Question

Integrate 1/(sqrt(1-x^2)) Explain please :D

Answer 1

\[\int\limits (\sqrt{1-x ^{2}} )^{-1}dx = (1/x )(\sqrt{1-x ^{2}})+c\]

Answer 2

i'm not sure for that..

Answer 3

look in your book and you will find a function whose derivative is
\[\frac{1}{\sqrt{1-x^2}}\]
hint,
it is section on derivatives of inverse trig functions

Answer 4

my book did not say anything about substituting inverse trigonometry

Answer 5

not a substitution

Answer 6

\[\frac{d}{dx}[\sin^{-1}(x)]=\frac{1}{\sqrt{1-x^2}}\] and so
\[\int\frac{1}{\sqrt{1-x^2}}dx =\sin^{-1}(x)+C\]

Answer 7

is it a formula?

Answer 8

i am not sure i would call it a formula. it is a fact though

Answer 9

it's a definition.

Answer 10

because i dont understand how the trigonometry get into this equation

Answer 11

if you take the derivative of any inverse function, you can do it via
\[\frac{d}{dx}[(f^{-1})'(x)]=\frac{1}{f'(f^{-1}(x))}\]

Answer 12

Put \[x=\sin \theta\] dx = cosθ dθ
\[\int\limits (\sqrt{1-x ^{2}})^{-1} dx = \int\limits (\sqrt{1-(\sin \theta)^2} )^{-1} \cos \theta d \theta = \int\limits d \theta = \theta+C =\sin^{-1} x+C\]

Answer 13

by trigo substitution sorry

Answer 14

\[\int \frac{1}{\sqrt{1-x^2}}dx ≡ \sin^{-1}(x)+C\]

Answer 15

therefore
\[\frac{d}{dx}[\sin^{-1}(x)]=\frac{1}{\cos(\sin^{-1}(x))}=\frac{1}{\sqrt{1-x^2}}\]

Answer 16

it works for sure, but i would not use a trig sub for this because you are going around in circles.
the derivative of arcsine is
\[\frac{1}{\sqrt{1-x^2}}\] so the anti derivative of
\[\frac{1}{\sqrt{1-x^2}}\] is arcsine

they say the same thing

Answer 17

thanks @satellite73 and @Callisto !!!!!! i understand it now!!

Answer 18

yw

Answer 19

sorry, can you explain why \[d/dx [\sin^{-1} x] = 1/\cos(\sin^{-1} x) \]
i don't understand this :(

Answer 20

sure

Answer 21

\[\frac{d}{dx}[(f^{-1})'(x)]=\frac{1}{f'(f^{-1}(x))}\]

Answer 22

\[f\circ f^{-1}(x)=x\]
\[f'(f^{-1}(x))(f^{-1}(x))'=1\] by the chain rule, therefore
\[(f^{-1}(x))'=\frac{1}{f'(f^{-1}(x))}\] by algebra

Answer 23

gives a method for finding derivatives of inveres function,
reciprocal of derivative of original function evaluated at the inverse

Answer 24

is one way to show that
\[\frac{d}{dx}\ln(x) =\frac{1}{x}\]

Answer 25

now you have
\[f(x)=\sin(x), f'(x) = \cos(x), f^{-1}(x)=\sin^{-1}(x)\] so
\[\frac{d}{dx}\sin^{-1}(x)=\frac{1}{\cos(\sin^{-1}(x))}\]

Answer 26

sorry to bother, but i don't understand the first 2 steps..

Answer 27

\[f\circ f^{-1}(x)=x\]

Answer 28

that one?

Answer 29

yup and the following step

Answer 30

that is true by definition of an inverse function
when you compose a function with its inverse you get the identity

Answer 31

\[f\circ f^{-1}(x)=x\] by what it means to be an inverse

Answer 32

\[e^{\ln(x)}=x\]
\[\sin(\sin^{-1}(x))=x\] etc

Answer 33

\[f'(f^{-1}(x))(f^{-1}(x))'=1\] is true because the derivative of the right hand side is clearly 1, and the derivative of the left hand side you get by the chain rule

Answer 34

thanks i understand this part now, but still have the other part, how to get \[[\cos (\sin^{-1} x)]^{-1} = (\sqrt{1-x ^{2}})^{-1}\]