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

[iRaid]

Homework Statement

$$\int (arcsin x)^{2}$$

Homework Equations

The Attempt at a Solution

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

 

[Dick]

Homework Statement

$$\int (arcsin x)^{2}$$

Homework Equations

The Attempt at a Solution

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

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

 

[iRaid]

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

Ok then I get:
$$xarcsinx-\int x/(\sqrt{1-x^{2}})$$

 

[Dick]

Ok then I get:
$$xarcsinx-\int x/(\sqrt{1-x^{2}})$$

The second term can be integrated with a simple substitution.

 

[iRaid]

The second term can be integrated with a simple substitution.

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

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.

 

[STEMucator]

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

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?

 

[iRaid]

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

 

[STEMucator]

That doesn't work or I'm not understanding

$\int \frac{x}{\sqrt{1-x^2}} dx$

$u = 1-x^2 → - \frac{1}{2} du = xdx$

$\int \frac{x}{\sqrt{1-x^2}} dx = - \frac{1}{2} \int \frac{1}{\sqrt{u}} du$

 

[Dick]

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

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.

 

[iRaid]

$\int \frac{x}{\sqrt{1-x^2}} dx$

$u = 1-x^2 → - \frac{1}{2} du = xdx$

$\int \frac{x}{\sqrt{1-x^2}} dx = - \frac{1}{2} \int \frac{1}{\sqrt{u}} du$

The integral I'm trying to solve is:
$$\int \frac{xarcsinx+\sqrt{1-x^2}}{\sqrt{1-x^{2}}}dx$$

 

[STEMucator]

The integral I'm trying to solve is:
$$\int \frac{xarcsinx+\sqrt{1-x^2}}{\sqrt{1-x^{2}}}dx$$

Oh my apologies, I thought :

Ok then I get:
$$xarcsinx-\int x/(\sqrt{1-x^{2}})$$

Was what you were having trouble with.

 

[iRaid]

Oh my apologies, I thought :



Was what you were having trouble with.

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

 

[iRaid]

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.

 

[Dick]

This just never ends omg.

You are almost there. It's downhill from here.

 

[iRaid]

I don't know the derivative of xarcsinx

 

[Dick]

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.

 

[iRaid]

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

Yeah I got it,
My answer comes out:
$$arcsinx(arcsinx+\sqrt{1-x^{2}})-(\frac{x(arcsinx)^{2}}{2}+x)$$

 

[Dick]

Yeah I got it,
My answer comes out:
$$arcsinx(arcsinx+\sqrt{1-x^{2}})-(\frac{x(arcsinx)^{2}}{2}+x)$$

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

 

[iRaid]

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.

 

[Dick]

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.

 

[iRaid]

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?

 

[Dick]

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.

 

[I like Serena]

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...

 

[STEMucator]

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?

 

[I like Serena]

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.

 

[iRaid]

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?

 

[Dick]

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?

 

[iRaid]

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.

 

[Dick]

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.

 

[iRaid]

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.