Help with integral arcsin(√(1-x^2))

[ilan12]

Hey,

I need help with computing the following integral:
[MATH]∫ arcsin(√(1-x^2))dx[/MATH]

 

[Steven G]

Fine. There are many helpers here to help! Can you please tell us where you are and what you need help with? It really is hard to help you if we do not know where you are stuck. Please post back with your work.

 

[pka]

I need help with computing the following integral:
[MATH]∫ arcsin(√(1-x^2))dx[/MATH]

Look here

 

[Dr.Peterson]

I'd first consider rewriting it as another inverse trig function to simplify it. Then I'd try integration by parts.

What have you tried? Where do you need help?

 

[ilan12]

thank you for your help,
well I tried to set [MATH]t=√(1−x^2)[/MATH]I got [MATH]∫-arcsin(t)*\frac{\sqrt{1−t^2}}{(t)}dt [/MATH]
then I set [MATH]p=arcsint[/MATH]and I got [MATH]∫\frac{p(1-sin^2p)}{sinp}dp[/MATH]and then i got stuck.

also tried to integrabe by parts
[MATH] u' = 1, v=arcsin\sqrt{1-x^2}[/MATH]But i seems like its not the way

 

[Dr.Peterson]

I'm not sure how important my first step is, but I started by thinking, if [MATH]\sin(\theta) = \sqrt{1-x^2}[/MATH], then what trig function has a simpler form; that turned out to be the cosine, so I could rewrite the integrand in terms of arccos.

Then I used parts, probably the way you stated, though you seem to be reversing the traditional roles of u and v. Please show the details of what you did there (with or without my first step).

 

[Deleted member 4993]

Hey,

I need help with computing the following integral:
[MATH]∫ arcsin(√(1-x^2))dx[/MATH]

We know:

arcsin(sin(A)) = A

So lets substitute:

\(\displaystyle x = cos(\theta)\)

\(\displaystyle dx = -sin(\theta)\ d\theta\)

continue....