Uploaded by TheIntegralCALC on 27.02.2011

Transcript:

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!

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!