Integral (arcsin x)^{2}=uv-\int vdu

  • Thread starter iRaid
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In summary: Because parts is the way to deal with the something like arcsin(x). You don't know how to integrate it, so you keep putting u=arcsin(x) and then the du won't have an arcsin(x) in it...which is why your answer was wrong.
  • #1
iRaid
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Homework Statement


[tex]\int (arcsin x)^{2}[/tex]

Homework Equations


The Attempt at a Solution


[tex]u=arcsin x[/tex] [tex]du=1/\sqrt{1-x^{2}}dx[/tex]
[tex]v=?[/tex] [tex]dv=arcsin x dx[/tex]
 
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  • #2


iRaid said:

Homework Statement


[tex]\int (arcsin x)^{2}[/tex]


Homework Equations





The Attempt at a Solution


[tex]u=arcsin x[/tex] [tex]du=1/\sqrt{1-x^{2}}dx[/tex]
[tex]v=?[/tex] [tex]dv=arcsin x dx[/tex]

To integrate arcsin(x)dx do parts again. u=arcsin(x) dv=dx.
 
  • #3


Dick said:
To integrate arcsin(x)dx do parts again. u=arcsin(x) dv=dx.

Ok then I get:
[tex]xarcsinx-\int x/(\sqrt{1-x^{2}})[/tex]
 
  • #4


iRaid said:
Ok then I get:
[tex]xarcsinx-\int x/(\sqrt{1-x^{2}})[/tex]

The second term can be integrated with a simple substitution.
 
  • #5


Dick said:
The second term can be integrated with a simple substitution.
Ok after all that:
[tex]\int arcsinx=xarcsinx+\sqrt{1-x^{2}}[/tex]
Then putting in for v, I end up with:
[tex](arcsinx(arcsinx+\sqrt{1-x^{2}})-\int \frac{(xarcsinx+\sqrt{1-x^{2}})}{(\sqrt{1-x^{2}})}[/tex]

Do I do another u-substition for the integral, u=arcsinx? But I don't know what to do with the x infront of it.
 
  • #6


iRaid said:
Ok after all that:
[tex]\int arcsinx=xarcsinx+\sqrt{1-x^{2}}[/tex]
Then putting in for v, I end up with:
[tex](arcsinx(arcsinx+\sqrt{1-x^{2}})-\int \frac{(xarcsinx+\sqrt{1-x^{2}})}{(\sqrt{1-x^{2}})}[/tex]

Do I do another u-substition for the integral, u=arcsinx? But I don't know what to do with the x infront of it.

No no!

Let u = 1 - x2 so that du = -2x dx which means -du/2 = x dx.

What does that second integral x / sqrt(1-x2) become now?
 
  • #7


Zondrina said:
No no!

Let u = 1 - x2 so that du = -2x dx which means -du/2 = x dx.

What does that second integral x / sqrt(1-x2) become now?

That doesn't work or I'm not understanding
 
  • #8


iRaid said:
That doesn't work or I'm not understanding

[itex]\int \frac{x}{\sqrt{1-x^2}} dx[/itex]

[itex] u = 1-x^2 → - \frac{1}{2} du = xdx[/itex]

[itex]\int \frac{x}{\sqrt{1-x^2}} dx = - \frac{1}{2} \int \frac{1}{\sqrt{u}} du[/itex]
 
  • #9


iRaid said:
Ok after all that:
[tex]\int arcsinx=xarcsinx+\sqrt{1-x^{2}}[/tex]
Then putting in for v, I end up with:
[tex](arcsinx(arcsinx+\sqrt{1-x^{2}})-\int \frac{(xarcsinx+\sqrt{1-x^{2}})}{(\sqrt{1-x^{2}})}[/tex]

Do I do another u-substition for the integral, u=arcsinx? But I don't know what to do with the x infront of it.

Split the integral into two parts. Looks like you need another round of integration by parts on the first piece.
 
  • #10


Zondrina said:
[itex]\int \frac{x}{\sqrt{1-x^2}} dx[/itex]

[itex] u = 1-x^2 → - \frac{1}{2} du = xdx[/itex]

[itex]\int \frac{x}{\sqrt{1-x^2}} dx = - \frac{1}{2} \int \frac{1}{\sqrt{u}} du[/itex]

The integral I'm trying to solve is:
[tex]\int \frac{xarcsinx+\sqrt{1-x^2}}{\sqrt{1-x^{2}}}dx[/tex]
 
  • #11


iRaid said:
The integral I'm trying to solve is:
[tex]\int \frac{xarcsinx+\sqrt{1-x^2}}{\sqrt{1-x^{2}}}dx[/tex]

Oh my apologies, I thought :

iRaid said:
Ok then I get:
[tex]xarcsinx-\int x/(\sqrt{1-x^{2}})[/tex]

Was what you were having trouble with.
 
  • #12


Zondrina said:
Oh my apologies, I thought :



Was what you were having trouble with.

Lol all good, I was like am I being stupid or what?
 
  • #13


Dick said:
Split the integral into two parts. Looks like you need another round of integration by parts on the first piece.

This just never ends omg.
 
  • #14


iRaid said:
This just never ends omg.

You are almost there. It's downhill from here.
 
  • #15


I don't know the derivative of xarcsinx
 
  • #16


iRaid said:
I made u=xarcsinx, but I don't know the derivative of that...

u=arcsin(x) dv=xdx/sqrt(1-x^2). Integrating that should look familiar.
 
  • #17


Dick said:
u=arcsin(x) dv=xdx/sqrt(1-x^2). Integrating that should look familiar.

Yeah I got it,
My answer comes out:
[tex]arcsinx(arcsinx+\sqrt{1-x^{2}})-(\frac{x(arcsinx)^{2}}{2}+x)[/tex]
 
  • #18


iRaid said:
Yeah I got it,
My answer comes out:
[tex]arcsinx(arcsinx+\sqrt{1-x^{2}})-(\frac{x(arcsinx)^{2}}{2}+x)[/tex]

Not right I'm afraid. I think you might have goofed up the last integration by parts. Check it.
 
  • #19


Dick said:
Not right I'm afraid. I think you might have goofed up the last integration by parts and mistracked some signs. Check it.

This is ridiculous so much work just to get the wrong answer lol.

I got what I did wrong, it was the last integration by parts.
 
  • #20


iRaid said:
This is ridiculous so much work just to get the wrong answer lol.

Now that you know the steps you just have to make sure everything is right. You're close.
 
  • #21


Dick said:
Now that you know the steps you just have to make sure everything is right. You're close.

Yeah I know. The only problem I have is, there is no way on an exam I would keep doing integration by parts for this problem (what was it 5 times). I would have ended up going to a different way of solving it.

How do I know that this problem must be solved using integration by parts?
 
  • #22


iRaid said:
Yeah I know. The only problem I have is, there is no way on an exam I would keep doing integration by parts for this problem (what was it 5 times). I would have ended up going to a different way of solving it.

How do I know that this problem must be solved using integration by parts?

Because parts is the way to deal with the something like arcsin(x). You don't know how to integrate it, so you keep putting u=arcsin(x) and then the du won't have an arcsin(x) in it anymore.
 
  • #23
Well, you could also have started with a straight forward substitution: ##y=\arcsin x##
This means that ##x=\sin y##.
You're rid of that pesky arcsin and get easier functions to integrate.

You'd have:
$$\int (\arcsin x)^2 dx = \int y^2 d(\sin y)$$

Now do integration by parts:
$$\int y^2 d(\sin y) = y^2 \sin y - \int 2y \cos y dy$$

Repeat integration by parts, and after that back substitute x for y...
 
  • #24
I like Serena said:
Well, you could also have started with a straight forward substitution: ##y=\arcsin x##
This means that ##x=\sin y##.
You're rid of that pesky arcsin and get easier functions to integrate.

You'd have:
$$\int (\arcsin x)^2 dx = \int y^2 d(\sin y)$$

Now do integration by parts:
$$\int y^2 d(\sin y) = y^2 \sin y - \int 2y \cos y dy$$

Repeat integration by parts, and after that back substitute x for y...

I'm sorry, but I found this interesting. I've never seen d(siny) in my life. Is that actually valid to do?
 
  • #25
Zondrina said:
I'm sorry, but I found this interesting. I've never seen d(siny) in my life. Is that actually valid to do?

Sure.

##d(f(x))## is just ##f'(x)dx##.

Or ##{dy \over dx}dx## if we define y=f(x).Integration by parts is often given as:
$$\int u dv = uv - \int v du$$
You can apply it literally.It becomes more interesting if we move on to things like d(f(x,y)) or d(f(x(u,v),y(u,v)).
It is called a total derivative.
 
  • #26
I like Serena said:
Well, you could also have started with a straight forward substitution: ##y=\arcsin x##
This means that ##x=\sin y##.
You're rid of that pesky arcsin and get easier functions to integrate.

You'd have:
$$\int (\arcsin x)^2 dx = \int y^2 d(\sin y)$$

Now do integration by parts:
$$\int y^2 d(\sin y) = y^2 \sin y - \int 2y \cos y dy$$

Repeat integration by parts, and after that back substitute x for y...

Sorry I just looked back at this thread, this is interesting, but I don't know how to follow it. Could someone explain?
 
  • #27
iRaid said:
Sorry I just looked back at this thread, this is interesting, but I don't know how to follow it. Could someone explain?

What part don't you get? Do you have a problem with ILS's clever substitution?
 
  • #28
Dick said:
What part don't you get? Do you have a problem with ILS's clever substitution?

I'm not sure on his notation with the d(siny) part particularly.
 
  • #29
iRaid said:
I'm not sure on his notation with the d(siny) part particularly.

ILS just substituted x=sin(y). So dx becomes d(sin(y)). arcsin(x)=arcsin(sin(y))=y. dx=d(sin(y))=cos(y)dy. You might have noticed in the way I led you through there was a lot of repetition in calculating the various integrals. ILS's idea compacts that a bit.
 
  • #30
Dick said:
ILS just substituted x=sin(y). So dx becomes d(sin(y)). arcsin(x)=arcsin(sin(y))=y. dx=d(sin(y))=cos(y)dy. You might have noticed in the way I led you through there was a lot of repetition in calculating the various integrals. ILS's idea compacts that a bit.

I see it, thanks. But I would probably never think of that lol.
 
  • #31
iRaid said:
I see it, thanks. But I would probably never think of that lol.

Well, I didn't either. Don't feel bad.
 
  • #32
Dick said:
Well, I didn't either. Don't feel bad.

:) Thanks for the help, I appreciate it.
 

What does the integral (arcsin x)^{2}=uv-\int vdu represent?

The integral (arcsin x)^{2}=uv-\int vdu represents the antiderivative of the function (arcsin x)^{2}. It is a way to find the original function when given its derivative.

How do you solve an integral equation like (arcsin x)^{2}=uv-\int vdu?

To solve an integral equation like (arcsin x)^{2}=uv-\int vdu, you need to use integration by parts. This involves choosing a u and v such that their product is equal to the original function, and then using the formula uv-\int vdu to find the antiderivative.

What is the purpose of using (arcsin x)^{2}=uv-\int vdu in mathematics?

The purpose of using (arcsin x)^{2}=uv-\int vdu in mathematics is to find the antiderivative of a function. This can be helpful in solving problems involving rates of change and optimization.

Can (arcsin x)^{2}=uv-\int vdu be used to find the area under a curve?

Yes, (arcsin x)^{2}=uv-\int vdu can be used to find the area under a curve. This is because the integral represents the sum of infinitely small rectangles that approximate the area under the curve.

Are there any limitations to using (arcsin x)^{2}=uv-\int vdu to solve integrals?

Yes, there are limitations to using (arcsin x)^{2}=uv-\int vdu to solve integrals. This method is only applicable to certain types of functions, and it may not always give an exact solution. Additionally, it can be difficult to determine the appropriate u and v for a given function.

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