Overlord Drakow
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Arcsin is the inverse function of sin, you might have seen it written as sin^-1.
e.g. sin pi/2 = 1, so arcsin 1 = pi/2.
Note that the range of arcsin x is -pi/2 to pi/2, so arcsin 1 = pi/2 and not, say, 5 pi/2 (even though sin 5 pi/2 = 1)
For me, instead of turning t back into x, I changed the limits and evaluated based on t:
Let x = sin t.
dx/dt = cos t
dx = cos t dt
First we change the upper and lower limits from x to t. arcsin 0 = 0 and arcsin 1 = pi/2.
= 0 - 0 + pi - 0
= pi
PROTIP
If you see sqrt(a^2-x^2), try to sub in x = a sin t.
For sqrt(x^2-a^2), sub in x = a sec t.
For sqrt(a^2+x^2), sub in x = a tan t.
@Drakow: Nitpicking here. It's incorrect to use 1 and 0 as the upper/lower limits after subbing x for t. You must either change the limits (as I did), or start off with the indefinite integral for x so that you don't need to write the limits for t.
What you say is true, however, because the substitution reverts back to x anyway, I did not bother changing the limits just to change them back to the original limits. Though you are right, I am mathematically wrong. Anyway, I think your solution is probably the one he was expected to give for his exam.
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