Trigonometric Substitution Example 7

Hi everyone! Welcome back to integralcalc.com. Today, we’re going to be doing another trigonometric
substitution example. In this one, we’re going to be working with this integral on
the range ten square root of three to ten. And, the first thing that we need to do as
with any trigonometric substitution problem is identify which kind of substitution we’re
going to be making. So, remember that there’s three kinds: a squared minus c squared, a
squared plus u squared, and u squared minus a squared. In our case, we have u squared
minus a squared, and the reason is because with trigonometric substitutions u represents
our variable which is x and you see that our variable comes first just like the u squared
here, and then a represents the constant and our constant here is sixty four. So, because
we have the variable first minus the constant second, we’re looking at u squared minus
a squared. So, along with this, u squared minus a squared, comes these two formulas
here that we’re going to be using later. So, let’s go ahead and do our trigonometric
substitution setup. If you haven’t watched the trigonometric substitution setup video,
it’s a good video to watch to get an idea of our first couple steps with every trigonometric
substitution problem, but let’s go ahead and through those now. So, the first thing
we need to do is identify what… what the value of u and a are. So, because we have
u squared minus a squared, and that matches up directly with x squared minus sixty four,
to find u, we take the square root of x squared which is x so u equals x, and to find a, we
take the square root of sixty four which is eight so a equals eight. So we’ve identified
those two and now we can plug them into second part of our formula. And when we do that,
we’ll get x equals eight secant of theta. We’ll leave this alone for now, this third
part, and we’ll come back to that later. So the very next that we’re going to do
is take the derivative of this equation here that we just found. So, we’ll get the derivative
dx equals to eight tangent of theta secant of theta and that’s just by the formula,
the derivative of secant of theta is tangent of theta times secant of theta. So, that’s
the derivative there and then we’re going to do a couple more things. First, we’re
going to solve for the identity secant of theta and we do that by dividing both sides
of this equation by eight and then we’re also going to solve for theta and we do that
by taking the inverse secant function of both sides of this equation. That cancels out secant
on the left, we’re just left with theta, and we get secant to the negative one or the
inverse secant function of x over eight on the right hand side. So, the last part of
our setup is to draw the secant trigonometric substitution triangle and this will be everything
we need. We do all of these steps ahead of time, even if we think we don’t need a couple
of them, so that substituting and solving in the end is easy. So, this is the triangle
for secant. When the…When the angle theta is right here, the opposite side is the square
root of u squared minus a squared which in our case is x squared minus sixty four, the
hypotenuse is u which in our case is x, and the adjacent side is a which in our case is
eight. So we’ve got that setup. We’ll use it later if we need it. This is all of
the setup. So, now that we’ve got the setup done, we’re
going to go ahead and start plugging back in to our integral. And, we’re going to
plug in for x, for x we’ll plug in eight secant of theta and for dx we’ll plug in
eight tangent of theta secant of theta d theta. So when we do that here, you can see we made
the substitution eight secant theta and we made a substitution for dx and this is our
setup for the integral. Now, from here, we’re going to start simplifying,
so here’s what that looks like. The first thing we’re going to do we’re going to
move this eight tangent theta secant of theta to the numerator because these two things
are multiplied together and we are going to distribute the squared exponent on the eight
secant theta which will give us sixty four secant squared of theta. Then, we will factor
what’s inside the square root, we’ll factor out a sixty four and we’ll get sixty four
times secant squared of theta minus one. Then, we will separate the two terms that we have
multiplied together here inside the square root. And these are standard steps for trigonometric
substitution that happened every time so you’ll start to get the hang of it as you go. But,
we’re able to do this operation here where we separate the square root of sixty four
from the square root of secant squared of theta minus one, so we do that. And then,
of course, the square root of sixty four is eight so we see eight there and at the same
time we’re going to make this substitution, we’re finally going to use this third part
of our formula. Notice we have secant squared of theta minus one and the third part of our
formula tells us that that is equal to tangent squared of theta, so we go ahead and we make
the substitution and put in tangent squared of theta. Then, of course, we’ll cancel
out our eights because we have one in the numerator and denominator so they go away.
And then, we have here the square root of something that squared so instead of the square
root of tangent squared of theta we’ll just end up with tangent squared of theta in its
place. As you can see, we can cancel out the tangent of theta in the numerator and denominator,
so, our integral, we’re just left with secant of theta d theta. Now, the formula for the
integral of secant of theta is the natural log of the absolute value of secant of theta
plus tangent of theta and that’s just a trigonometric integral formula. So, helpful
to know that when we are obviously… but, we take the natural log of secant of theta
plus tangent of theta and then we’ll be evaluating from ten square root of three to
ten. So let’s go ahead and start evaluating.
We’re going to be using these 2 formulas. Remember, when we setup this problem, we solved
for theta and we got the inverse secant function of x over eight, we’re also going to need
this formula here which I’ll explain in a second.
So, the first thing we’ll do is substitute back for theta. Remember, we solved for theta
here so we’ll go ahead and plug in the inverse secant function of x over eight for theta,
both here and here, and that’s what it looks like when we replace everything. Then, our
next step, here, for this first term, secant and the inverse secant function cancel one
another so these go away and we’re just left with x over eight, so that’s the first
term. The second term, we will need this formula up here. Secant or the inverse secant function,
secant to the negative one, a variable is the same thing as the inverse cosine function
of one divided by whatever what’s inside of the inverse secant function. So, instead
of secant to the negative one of x over eight, we’ll take cosine to the negative one of
one divided by x over eight which just turns into eight over x. So, we now have the inverse
cosine function of eight over x and the reason that’s helpful is because we can much more
easily work with the inverse cosine function than we can with the inverse secant function.
So our next step, we leave the first term alone but, this second term here, this is
what we can turn it into. Tangent is sine over cosine so we break this apart into a
fraction sine over cosine. So this first right here represents the tangent function and then
each of those we’ll take sine of what’s inside here and then cosine of what’s inside
here. So we just broke apart the tangent function but maintained this inner part with both sine
and cosine. So, now that we’ve done that, we can use this formula here to simplify the
numerator. The denominator, let’s start there… Here, the denominator, of course
we have cosine and the inverse cosine function and those two will cancel one another so we’ll
just be left with eight over x in the denominator as you see down here. For the numerator, we
have sine of the inverse cosine function of something and we’re going to go ahead and
use this formula here which tells that when we have that same arrangement, we take whatever
is inside the inverse cosine function and we put it in to this square root of one minus
x squared formula. So you can see our x here in the formula is eight over x so we take
eight over x and we put it in place right here. And so, in the numerator, we end up
with the square root of one minus eight over x squared. So that’s how we get out of that
one and simplify. So now that we’ve got all our trigonometric
identities eliminated, it’s going to be a lot easier from here. So, remember that
when we have a fraction within a fraction, here we have something divided by a fraction,
it’s going to be the numerator, instead of divided by the fraction multiplied by the
inverse of the fraction, and that’s just straight algebra. So, we multiply by the inverse,
x over eight, and we are left with the square root. We… we went ahead and distributed
this squared exponent here and we got sixty four over x squared so we end up with sixty
four over x squared right there. Now, we’re going to go ahead and start plugging
in our limits here, our range, because we’ve simplified really as much as we can. So first,
remember when we’re evaluating our range, we always plug in this top number here, the
first number, ten, so we plug in ten for x. Notice here we’ll get one hundred in the
denominator so that’s why we get one hundred right there, ten squared. And then, we’ll
plug in ten square root of three so we get ten square root of three, ten squared root
of three, and then in the denominator, ten squared is a hundred and then this square
root of three squared is three so we end up with a hundred times three.
So now, let’s go ahead and simplify further. We can divide through all of these fractions
by two, ten over eight reduces to five over four. If we simplify what’s inside the square
root here, we’ll take… we’ll make a common denominator and we’ll have one hundred
over one hundred. So when we do one hundred minus sixty four, we get thirty six over one
hundred. Again, dividing through by two for these first two fractions and then we’ll
get sixty four over three hundred, one minus sixty four over three hundred simplifies to
two hundred and thirty six over three hundred. So, now, let’s go ahead and take the square
root of thirty six over one hundred, we get six tenths exactly so we’re multiplying
here by six tenths. And then, this square root over here reduces to two times the square
root of fifty nine… Remember that when you’re taking the square root of a fraction, you
can take the square root of the numerator and the denominator separately, so, you don’t
have to worry about them together. So, we’ll pretend just we have the square root of two
thirty six, we get two times the square root of fifty nine, when we take the square root
of three hundred we get ten times the square root of three. So, that’s as far as we can
simplify our square roots. Now here, inside the first natural log function,
we’ll end up with five fourths and then this five fourths here times six tenths gives
us thirty over forty. This second term here, notice that we have a square root of three
in the numerator and one in the denominator, both of those will cancel and we’ll be left
with ten times the square root of fifty nine over forty. This first natural log, if we
combine these two here, the second term reduces to three fourths so we have five fourths plus
three fourths gives us eight fourths. And then, here, we’ll find a common denominator
with this first term so we’ll make this fifty times the square root of three over
forty and when we combine those we’ll get fifty square root of three plus ten square
root of fifty nine all over forty, we combined these two fractions. So now, we’ll reduce…
we’ll divide through this fraction here by ten so this term… this fifty becomes
five, the ten becomes one, and the forty becomes four.
And now, we’re about to use a property of natural logs. When we have one natural log
minus another natural log, what we do is we put the first one over the second one, we
combine them into one fraction. So the first one goes to the numerator and the second one,
the entire thing, goes to the denominator and we have natural log of one giant fraction.
But again, remember that instead of dividing by a fraction, we can multiply by its inverse.
So we have eight fourths, the numerator, and then instead of dividing by everything here,
we multiply by the inverse, so we flip it upside down, we put this four here that was
in the denominator in the numerator and this numerator becomes the denominator and we’re
multiplying now instead of dividing. And now, you can see that when we multiply these together,
the fourths will cancel and then our final answer will be left with just eight divided
by five square root of three plus the square root of fifty nine and of course we’re taking
the natural log of that whole thing. So, I know that was quite a problem but I
do hope it helped you guys and I will see you in the next video. Bye!