Question

(integral sign) 2/sqrt(4-x^2) dx using the substitution x= 2sin theta

Answer 1

get
\[u=2\sin(\theta)\]
\[du=2\cos(\theta)d\theta\] and get
\[\int\frac{\cos(\theta)}{\sqrt{4-4\sin^2(\theta)}}\]

Answer 2

denominator is
\[\sqrt{4(1-\sin^2(\theta))}=2\cos(\theta)\]

Answer 3

since
\[1-\sin^2(\theta)=\cos^2(\theta)\]

Answer 4

making your integral
\[\frac{1}{2}\int1d\theta\]

Answer 5

pretty easy integral you get
\[\frac{1}{2}\theta\]

Answer 6

now since
\[x=2\sin(\theta)\] you have
\[\frac{x}{2}=\sin(\theta)\] and therefore
\[\theta =\sin^{-1}(\frac{x}{2})\]

Answer 7

so answer is
\[\frac{1}{2}\sin^{-1}(\frac{x}{2})\] oh and don't forget plus C

Answer 8

no i screwed up. i am off by the constant. it is
\[2\sin^{-1}(\frac{x}{2})+C \]sorry