- #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]
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]
Dick said:To integrate arcsin(x)dx do parts again. u=arcsin(x) dv=dx.
iRaid said:Ok then I get:
[tex]xarcsinx-\int x/(\sqrt{1-x^{2}})[/tex]
Ok after all that:Dick said:The second term can be integrated with a simple substitution.
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.
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?
iRaid said:That doesn't work or I'm not understanding
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.
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]
iRaid said:The integral I'm trying to solve is:
[tex]\int \frac{xarcsinx+\sqrt{1-x^2}}{\sqrt{1-x^{2}}}dx[/tex]
iRaid said:Ok then I get:
[tex]xarcsinx-\int x/(\sqrt{1-x^{2}})[/tex]
Zondrina said:Oh my apologies, I thought :
Was what you were having trouble with.
Dick said:Split the integral into two parts. Looks like you need another round of integration by parts on the first piece.
iRaid said:This just never ends omg.
iRaid said:I made u=xarcsinx, but I don't know the derivative of that...
Dick said:u=arcsin(x) dv=xdx/sqrt(1-x^2). Integrating that should look familiar.
iRaid said:Yeah I got it,
My answer comes out:
[tex]arcsinx(arcsinx+\sqrt{1-x^{2}})-(\frac{x(arcsinx)^{2}}{2}+x)[/tex]
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.
iRaid said:This is ridiculous so much work just to get the wrong answer lol.
Dick said:Now that you know the steps you just have to make sure everything is right. You're close.
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?
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...
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?
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...
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?
Dick said:What part don't you get? Do you have a problem with ILS's clever substitution?
iRaid said:I'm not sure on his notation with the d(siny) part particularly.
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.
iRaid said:I see it, thanks. But I would probably never think of that lol.
Dick said:Well, I didn't either. Don't feel bad.
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.
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.
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.
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.
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.