Question

Inverse trigonometric functions,

Can you please help me where am I wrong in this problem ?
http://screencast.com/t/mQtujojz4F

Answer 1

It's not clear what it is you are trying to do. You can't treat separately the two terms under the radical. You must take them together.

Completing the Square
\(4x - x^{2} = -(x^{2} - 4x + 4) + 4 = 4 - (x-2)^{2}\)

Now try it. Maybe \(x-2 = 2\sin(u)\)?

Answer 2

well in our formula of \[∫1/√a2−x2=\sin−1(x/a)\] a is constant not a variable

Answer 3

you have considered it as variable

Answer 4

that's wrong why @prince90p ?

Answer 5

because constant doesn't change its value but variable does .. okay let's do an example

let k be constant and x be variable

\[\int\limits_{}^{}k.dx=kx \]and

\[\int\limits_{}^{}x=x ^{2}/2\]

so my point is integration treats constant and variable in different ways got it?

you will get wrong answer if you consider variable x as a constant