int(r

=int r(1-

sub x/r=sin&

=r

cos2&=

=r

sub u=2&

1/2du=d&

=r

=r

=r

=r

besides the questions in red about changing limits, here is the main question;

2sin

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 (r

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

that is sin(sin

^{2}-x^{2})^{1/2}dx, limits x=x, 0 (x is upper limit, 0 is lower limit)=int r(1-

^{x2}/_{r2})^{1/2}dxsub x/r=sin&

=r

^{2}int cos^{2}&d&, limits &=sin^{-1}(x/_{r}), 0 ??cos2&=

^{1}/_{2}(1+cos2&)=r

^{2}/2 int(1+cos2&)d&sub u=2&

1/2du=d&

=r

^{2}/4 int(1+cosu)du limits u=2sin^{-1}(x/_{r}),0 ??=r

^{2}/4 (u+sinu) evaluated at limits u=2sin^{-1}(x/_{r}),0=r

^{2}/4 ((2sin^{-1}(x/_{r}) + sin 2sin^{-1}(x/_{r}))-(0+sin0))=r

^{2}/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 (r

^{2}/4), you get inches^{2}, or mm^{2}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&].
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