So one of my classmates asked me about this question and I told her what I\'ve d

Question

So one of my classmates asked me about this question and I told her what I've done, basically factoring out a 4 from the (4-x2) and made it 1/2 (1/sqrt 1-(x/2)2) Using this I got to the answer (which should be the right answer pi/3) from sin-1(x/2). She asked me why I didn't plug in the 1/2, I answered that it doesn't get involved in the integration. But when I did another problem, I actually had to plug in the number to get the right answer. So I'm really confused now.

Hope someone could help me by giving me a detailed solution.

Explanation / Answer

Note that the derivative of arcsin(x) = 1/sqrt(1-x^2).

So, if you let y = 2u, then y^2 = 4u^2, dy = 2 du.
Our integrand becomes
2*(1/sqrt(4 - 4u^2)) du
Factor out 4 in the square root and then take the square root, you get
2*(1/(2sqrt(1-u^2))) du
The 2s in the numerator and denominator will reduce leaving
1/(sqrt(1 - u^2)) du. The integral is arcsin(u) + 1.

Back-substituting yields and put the limit

we get arcsin(4/2)/2 - arcsin(0/2)/2

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