in this video we're introducing the inverse hyperbolic tangent function and establishing
why it's useful to know about this function as a first-year calculus physics or engineering
student so the video is split up into chapters if you want to skip around and we start with a basic
definition of the hyperbolic tangent function and its expression in terms of exponential
functions next we figure out how to express the inverse hyperbolic tangent function in terms of
natural logs and because the hyperbolic tangent is expressible in terms of exponential functions
is not too surprising that this is possible then we start to get into why it matters the inverse
hyperbolic tangent function is a solution to a common type of integral that we see
occasionally in first year calculus and physics courses and finally we're going to generalize
our basic inverse hyperbolic tangent integral by expanding to integrands that have different
constants floating around so before we talk about the inverse hyperbolic tangent function
we have to start with the ordinary hyperbolic tangent function and that begins with a reminder
of our basic hyperbolic trig functions cinch and Kosh so cinch X is one half times e to
the x minus E to the negative X and Cosh X is one half e to the X Plus e to the negative X
now it's not surprising to learn that the hyperbolic tangent function is given
by cinch divided by Cosh so that's going to be our basic definition of the hyperbolic tangent
but if we're going to attempt to invert this thing we want to get more fundamental
and replace the sension Cosh functions with their expression in terms of exponentials
Okay so we've got that done and clearly the factors of one-half are going to cancel out
of this and we arrive at a definition of the hyperbolic tangent function purely in terms
of exponential functions next we're going to invert this thing so how do we get a grip on
the inverse hyperbolic tangent of X like any inverse function we start by giving it a name
so Y is the inverse hyperbolic tangent of X the reason that name is useful is it allows us to
invert this inverse function and rewrite this equation as hyperbolic tangent of Y
is equal to X now we can plug in our definition of hyperbolic tangent in terms of exponentials
and the goal here is to solve for y because Y is the name for the inverse hyperbolic tangent
of X and it looks like we can make progress by multiplying both sides by that denominator this
gives us e to the Y minus E to the negative I equals x times the quantity e to the Y plus e
to the negative Y and we can distribute that X and combine like terms and I'm going to combine
all the terms on the left hand side so when I distribute the X to the first exponential
that's x e to the Y I'm going to subtract that from both sides and on the left hand side I
end up with a 1 minus x times e to the Y when we multiply the X into the second exponential
we end up with an x e to the negative Y and I subtract that from both sides this gives me a
minus the quantity 1 plus x e to the negative Y and this is all equal to zero the next thing we
can do here is get down to just one exponential function by multiplying both sides by e to the Y
and using exponent properties Distributing e to the Y to the first term gives me 1 minus x e to
the 2y and to the second term I end up with an e to the negative y plus y or e to the 0 which
is just one so I have a minus 1 plus X there and this is still all equal to zero now I'm going to
add the 1 plus X to both sides and divide by the 1 minus X in front of that exponential and I get
e to the 2y is equal to 1 plus x divided by 1 minus X and now we're getting really close to
a solution we take the natural log of both sides and this gives me a 2y is equal to the natural log
of 1 plus x divided by 1 minus X finally we divide by 2 and we find that Y is equal to
one-half natural log 1 plus x divided by one minus X and because y was the inverse hyperbolic tangent
of X that means the inverse hyperbolic tangent of X is one half times the natural log of one plus
x divided by one minus X now there's a slight complication to this the natural log function
is only defined if its argument is positive so we have to think about what x values make
this argument positive and if I'm thinking about positive values of X well X better be less than
one otherwise that denominator becomes negative while the numerator is positive so we have that
X is less than one if x is a positive number zero doesn't cause any problems and then we think about
what happens if x is a negative number well the denominator is always going to be positive in
that case but the numerator could become negative if x was less than negative one so it's got to be
bigger than that so it turns out that our domain for the inverse hyperbolic tangent function is
negative one to one now I can tell you as a physics student I never once thought about the
domain of the inverse hyperbolic tangent function when this function crops up in a physical setting
the X always just turns out right so that you're within this domain but as a math student you might
have to worry about it from time to time so why should we care about such an exotic function the
answer is that it arises as a solution to a very common type of integral that we see both
in first year calculus and physics courses I'm currently teaching first semester engineering
physics and second semester calculus and in the last few weeks integrals of this type have popped
up multiple times in both courses in my Physics course it was a free fall with quad drag problem
and in my calculus course integrals like this have popped up in the trig substitution section and the
partial fractions decomposition section so to get this integral done we basically have two options
and option number one is to do a trigonometric substitution where we let x equal sine Theta or
cosine Theta we apply a Pythagorean identity which simplifies the denominator and we go from there
but the easiest way to solve this integral is with partial fractions decomposition we just have to
recognize that that denominator is a difference of two squares so we could write it as 1 minus x
times 1 plus X and we propose a partial fraction's decomposition a over one minus X plus b over 1
plus X multiply both sides by the least common denominator and we get a 1 on the left hand side
the 1 minus X cancels on the a containing term so I have a times 1 plus X and the opposite
thing happens with the B containing term I have a b times 1 minus X and we can use substitution
to get a and b really quick if I sub in x equals negative 1 the a containing term vanishes and I
end up with one equals B times 1 minus negative 1 or B times two so B is equal to one half if
we Sub in x equals positive one the B containing term vanishes and I end up with one equals a times
two so a is one half this allows us to rewrite the integrand as a one-half over one minus X
plus one half over one plus X now each of those terms is going to integrate to a natural log
sort of thing we end up with the natural log of the absolute value of the denominator the
first one has a bit of a complication because the derivative of 1 minus X is negative one
to account for the chain rule we end up with a negative one-half natural log absolute value 1
minus X the second one is the easier one there's no complication from the chain rule because the
derivative of the denominator is just one and I end up with a one-half natural log absolute
value One Plus X and we'll put a plus c on there now we can use log properties to combine these
and I end up with a one-half out in front of a natural log of one plus x minus a natural log
of 1 minus X well that becomes a natural log of 1 plus X over 1 minus X and we're putting absolute
value bars on this plus C and in many calculus books this is the formula that you're going to
see in the integration tables for the solution to this integral but some of the textbooks use the
inverse hyperbolic tangent instead and the same is true of physics books and some computer algebra
systems random blog articles that you run into on the internet Wikipedia entries so it's important
that you're able to express this as an inverse hyperbolic tangent so you get what you're looking
at when you run into an article like this so I'm wearing my math hat right now and I'll be extra
careful about the domain issues I'm going to say if x belongs to the interval negative one to one
in other words it's between negative one and one so that this argument of the log function
is always positive then I can say the result of this integral one over one minus x squared DX
is one half natural log of 1 plus X over one minus X without the absolute value bars because
that argument is taken to be positive and I'll put a plus c on there but we just figured out
that's the inverse hyperbolic tangent function so I end up with an inverse hyperbolic tangent of x
plus c so the last thing I want to do is just generalize that result a little bit to deal with
the case where the constant in the denominator is not one so it's convenient to express that as a
square I'm going to call it a squared and one way to figure out how this modifies the formula is to
just redo the partial fractions decomposition from scratch it was a pretty quick one so it's not that
bad we start by factoring the difference of two squares a minus x a plus X we propose a partial
fractions decomposition a over a minus X well that was a little a in the denominator plus b
over little a plus X you multiply by the least common denominator that gives us one on the left
hand side and a times the quantity little a plus X plus b times the quantity little a minus X on
the right hand side and we use substitution again let x equal negative a that's going to kill the
big a containing term and I get 1 equals B times a minus negative a which is a 2A so B turns out to
be one over two a if we let x equal positive a the B containing term is going to vanish and I get a 1
equals Big A times 2 little a and I find again that big a and Big B are the same they're both
1 over 2A so we re-express our integral with the partial fractions decomposition each of the terms
has a factor of one over two a in the numerator so I have a 1 over 2A over little a minus X plus
a one over two a over little a plus X DX and again each of these integrates to a natural log of the
denominator and again the one with the minus sign in it has a chain rule issue that produces
a factor of negative one so I end up with a one over two a times a negative natural log absolute
value a minus X Plus natural log absolute value a plus X plus C which can be combined into a single
logarithm as a natural log absolute value a plus X over a minus X and that's a fine answer to this
integral and in fact it's the one I'll usually write down but again occasionally I have my
nose in a book that just loves inverse hyperbolic tangents so we want to be able to understand what
that's about so we're going to have to think about the domain of this thing like when can I disappear
the absolute value bars it requires that X is less than a for that denominator to be positive and
again it requires that X is bigger than negative a for that numerator to be positive so I could
say when X belongs to the open interval negative a to a we can disappear the absolute value bars we
find that our integral of 1 over a squared minus x squared DX is equal to 1 over 2A natural log
a plus X over a minus x with no absolute value bars plus C and now we're trying to match our
definition of the inverse hyperbolic tangent function in terms of logarithms I need to have
a 1 plus a variable and a 1 minus a variable in that logarithm so I divide the top and bottom
by a and I end up with a 1 over 2A natural log 1 plus X over a over 1 minus X over a plus C so the
one-half natural log one plus something divided by one minus something that's our inverse hyperbolic
tangent and there's still a factor of one over a out in front so I'm going to write 1 over a
inverse hyperbolic tangent X over a plus C so if I've accomplished my goal here you'll be a little
less frightened the next time a computer algebra system spits out an inverse hyperbolic tangent
remember if it helps you to get a grip on the problem it's always expressible as a natural log
of a ratio of these two linear factors let me know in the comments if you'd like me to do a video on
some other exotic function and I'll give it a shot see you next time if you enjoyed this video or at
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