integral [from x to 0] (r^2 - x^2)^{1/2} dx : different units

[sinx]

int(r²-x²)^1/2dx, limits x=x, 0 (x is upper limit, 0 is lower limit)
=int r(1-^x2/_r²)^1/2dx

sub x/r=sin&
=r²int cos²&d&, limits &=sin^-1(x/_r), 0 ??
cos2&=¹/₂(1+cos2&)
=r²/2 int(1+cos2&)d&

sub u=2&
1/2du=d&
=r²/4 int(1+cosu)du limits u=2sin^-1(x/_r),0 ??
=r²/4 (u+sinu) evaluated at limits u=2sin^-1(x/_r),0
=r²/4 ((2sin^-1(x/_r) + sin 2sin^-1(x/_r))-(0+sin0))
=r²/4 (2sin^-1(x/_r) + sin (2sin^-1(x/_r)))

besides the questions in red about changing limits, here is the main question;
2sin^-1(x/_r) is an angle; sin (2sin^-1(x/_r) is a number, the units are not the same.
in fact, the expression u+sinu gives units are that are not the same (whether or not my limits are correct).
when multiplied by the constant in front (r²/4), you get inches², or mm² or whatever r is.
this should be the units of the integral, since it is an area.
I know radians are 'unitless', but do i ignore the fact that i am adding an angle to the sin of an angle?
is sin& 'unitless' also? (of course it is, inches/inches, answered my own question here)
still do you add radians to a dimentionless number?

Is there a way to determine sin(2sin^-1&)?,
that is sin(sin^-1&) is just &, but what about sin(2sin^-1&)?, [I'm sure it is not 2&].

 

[Dr.Peterson]

int(r²-x²)^1/2dx, limits x=x, 0 (x is upper limit, 0 is lower limit)
=int r(1-^x2/_r²)^1/2dx

sub x/r=sin&
=r²int cos²&d&, limits &=sin^-1(x/_r), 0 ??
cos2&=¹/₂(1+cos2&)
=r²/2 int(1+cos2&)d&

sub u=2&
1/2du=d&
=r²/4 int(1+cosu)du limits u=2sin^-1(x/_r),0 ??
=r²/4 (u+sinu) evaluated at limits u=2sin^-1(x/_r),0
=r²/4 ((2sin^-1(x/_r) + sin 2sin^-1(x/_r))-(0+sin0))
=r²/4 (2sin^-1(x/_r) + sin (2sin^-1(x/_r)))

besides the questions in red about changing limits, here is the main question;
2sin^-1(x/_r) is an angle; sin (2sin^-1(x/_r) is a number, the units are not the same.
in fact, the expression u+sinu gives units are that are not the same (whether or not my limits are correct).
when multiplied by the constant in front (r²/4), you get inches², or mm² or whatever r is.
this should be the units of the integral, since it is an area.
I know radians are 'unitless', but do i ignore the fact that i am adding an angle to the sin of an angle?
is sin& 'unitless' also? (of course it is, inches/inches, answered my own question here)
still do you add radians to a dimentionless number?

Is there a way to determine sin(2sin^-1&)?,
that is sin(sin^-1&) is just &, but what about sin(2sin^-1&)?, [I'm sure it is not 2&].

First, all your limits are correct (though you should know that limits are read from bottom to top, so it's "from 0 to x"; and it's really a little wrong to use x both as the variable of integration and in the limit). It would have been easier to read if you'd used either an actual theta (θ) or a letter like t, but that's okay.

As far as calculus is concerned, trig functions take mere numbers as argument, and produce mere numbers -- all dimensionless quantities. We don't think in terms of radians as a unit in this context. That is, the sine or inverse sine is just a function from the real numbers to the real numbers. A good presentation of the unit circle model will make this clear; we just wrap a number line around the unit circle, with no units involved.

To evaluate sin(2sin^-1u) [you didn't mean theta there, did you?], you can just plug numbers into your calculator. But if you had a reason to want to simplify the expression, you can use the double-angle formula, sin(2θ) = 2sin(θ)cos(θ). So you get sin(2sin^-1u) = 2sin(sin^-1u)cos(sin^-1u) = 2u√(1-u²). [This is valid without worrying about the sign of the cosine because the inverse sine is always in a quadrant where the cosine is positive.]