Question

@solomonzelman

Answer 1

I have to prove that it's true

Answer 2

tri sub
u=sin(x)

seeems to be the first thing

Answer 3

trig* sub

Answer 4

Or you can differentiate the inverse sin

Answer 5

okay, so if we substitute, we get u^-1, then we differentiate to get ln|x| + C

Don't we?

Answer 6

what do you substitute? ??

Answer 7

ln|u| + C I meant


substitute u for sin

Answer 8

if your expression is a quadratic on the bottom, you don't have ln(u)

Answer 9

alert... I got to go...
please tag someone.

Answer 10

aww...ok. Thanks for your help tho

Answer 11

@ParthKohli @mathmale @Hero

Answer 12

@zepdrix

Answer 13

I'd suggest that you differentiate \[\sin ^{-1}u \] as your first step towards obtaining the integrand \[\frac{ 1 }{ 1-u^2 }.\]

Answer 14

the integral formula gives you the inverse sine of u.
Differentiating this result should take you back to the original integrand.

Answer 15

okay, so that equals -csc u cot u, right?

Answer 16

the derivative of sin^-1 I mean

Answer 17

mmmm

is that arcsin or cosec :-))

Answer 18

cosecant i believe

Answer 19

Now, using that, I got -cos u/cos^2 u

Is that right? And what do I do next?

Answer 20

1- cos^2 u I mean

Answer 21

Sorry, but no. Please look up "Derivatives of the Inverse Trig Functions."

Answer 22

i think there might be something wrong with this question

Answer 23

No, not at all.

Answer 24

\[\sin ^{-1}x \]

Answer 25

... is the "inverse sine" function. what is its derivative? Hint: the derivative does NOT involve cos x.

Answer 26

there's a square root missing

\[\dfrac{d}{du} [sin^{-1}(u)] = \dfrac{1}{\sqrt{1-u^2}}\]

Answer 27

Now that provides some useful insight. Thank you.

Answer 28

I have verified this through an Internet search. You're right.

Thus, you may also be right about there being something missing or wrong about the integral that we're supposed to prove.

Answer 29

\[\int du \; \dfrac{1}{1-u^2} = ln \sqrt{\dfrac{1+u}{1-u}}+C\]

partial fractions or even a trig or hyperbol sub.....

Answer 30

but i'd be a lot happier if you too agreed that the question is wrong.........😖

Answer 31

First: partial fractions don't apply here, I believe, because we're told to use differentiation (not integration) to prove the given supposed identity.

I agree with the integral you've shared, but must point out that it does not help us to prove the so-called integration identity.

I won't say more, since I have no sound basis for saying more!

Answer 32

PF's ??

\(\dfrac{1}{1-u^2} = \dfrac{1}{(1-u)(1+u)}\)


that's where the ln solution comes from 😋

Answer 33

in terms of diff'ing the RHS

**if** we have an arcsin, we can say that \(z = \arcsin u\)

\(\sin z = u\)

\(\cos z \frac{dz}{du} = 1\)

\( \dfrac{dz}{du} = \dfrac{1}{\cos z} = \dfrac{1}{\sqrt{1-u^2}} = \dfrac{d}{du} \sin^{-1} u\)

Answer 34

I follow this, but wonder what we can conclude from it.

Answer 35

we can conclude that the question is wrong 😞

Answer 36

as in flawed

Answer 37

But doesn't sin^-1 = csc?

And my book says the derivative of csc is -csc x cot x

Answer 38

@SolomonZelman @zepdrix @mathmale

Answer 39

Sup? :)

Answer 40

wow long thread lol

Answer 41

yea, lol

Answer 42

no no no,\[\large\rm \sin^{-1}x\quad \ne \quad\left(\sin x\right)^{-1}\]
The negative 1 on the sine is reserved for inverse function.
It doesn't behave the same way as other powers on sine.

Answer 43

Oh...

Answer 44

Still need help on this one? :)
This is a fun question.

Answer 45

But doesn't sin^-1 = csc? No

Answer 46

Okay, well I did just check the back of my book and there is a formula that says the derivative of sin^-1 x = \(\Large\frac{1}{\sqrt{1-x^2}}\)

Answer 47

Ya, small typo in the problem it looks like.

Answer 48

but how do I get from that to the answer?

Answer 49

\[\large\rm y=\arcsin(x)\]You start here.

Answer 50

We can rewrite this as\[\large\rm x=\sin(y)\]are you ok with that step?

Answer 51

Please write both possibilities in careful detail:\[\frac{ 1 }{ \sin x }=\csc x.\]

\[\sin ^{-1}x \] id quite different; it's called the "inverse sine function."

Answer 52

does x = u, or does y = u in your situation @zepdrix

Answer 53

You might want to look up the derivative of the inverse sine function.

Answer 54

already did^

Answer 55

Oh sorry, ya let's start here then, \(\large\rm y=arcsin(u)\) so the variable matches up.

Answer 56

k

Answer 57

\[\large\rm y=\arcsin(u)\qquad\to\qquad u=\sin(y)\]Ok with that? :d

Answer 58

yeah

Answer 59

Let's draw a triangle with angle y, to describe this situation.