) \(\ds \frac {x \sqrt {x^2 - a^2} } 2 - \frac {a^2} 2 \paren {\map \ln {x + \sqrt {x^2 - a^2} } - \ln a} + C\) $\arcosh \dfrac x a$ in Logarithm Form
\(\ds \) \(=\) \(\ds \frac {x \sqrt {x^2 - a^2} } 2 - \frac {a^2} 2 \map \ln {x + \sqrt {x^2 - a^2} } + \frac {a^2} 2 \ln a + C\) simplifying
\(\ds \) \(=\) \(\ds \frac {x \sqrt {x^2 - a^2} } 2 - \frac {a^2} 2 \map \ln {x + \sqrt {x^2 - a^2} } + C\) subsuming $\dfrac {a^2} 2 \ln a$ into arbitrary constant
\(\ds \) \(=\) \(\ds \frac {x \sqrt {x^2 - a^2} } 2 - \frac {a^2} 2 \ln \size {x + \sqrt {x^2 - a^2} } + C\) Definition of Absolute Value

Now suppose $x \le -a$.

Let $z = -x$.

Then:

$\d x = -\d z$

and we then have:

\(\ds \int \sqrt {x^2 - a^2} \rd x\) \(=\) \(\ds -\int \sqrt {\paren {-z}^2 - a^2} \rd z\) Integration by Substitution
\(\ds \) \(=\) \(\ds -\int \sqrt {\paren z^2 - a^2} \rd z\) simplifying
\(\ds \) \(=\) \(\ds -\frac {z \sqrt {z^2 - a^2} } 2 + \frac {a^2} 2 \map \ln {z + \sqrt {z^2 - a^2} } + C\) from above
\(\ds \) \(=\) \(\ds -\frac {z \sqrt {z^2 - a^2} } 2 - \frac {a^2} 2 \paren {\map \ln {z - \sqrt {z^2 - a^2} } - \map \ln {a^2} } + C\) Negative of $\map \ln {z + \sqrt {z^2 - a^2} }$
\(\ds \) \(=\) \(\ds -\frac {z \sqrt {z^2 - a^2} } 2 - \frac {a^2} 2 \map \ln {z - \sqrt {z^2 - a^2} } + C\) subsuming $-\dfrac {a^2 \map \ln {a^2} } 2$ into constant
\(\ds \) \(=\) \(\ds -\frac {\paren {-x} \sqrt {\paren {-x}^2 - a^2} } 2 - \frac {a^2} 2 \map \ln {\paren {-x} - \sqrt {\paren {-x}^2 - a^2} } + C\) substituting back for $x$
\(\ds \) \(=\) \(\ds \frac {x \sqrt {x^2 - a^2} } 2 - \frac {a^2} 2 \map \ln {-x - \sqrt {x^2 - a^2} } + C\) simplifying
\(\ds \) \(=\) \(\ds \frac {x \sqrt {x^2 - a^2} } 2 - \frac {a^2} 2 \ln \size {x + \sqrt {x^2 - a^2} } + C\) as $-x - \sqrt {x^2 - a^2} > 0$: Definition of Absolute Value

The result follows.

$\blacksquare$


Sources

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