Question

integrate sqrt(1-x^2)
Let x=cos theta

Answer 1

ok let \(x=\cos(\theta)\) then \(\sqrt{1-x^2}=\sin(\theta)\) and \(dx=-\sin(\theta)\) your integral is
\[-\int \sin^2(\theta)d\theta\]

Answer 2

alright i am past that

Answer 3

where you at?

Answer 4

i have already integrated with respect to theta and have trouble substituting in

Answer 5

ooh ok
did you get to
\[-\int\sin^2(\theta)=\frac{1}{2}(\sin(x)\cos(x)-1)\]?

Answer 6

so i have (sin(2 theta)/4)-theta/2

Answer 7

ok we can work with that one too, makes no difference

Answer 8

how do i substitute in for theta

Answer 9

sin(2theta)=2 sin theta cos theta=xsqrt(1-x^2)

Answer 10

but what about theta

Answer 11

\[\frac{x}{2}+\frac{1}{4}\sin(2\theta)\] want to work with that one?

Answer 12

oooh you really did everything!

Answer 13

you got the \(x\sqrt{1-x^2}\) part, that is the hard part
\(\theta\) is the easy part
\[x=\cos(\theta)\\
\theta=\cos^{-1}(x)\]

Answer 14

but then many times i see arcsin(x) being used for theta instead of arccos(x)

Answer 15

that is because you chose the substitution \(x=\cos(\theta)\)
which makes \(\theta =\arccos(x)\)

Answer 16

you could have chosen \(x=\sin(\theta)\) instead and done the integral that way

Answer 17

\[x=\sin(\theta), dx=\cos(\theta)d\theta, \sqrt{1-x^2}=\cos(\theta)\] your integral is
\[\int \cos^2(\theta)d\theta\]

Answer 18

in this case you would use
\[\cos^2(\theta)=\frac{1}{2}(1+\cos(2\theta))\]

Answer 19

everything works out exactly the same
and don't forget the the derivative of \(\arcsin(x)=\frac{1}{\sqrt{1-x^2}}\) while the derivative of \(\arccos(x)=-\frac{1}{\sqrt{1-x^2}}\) so it all works out exactly the same

Answer 20

thanks

Answer 21

yw