Uploaded by TheIntegralCALC on 23.08.2010

Transcript:

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

substitution problem today. This one is going to be the integral of 1 over (x2 + 1)2 dx.

So the first thing that we need to do is identify which of our three basic trogonometric substitutions

we're going to make. In this case because we have addition here, we've got x2 and

1 we have addition between the two so we're now going to be using the u2 plus a2 subsitution

which is u equals a tan theta. So we're going to correlate this to u2 plus

a2 which comes along with this substitution u equals a tan theta. So in order to solve

for u and a, we'll set u2 equal to x2 and we will set a2 equal to 1.

We'll take the square root of both sides here with the u2 equals x2 and we'll get u equals

x. Again take the square root here of both sides

and get a equals 1. So now we've solved for u and a and we'll

go ahead and plug those into u equals a tan theta. We will get x equals 1 times tan theta.

So we've done that. The couple last things we need to do before we start substituting

back into our problem is find the derivative of x, dx and the derivative of x is going

to be sec2 theta d theta because the derivative of tan theta is sec2 theta.

Remember: And that's a formula that you should memorize.

And the last thing that we need to do is solve for tangent of theta, which is really simple

to do. We’ll get tan theta, we divide both sides by 1 and we get tan theta equals x over

1. I'm going to leave it in this form because

it's helpful later on. But we can also simplify that over here and say tan theta equals x

because we'll probably want both. So we've done everything we need to do to

set up for our problem. So now we go ahead and substitute back into our original integral

for x and our dx. So we will end up with the integral of 1 over

tan2 theta because we have an x2 here plus 1 and then that is squared.

And then substituting for dx, we'll put in tan2 theta d theta.

So now we've substituted and all we need to do is simplify. So the first thing we need

to do, this tan2 of theta plus 1 coming along with this substitution or this identity here

that was in our problem. And this substitution tells us to use the identity tan2 theta plus

1 equals sec2 theta which is a formula. Since we have tan2 theta in our problem, we

can change that to 1 over sec2 theta, which we plugged in for this tan2 theta plus 1 and

then we'll leave the squared because that's there.

So, we have that and then we leave this sec2 theta d theta. What we'll end up with is the

integral of sec2 theta over, since we have sec2 and then that is also squared, we end

up with sec4 theta and then d theta here at the end.

And of course we could cancel out the sec2 theta from the numerator and denominator so

we'll end up with the integral of 1 over sec2 theta d theta.

So 1 over sec2 is the same thing as cos2 so that would simplify to the integral of cos2

theta d theta. Which is about as simple as we can get our

integral before we go ahead and integrate. What we'll have to do is use a formula for

the integral of cos2 of a variable. The formula for that integral is going to be 1/2 theta

plus 1/4 sin of 2 theta plus c. So this is the integration formula when you

have cos2 of the variable. Now what we'll need to do is use the double angle formula

for sin of 2 theta. sin 2 theta is the same thing as 2 times sin theta times cos theta.

That is in double angle formula because this sin 2 theta means twice the angle theta. So

2 sin theta cos theta and then of course plus c.

So now we can simplify this. If we multiply out 2 times 1/4 here, we'll get 1/2 and then

this parenthesis will go away. And then I can go ahead and factor out the

1/2, 1/2 times theta times sin theta cos theta plus c.

Now that weve simplified it to this point, what we ned to do is go ahead and convert

this from thetas back to xs. And the way that we do that, we have to solve for theta. We

need to go back to this formula here. We don't need the 1. And the way that we do

that is by raising both sides, the base tangent to the negative 1 so if we do tan to the negative

1 of x equals tan to the negative 1 of tan theta, that tan to the negative 1 and tan

will cancel and we'll just be left with theta. So we've solved for theta so we can plug in

tan to the negative 1 of x for this theta here so we'll have 1 times tan to the negative

1 of x. And then for sin and cos, we'll need to use

a reference triangle for this substitution here. And our reference triangle, we can draw

down here. So we know that tan is opposite over adjacent

from trigonometry so if we draw our reference triangle and we they always draw the reference

triangle the same way, theta is here. tan we've already determined is opposite over

adjacent and we have x over 1. So that means x is opposite and 1 is adjacent

and then the hypotenuse is the square root of u2 plus a2 which means it's the square

root of x2 plus 1. So this is our reference triangle. And now

that we have this, we can use it to plug in for sin and cos. Because we know that sin

is opposite over hypotenuse, so opposite, x over hypotenuse, the square root of x2 plus

1 times cos theta and cos is adjacent over hypotenuse.

So 1 over the square root of x2 plus 1 is for cos and then plus c.

Therefore: Now all we need to do is simplify and it looks

like our final answer will be 1/2 times tan to the negative 1 of x plus and then we'll

have x on the top. And since we have a square root times a square

root, the square root will go away and we'll just be left with x2 plus 1 on the bottom

and then plus c. And that is our final answer. Thanks, guys.

I’ll see you next time. Bye!

substitution problem today. This one is going to be the integral of 1 over (x2 + 1)2 dx.

So the first thing that we need to do is identify which of our three basic trogonometric substitutions

we're going to make. In this case because we have addition here, we've got x2 and

1 we have addition between the two so we're now going to be using the u2 plus a2 subsitution

which is u equals a tan theta. So we're going to correlate this to u2 plus

a2 which comes along with this substitution u equals a tan theta. So in order to solve

for u and a, we'll set u2 equal to x2 and we will set a2 equal to 1.

We'll take the square root of both sides here with the u2 equals x2 and we'll get u equals

x. Again take the square root here of both sides

and get a equals 1. So now we've solved for u and a and we'll

go ahead and plug those into u equals a tan theta. We will get x equals 1 times tan theta.

So we've done that. The couple last things we need to do before we start substituting

back into our problem is find the derivative of x, dx and the derivative of x is going

to be sec2 theta d theta because the derivative of tan theta is sec2 theta.

Remember: And that's a formula that you should memorize.

And the last thing that we need to do is solve for tangent of theta, which is really simple

to do. We’ll get tan theta, we divide both sides by 1 and we get tan theta equals x over

1. I'm going to leave it in this form because

it's helpful later on. But we can also simplify that over here and say tan theta equals x

because we'll probably want both. So we've done everything we need to do to

set up for our problem. So now we go ahead and substitute back into our original integral

for x and our dx. So we will end up with the integral of 1 over

tan2 theta because we have an x2 here plus 1 and then that is squared.

And then substituting for dx, we'll put in tan2 theta d theta.

So now we've substituted and all we need to do is simplify. So the first thing we need

to do, this tan2 of theta plus 1 coming along with this substitution or this identity here

that was in our problem. And this substitution tells us to use the identity tan2 theta plus

1 equals sec2 theta which is a formula. Since we have tan2 theta in our problem, we

can change that to 1 over sec2 theta, which we plugged in for this tan2 theta plus 1 and

then we'll leave the squared because that's there.

So, we have that and then we leave this sec2 theta d theta. What we'll end up with is the

integral of sec2 theta over, since we have sec2 and then that is also squared, we end

up with sec4 theta and then d theta here at the end.

And of course we could cancel out the sec2 theta from the numerator and denominator so

we'll end up with the integral of 1 over sec2 theta d theta.

So 1 over sec2 is the same thing as cos2 so that would simplify to the integral of cos2

theta d theta. Which is about as simple as we can get our

integral before we go ahead and integrate. What we'll have to do is use a formula for

the integral of cos2 of a variable. The formula for that integral is going to be 1/2 theta

plus 1/4 sin of 2 theta plus c. So this is the integration formula when you

have cos2 of the variable. Now what we'll need to do is use the double angle formula

for sin of 2 theta. sin 2 theta is the same thing as 2 times sin theta times cos theta.

That is in double angle formula because this sin 2 theta means twice the angle theta. So

2 sin theta cos theta and then of course plus c.

So now we can simplify this. If we multiply out 2 times 1/4 here, we'll get 1/2 and then

this parenthesis will go away. And then I can go ahead and factor out the

1/2, 1/2 times theta times sin theta cos theta plus c.

Now that weve simplified it to this point, what we ned to do is go ahead and convert

this from thetas back to xs. And the way that we do that, we have to solve for theta. We

need to go back to this formula here. We don't need the 1. And the way that we do

that is by raising both sides, the base tangent to the negative 1 so if we do tan to the negative

1 of x equals tan to the negative 1 of tan theta, that tan to the negative 1 and tan

will cancel and we'll just be left with theta. So we've solved for theta so we can plug in

tan to the negative 1 of x for this theta here so we'll have 1 times tan to the negative

1 of x. And then for sin and cos, we'll need to use

a reference triangle for this substitution here. And our reference triangle, we can draw

down here. So we know that tan is opposite over adjacent

from trigonometry so if we draw our reference triangle and we they always draw the reference

triangle the same way, theta is here. tan we've already determined is opposite over

adjacent and we have x over 1. So that means x is opposite and 1 is adjacent

and then the hypotenuse is the square root of u2 plus a2 which means it's the square

root of x2 plus 1. So this is our reference triangle. And now

that we have this, we can use it to plug in for sin and cos. Because we know that sin

is opposite over hypotenuse, so opposite, x over hypotenuse, the square root of x2 plus

1 times cos theta and cos is adjacent over hypotenuse.

So 1 over the square root of x2 plus 1 is for cos and then plus c.

Therefore: Now all we need to do is simplify and it looks

like our final answer will be 1/2 times tan to the negative 1 of x plus and then we'll

have x on the top. And since we have a square root times a square

root, the square root will go away and we'll just be left with x2 plus 1 on the bottom

and then plus c. And that is our final answer. Thanks, guys.

I’ll see you next time. Bye!