Uploaded by TheIntegralCALC on 09.10.2010

Transcript:

Hi everyone. Welcome back to integralcalc.com. We're going to be doing another integral problem

today. This one's actually going to involve integration by parts as well as integration

with u-substitution and even a trigonometric integral so a lot of thing's going on in

this problem. It was actually submitted by one of our viewers on youtube. So the problem

is the integral of arc tan of x dx. So whenever I have a trigonometric integrals

like this, to me arc tan's pretty rare. This is not something that's readily apparent.

I don't look at this and say ,"oh, obviously the integral is... whatever". So in that case,

with trigonometric integrals, often times a good place to start is integration by parts.

It works well with a lot of trigonometric integrals and the parts that you dissolve

this function into, the first one is arc tan x and the second one is dx. So let's go ahead

and use integration by parts. And remember the integration by parts formula, u dv,

this part, of course representing our function over here, equals u v minus the integral of

v du. Remember when we're using integration by parts,

we need to first assign u and dv to values in our problem. So one of the things that's

important in this problem, we're going to assign u to arc tan of x, which means that

we're going to need to know the derivative of arc tan of x and that is a formula, obviously

you can derive it with a proof but it's pretty complicated and kind of beyond the level of

calc one and two, so my recommendation would be to take sine, co-sine, tangent, secant,

arc tan, each one of the trigonometric functions and put in your calculator or put in a formula

sheet, the integral of each one and the derivative of each one so that you have to reference

because otherwise, there's no apparent way to just figure out what the derivative of

arc tan of x is. So that's what I would recommend and we're going to go ahead and write it here.

The derivative of arc tan of x is one over x squared plus one.

That is the derivative of arc tan of x. It's pretty complicated to figure out just how

to find this so, this is something to memorize, put in a calculator or write down in a formula

sheet before an exam. So given that, let's go ahead and use our integration by parts

formula. So we're going to assign, because this part is representing our function, we're

going to assign u and dv to values in our problem. So u, we're going to assign to arc

tan of x and dv we're going to assign to dx or one dx which when we're taking, when we're

trying to find the integrals of trigonometric functions and we're using integration by parts,

we'll end up right here because it will just be, you know, it will be just sine of x or

secant of x or something like that so that means that you're dv is just going to end

up being dx and an easier way to remember that is this is dx times one obviously is

the same thing as dx but having that one here cues you in to the fact that v is x. We

take the integral of dv to get v and of course the integral of one is just x, and that's

how we get v. du of course is the derivative of arc tan of x which we're going to grab

from our formula over here. It's one over x squared plus one.

So we assigned u and dv to values in our problem and then we took the derivative of u to get

du and took the integral of dv to get v. So now that we have all four of these components,

we go ahead and plug them back in to the right hand side of this equation here. Starting

of course with u. So u, in this case, right is arc tan of x. v is x, so we multiply by

x and then we have minus the integral of v so x times du, which in this case is one x

squared plus one and then we always just add dx here because dx goes with this integral

notation so it's notation. You just tap it off at the end there.

So let's go ahead and simplify this. I don't think that we would need any of these anymore

so we can go ahead and simplify. So simplifying this, we'll move the x in front. x arc tan of x minus the integral and it will

be x over x squared plus one dx. Alright, so there's our problem after we've

used integration by parts. Normally the goal of integration by parts is you know is to

get this part here, what's inside the new integral to be more simple than what we started

out with because this function, we can't just integrate. But what we want to do is get to

a point where this is something we can integrate. That's why we used integration by parts.

And by the way, if you're having problem with integration by parts specifically, like u-substitution,

anything, I have those sections in my website so you can go there.

So since this is still not simple enough to just see easily as something that we can simply

integrate, we need to perform some other kind of operation to make this simpler. The thing

that pops out is most obvious. You know, first thing you should always look at. If you need

to make the integral simpler, the first three options you should go for are u-substitution,

integration by parts, and integration by partial fractions. Integration by parts, we don't

usually like to use with fractions because it's difficult so I wouldn't use integration

by parts to simplify this again. I won't use integration by partial fractions because you

can't factor the denominator here and that's the dead give-away for integration by partial

fractions. So what I love with this u-substitution, which

might seem a good candidate because if I assign u to x squared plus one, then you'll end up

with something like x over u and that might end up being simple. So that's my first guess

and what we'll do is go ahead and try it. So we're going to use, we're going to leave

this alone for now and just focus on evaluating this integral with u-substitution. And what

we'll do is assign u to the most complicated part which is x squared plus one. So say u

equals x squared plus one and then as always with u-substitution, we always take the derivative

of u which is du and du is going to be two x and then when we do this, we always just

have to tap on dx to the end of this. The reason being we need to solve this equation

here for dx. And we can't solve for dx if we don't have dx in it. So we just tap on

to dx in the end and then solve for dx by dividing both sides by two x, which will leave

us with dx with du over two x. So now we've solved for dx, we can go ahead and plug back

in to our function here. So we'll have x arc tan of x which we're just leaving alone minus

the integral x will be left alone of course we assigned u to x squared plus one here so

this is u and then we solved for dx here and we said that it was du over two x, so we plug

du over two x in for dx, so we get du over two x.

And looks like this is going to be great news. It's going to turn out well for us. I bet

you're shocked so what we need to do, you can see, we have an x in the numerator and

an x in the denominator so those are both going to cancel and we have a two in the denominator

or a coefficient of one-half. Since everything inside the integral here is multiplied together,

this is x over u times du over two x, right? They’re multiplied together. Since there's

just one term inside the integral where everything's multiplied together, that means there's a

coefficient on this integral that can be moved outside the integral like this. x arc tan

of x minus the two comes, the one-half, sorry comes out in front. One-half times the integral,

we cancelled the x 's so we're left with one over u du.

So now that we've gotten to this point, the good news is that one over u du is finally

something simple enough for us to integrate. So the integral, here’s another rule for

you. I know there are a lot of them in this problem. The integral of one over x equals

natural log or ln of the absolute value of x plus c. This is absolutely something you

should have memorized. Suppose just plug it into your calculator or writing it down in

a formula sheet because it comes up so often and it's just important to kind of know

it off the top of your heads so you can be quick. Natural log of the absolute value of

x. So let's go ahead and take the integral. We have x arc tan of x minus one-half and

then we just draw a big parenthesis because that symbolizes that okay, here's the beginning

of whatever is going to come out of this integral here. And the integral of this is of course

natural log of the absolute value of u and then we have to always add c to account for

the constant. Remember that we had said that u was equal

to x squared plus one, before we had done the u-substitution problem, all we need to

do is plug it back in, x squared plus one for u and simplify this which there actually

is a lot of simplifying to be done. So it's actually going to come out to be x arc tan

of x minus one-half natural or ln of the absolute value of x squared plus one plus c and now

that we've used integration by parts and u-substitution, back to back to integrate this arc tan of

x, we're done. And that's our final answer. Thanks, guys!

today. This one's actually going to involve integration by parts as well as integration

with u-substitution and even a trigonometric integral so a lot of thing's going on in

this problem. It was actually submitted by one of our viewers on youtube. So the problem

is the integral of arc tan of x dx. So whenever I have a trigonometric integrals

like this, to me arc tan's pretty rare. This is not something that's readily apparent.

I don't look at this and say ,"oh, obviously the integral is... whatever". So in that case,

with trigonometric integrals, often times a good place to start is integration by parts.

It works well with a lot of trigonometric integrals and the parts that you dissolve

this function into, the first one is arc tan x and the second one is dx. So let's go ahead

and use integration by parts. And remember the integration by parts formula, u dv,

this part, of course representing our function over here, equals u v minus the integral of

v du. Remember when we're using integration by parts,

we need to first assign u and dv to values in our problem. So one of the things that's

important in this problem, we're going to assign u to arc tan of x, which means that

we're going to need to know the derivative of arc tan of x and that is a formula, obviously

you can derive it with a proof but it's pretty complicated and kind of beyond the level of

calc one and two, so my recommendation would be to take sine, co-sine, tangent, secant,

arc tan, each one of the trigonometric functions and put in your calculator or put in a formula

sheet, the integral of each one and the derivative of each one so that you have to reference

because otherwise, there's no apparent way to just figure out what the derivative of

arc tan of x is. So that's what I would recommend and we're going to go ahead and write it here.

The derivative of arc tan of x is one over x squared plus one.

That is the derivative of arc tan of x. It's pretty complicated to figure out just how

to find this so, this is something to memorize, put in a calculator or write down in a formula

sheet before an exam. So given that, let's go ahead and use our integration by parts

formula. So we're going to assign, because this part is representing our function, we're

going to assign u and dv to values in our problem. So u, we're going to assign to arc

tan of x and dv we're going to assign to dx or one dx which when we're taking, when we're

trying to find the integrals of trigonometric functions and we're using integration by parts,

we'll end up right here because it will just be, you know, it will be just sine of x or

secant of x or something like that so that means that you're dv is just going to end

up being dx and an easier way to remember that is this is dx times one obviously is

the same thing as dx but having that one here cues you in to the fact that v is x. We

take the integral of dv to get v and of course the integral of one is just x, and that's

how we get v. du of course is the derivative of arc tan of x which we're going to grab

from our formula over here. It's one over x squared plus one.

So we assigned u and dv to values in our problem and then we took the derivative of u to get

du and took the integral of dv to get v. So now that we have all four of these components,

we go ahead and plug them back in to the right hand side of this equation here. Starting

of course with u. So u, in this case, right is arc tan of x. v is x, so we multiply by

x and then we have minus the integral of v so x times du, which in this case is one x

squared plus one and then we always just add dx here because dx goes with this integral

notation so it's notation. You just tap it off at the end there.

So let's go ahead and simplify this. I don't think that we would need any of these anymore

so we can go ahead and simplify. So simplifying this, we'll move the x in front. x arc tan of x minus the integral and it will

be x over x squared plus one dx. Alright, so there's our problem after we've

used integration by parts. Normally the goal of integration by parts is you know is to

get this part here, what's inside the new integral to be more simple than what we started

out with because this function, we can't just integrate. But what we want to do is get to

a point where this is something we can integrate. That's why we used integration by parts.

And by the way, if you're having problem with integration by parts specifically, like u-substitution,

anything, I have those sections in my website so you can go there.

So since this is still not simple enough to just see easily as something that we can simply

integrate, we need to perform some other kind of operation to make this simpler. The thing

that pops out is most obvious. You know, first thing you should always look at. If you need

to make the integral simpler, the first three options you should go for are u-substitution,

integration by parts, and integration by partial fractions. Integration by parts, we don't

usually like to use with fractions because it's difficult so I wouldn't use integration

by parts to simplify this again. I won't use integration by partial fractions because you

can't factor the denominator here and that's the dead give-away for integration by partial

fractions. So what I love with this u-substitution, which

might seem a good candidate because if I assign u to x squared plus one, then you'll end up

with something like x over u and that might end up being simple. So that's my first guess

and what we'll do is go ahead and try it. So we're going to use, we're going to leave

this alone for now and just focus on evaluating this integral with u-substitution. And what

we'll do is assign u to the most complicated part which is x squared plus one. So say u

equals x squared plus one and then as always with u-substitution, we always take the derivative

of u which is du and du is going to be two x and then when we do this, we always just

have to tap on dx to the end of this. The reason being we need to solve this equation

here for dx. And we can't solve for dx if we don't have dx in it. So we just tap on

to dx in the end and then solve for dx by dividing both sides by two x, which will leave

us with dx with du over two x. So now we've solved for dx, we can go ahead and plug back

in to our function here. So we'll have x arc tan of x which we're just leaving alone minus

the integral x will be left alone of course we assigned u to x squared plus one here so

this is u and then we solved for dx here and we said that it was du over two x, so we plug

du over two x in for dx, so we get du over two x.

And looks like this is going to be great news. It's going to turn out well for us. I bet

you're shocked so what we need to do, you can see, we have an x in the numerator and

an x in the denominator so those are both going to cancel and we have a two in the denominator

or a coefficient of one-half. Since everything inside the integral here is multiplied together,

this is x over u times du over two x, right? They’re multiplied together. Since there's

just one term inside the integral where everything's multiplied together, that means there's a

coefficient on this integral that can be moved outside the integral like this. x arc tan

of x minus the two comes, the one-half, sorry comes out in front. One-half times the integral,

we cancelled the x 's so we're left with one over u du.

So now that we've gotten to this point, the good news is that one over u du is finally

something simple enough for us to integrate. So the integral, here’s another rule for

you. I know there are a lot of them in this problem. The integral of one over x equals

natural log or ln of the absolute value of x plus c. This is absolutely something you

should have memorized. Suppose just plug it into your calculator or writing it down in

a formula sheet because it comes up so often and it's just important to kind of know

it off the top of your heads so you can be quick. Natural log of the absolute value of

x. So let's go ahead and take the integral. We have x arc tan of x minus one-half and

then we just draw a big parenthesis because that symbolizes that okay, here's the beginning

of whatever is going to come out of this integral here. And the integral of this is of course

natural log of the absolute value of u and then we have to always add c to account for

the constant. Remember that we had said that u was equal

to x squared plus one, before we had done the u-substitution problem, all we need to

do is plug it back in, x squared plus one for u and simplify this which there actually

is a lot of simplifying to be done. So it's actually going to come out to be x arc tan

of x minus one-half natural or ln of the absolute value of x squared plus one plus c and now

that we've used integration by parts and u-substitution, back to back to integrate this arc tan of

x, we're done. And that's our final answer. Thanks, guys!