Of four minus X squared, this looks like something that I might get if I'm dealing with the How do you tackle this? And an insight that you might have is well this thing, this square root You've got the square root of four mins X squared in the denominator, you could try to use substitution, but it really doesn't simplify this in any reasonable way. That we want to evaluate this indefinite integral right over here.

We have just swapped x and theta, but I have solved the problem so that you can see that it will be the same. Scenario TWO: This is identical to how Sal solves the problem in the video. Regardless, the curves for -arccos(x / sqrt(u)) and arcsin(x / sqrt(u)) are identical except that one is vertical translation of the other, and that is all we are trying to prove with integration, anyway. Since after we integrate we are left with a constant of integration (I have used c to denote said constant), this 'absorbs' the extra pi / 2 that we would need to add for the graphs to match exactly. Although this appears different, if you look at this graph (drag the slider at the top-left at your own leisure) => <= then you will see that -arccos(x / sqrt(u)) + pi/2 is actually equal to arcsin(x / sqrt(u)). If we choose x as the side adjacent to theta, then we will end up with -arccos(x / sqrt(u)) + c. Scenario ONE: this one is comparatively complex, but it still does make sense. There are two scenarios, and these are as follows:

Inverse trig integrals cheat sheet