# Thread: Evaluating this indefinite integral involving arc sin

1. ## Evaluating this indefinite integral involving arc sin

The indefinite integral of sin^-1(x) or arc sin (x) divided by the square root of 1-x^2. All I can say is I have no idea what to do with this, I've never really worked with arcsin, I looked up derivatives and it doesn't seem like I can make one part of the equation's derivative cancel out another so I don't know if substitution would be a good route here. Could somebody help me out???

2. Originally Posted by fattydq
The indefinite integral of sin^-1(x) or arc sin (x) divided by the square root of 1-x^2. All I can say is I have no idea what to do with this, I've never really worked with arcsin, I looked up derivatives and it doesn't seem like I can make one part of the equation's derivative cancel out another so I don't know if substitution would be a good route here. Could somebody help me out???
If you're dealing with $\int\frac{\sin^{-1}x}{\sqrt{1-x^2}}\,dx$, make the substitution $u=\sin^{-1}x$. Things will work out really nice from there.

Can you continue?

3. So wouldn't I just be left with the indefinite integral of u? But then how does that simplify? Because the antideriv. would be u^2/2 so my final answer would be (sin^-1(x))2/2? Or am I mistaken?

4. Originally Posted by fattydq
So wouldn't I just be left with the indefinite integral of u? But then how does that simplify? Because the antideriv. would be u^2/2 so my final answer would be (sin^-1(x))2/2? Or am I mistaken?
You're not mistaken. The answer would be $\frac{\left(\sin^{-1}x\right)^2}{2}+{\color{red}C}$

5. Originally Posted by fattydq
So wouldn't I just be left with the indefinite integral of u? But then how does that simplify? Because the antideriv. would be u^2/2 so my final answer would be (sin^-1(x))2/2? Or am I mistaken?
Yes, you've got it (assuming the first two is an exponent). Don't forget the constant.