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

From ProofWiki
Jump to navigation Jump to search

Theorem

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

for $0 < a < \size x$.


Corollary

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


Proof

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

$x > a$

or:

$x < -a$

where it is assumed that $a > 0$.


Consider the arcsecant substitution:

$u = \arcsec {\dfrac x a}$

which is defined for all $x$ such that $\size {\dfrac x a} \ge 1$.

That is:

$\size x \ge a$

Hence from Shape of Secant Function, this substitution is valid for all for all $x$ such that $\size {\dfrac x a} > 1$.


\(\ds u\) \(=\) \(\ds \arcsec {\frac x a}\)
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds a \sec u\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d x} {\d u}\) \(=\) \(\ds a \sec u \tan u\) Derivative of Secant Function


Let $x > a$.

Then:

\(\ds \int \frac 1 {\sqrt {x^2 - a^2} } \rd x\) \(=\) \(\ds \int \frac {a \sec u \tan u} {\sqrt {a^2 \sec^2 u - a^2} } \rd u\) Integration by Substitution
\(\ds \) \(=\) \(\ds \frac a a \int \frac {\sec u \tan u} {\sqrt {\sec^2 u - 1} } \rd u\) simplifying
\(\ds \) \(=\) \(\ds \int \sec u \frac {\tan u} {\sqrt {\tan^2 u} } \rd u\) Difference of Squares of Secant and Tangent
\(\ds \) \(=\) \(\ds \int \sec u \rd u\) as $x > 1$ both $\sec u$ and $\tan u$ are greater than $0$
\(\ds \) \(=\) \(\ds \ln \size {\sec u + \tan u} + C\) Primitive of $\sec x$: Secant plus Tangent Form
\(\ds \) \(=\) \(\ds \map \ln {\sec u + \tan u} + C\) as both $\sec u$ and $\tan u$ are greater than $0$
\(\ds \) \(=\) \(\ds \map \ln {\frac x a + \sqrt {\frac {x^2} {a^2} - 1} } + C\) substituting back: $\sec u = \dfrac x a$ and Difference of Squares of Secant and Tangent
\(\ds \) \(=\) \(\ds \map \ln {\frac x a + \sqrt {\frac {x^2 - a^2} {a^2} } } + C\) rearrangement
\(\ds \) \(=\) \(\ds \map \ln {\frac {x + \sqrt {x^2 - a^2} } a} + C\) rearrangement
\(\ds \) \(=\) \(\ds \map \ln {x + \sqrt {x^2 - a^2} - \ln a} + C\) Difference of Logarithms
\(\ds \) \(=\) \(\ds \map \ln {x + \sqrt {x^2 - a^2} } + C\) subsuming $-\ln a$ into constant
\(\ds \) \(=\) \(\ds \ln \size {x + \sqrt {x^2 - a^2} } + C\) as $x + \sqrt {x^2 - a^2} > 0$ for $x > a$


Now suppose $x < -a$.

Let $z = -x$.

Then:

$\d x = -\d z$

and we then have:

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

The result follows.

$\blacksquare$


Also presented as

Some sources present this in the form:

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

which is the same as above, except that the constant $a$ has not been subsumed into the arbitrary constant $C$.


Also see


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Primitive_of_Reciprocal_of_Root_of_x_squared_minus_a_squared/Logarithm_Form&oldid=675134"