Question

y^{'}=sqrt{\frac{ 1-y^{2} }{ 1-x ^{2} }}

Answer 1

\[y^{'}=\sqrt{\frac{ 1-y^{2} }{ 1-x ^{2} }}\]

Answer 2

yes...i hve open the square root to equal 1/2

Answer 3

separate variables
\[\frac{dy}{\sqrt{1-y^{2}}} = \frac{dx}{\sqrt{1-x^{2}}}\]
integrate
\[\sin^{-1} y = \sin^{-1} x +C\]
\[y = \sin (\sin^{-1} x + C)\]

Answer 4

can explain more how u can get sin\[\sin^{-1} x\]

Answer 5

it use subtitude u

Answer 6

sure
use trig substitution
\[x = \sin \theta\]
\[dx = \cos \theta\]
\[\int\limits \frac{dx}{\sqrt{1-x^{2}}} = \int\limits \frac{\cos \theta}{\sqrt{1-\sin^{2} \theta}} = \int\limits d \theta\]

Answer 7

still not understand...its the equation t
\[\int\limits d \theta = \sin ^{-1} x\]

Answer 8

\[\int\limits d \theta = \theta\]
we know
\[x = \sin \theta\]
to solve for theta in terms of "x" you must use inverse sine function
\[\theta = \sin^{-1} x\]

Answer 9

ohh ok undertsnd,,thanks for helping,,

Answer 10

yw

Answer 11

its \[\sqrt{1-\sin ^{2} \theta}= \cos \theta \]
from trig identity..
so we just simplify get just
\[\int\limits d \theta\]

Answer 12

to confirm the answer it true or not...and to more undertsand..TQQQ

Answer 13

oh yes correct...sorry i skipped some steps in my explanation
from pythagorean thm
\[\sin^{2} x + \cos^{2} x = 1\]
we know
\[\cos x = \sqrt{1-\sin^{2} x}\]
then i simplified
\[\frac{\cos \theta}{\cos \theta} d \theta = d \theta\]

Answer 14

it okay u are lot helping just right know...hopely if anything i cannot understand i can ask u again???

Answer 15

sure if im on and available i'll help you out