Question

help, is this true?
sin(theta)=x-2
the same as
(theta)=arcsin(x-2)

Answer 1

or how do i solve for theta?

Answer 2

no not the same sin(theta)=x-2 gives me any solutions to theta
theta=arcsin(x-2) gives only one value for theta

Answer 3

how would i solve it? im doing trig substitution but i got a theta and i dont know what to substitute, i know i have sintheta=x-2

Answer 4

sin(theta)=sin(theta+2 pi *n) for integer n right?

Answer 5

trig sub you only use one value

Answer 6

\[\sin(\theta+2 n \pi) =x-2 \]

Answer 7

So now solve for theta like you were

Answer 8

i bet if you post the original problem myininaya would be all over it.

Answer 9

maybe post the original problem and say what the substitution was
then we can go backwards

Answer 10

im so lost :( when i integrate i get -cos(theta)+(1/4)cos^4(theta)+(1/3)cos^3(theta)+2theta

Answer 11

\[\int\limits_{1}^{3}x \sqrt{1-(x-2)^2}dx\]

Answer 12

i will brb
i have food
@satellite73

Answer 13

lord what a pain
but that is ok, you have the answer it looks like
the last term will just be
\[\sin^{-1}(x-2)\] there is nothing you can do about it

Answer 14

so that is what theta equals?

Answer 15

fortunately for you this is easy to evaluate at 3 and at 1

Answer 16

yup

Answer 17

so i should get pi?

Answer 18

as my final?

Answer 19

3 gets you
\[\sin^{-1}(1)=\frac{\pi}{2}\] and 1 gets you
\[\sin^{-1}(-1)=-\frac{\pi}{2}\] so it is not a big deal

Answer 20

yes subtract and get
\[\pi\]

Answer 21

not sure about the first part though

Answer 22

actually yes, the whole thing is pi
that is it!

Answer 23

sweet

Answer 24

i think (although i am not sure) you can get the answer most easily by changing the limits of integration twice

Answer 25

ok so i'm back

Answer 26

first by
\[u=x-2\] change to
\[\int_{-1}^1\] and then when you make
\[\sin(\theta)=u\] change to
\[\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\]

Answer 27

that way you do not have to go backwards once you get your answer.
myininaya knows more about this than i do. maybe i am wrong.
your answer is right however

Answer 28

thanks a lot

Answer 29

\[\int\limits\limits_{1}^{3}x \sqrt{1-(x-2)^2}dx \]
\[\text{ Let } x-2 = \cos(\theta)\]
=> \[dx=-\sin(\theta) d \theta\]
So we have x=1 to x=3
\[\text{We this in terms of } \theta \text{ since we are putting the expression in terms of } \theta\]
Ok so we have:
\[1-2=\cos(\theta) \text{ to } 3-2=\cos(\theta)\]
\[=>-1=\cos(\theta) \text{ to } 1= \cos(\theta)\]
\[=>\theta=\pi \text{ to } \theta=0\]
So we have
\[\int\limits_{\pi}^{0} \sqrt{1-\cos^2(\theta)} (\cos(\theta)+2) (-\sin(\theta)) d \theta \]

and by the way you could have gone with the sin sub if you wanted to