Question

How is the answer to this 2pi? I remember this prior to learning the integral techniques, but it is not coming to my myind as to how to do it.

Answer 1

this is a substitution i think

Answer 2

Trig? But then I am confused with what to do with the limits of integration. I mean, what is sin(1)?

Answer 3

\[\int\limits{\sqrt{4-x^2}}dx = \int\limits{2\sqrt{1-\frac{x^2}{4}}}dx\]

Answer 4

\[x = 2\sin(\theta) \]

Answer 5

\[\frac{dx}{d \theta} = 2\cos(\theta)\]

\[dx = d \theta (2\cos(\theta))\]

Answer 6

hmm good point with limits.. lets see

Answer 7

\[\theta = \arcsin{\frac{x}{2}}\]

so subbing our x limits in we get

\[\theta = \arcsin(\frac{2}{2}) = \arcsin{1} = \frac{\pi}{2}\]

Answer 8

i think you confused yourself by trying to sub in the x limits as theta

Answer 9

REALLY? How is arcsin(1) = pi/2? What does arcsin mean?

Answer 10

sorry, some people use this notation and some use the other, does this help:

\[\arcsin = \sin^{-1}\]

Answer 11

so its the inverse sin function

Answer 12

if sin x = y then arcsiny = x

(of course we have to limit the domain/range to get a proper inverse function thingy)

Answer 13

Yes ok I will try it again. Thanks