integration problem

In fact we should consider the range of the corresponding inverse trigonometric functions as follows:
(1) Sub x = a sin θ
Then θ = sin-1 (x/a) and with reference to the curve of arc sin below:

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Oct07/arcsin.jpg

We can see that the range is between -π/2 and π/2 (close interval) and therefore:
-π/2 ≦ θ ≦ π/2
(2) Sub x = a tan θ
Then θ = tan-1 (x/a) and with reference to the curve of arc sin below:

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Oct07/arctan.jpg

We can see that the range is between -π/2 and π/2 (open interval) and therefore:
-π/2 < θ < π/2
(3) Sub x = a sec θ
Then θ = sec-1 (x/a) and with reference to the curve of arc sin below:

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Oct07/arcsec.jpg

We can see that the range is between 0 and π (close interval) except π/2 and therefore:
π ≧ θ > π/2 or 0 ≦ θ < π/2

1, square 入面個數要係正數
2. arc sin, arc cos, arc tan的取值範圍係規定左 ,冇proof 可言

For example, sin x = 1/2, arc sin(1/2) = 30度,唔會出150度