Which method is correct to solve this question, my textbook says method two but i don't know why method one wouldn't work.

Any help would be appreciated.

(Question and methods below)

Any help would be appreciated.

(Question and methods below)

Original post by firestudent

Method 2 gives arcsin(x).

Method 1 you seem to have

dy/dx = x/sqrt(1-x^2)

so

1/x dy = 1/sqrt(1-x^2) dx

so

I = Int 1/x dy = Int 1/sqrt(1-y^2) dy

so, somewhat appropriately, youve gone round in circles. Not sure how you wrote down

I = ...

on the following line.

(edited 4 months ago)

Ok i've just seen my mistake but either way would I be able to use this method of integration instead of the substitution?

here is the corrected (i think) working

Original post by firestudent

Ok i've just seen my mistake but either way would I be able to use this method of integration instead of the substitution?

First off, there is a fair bit of interesting history behind integrating sqrt(1-x^2) and 1/sqrt(1-x^2)

https://math.stackexchange.com/questions/3759552/purely-geometric-proof-of-inverse-trigonometric-functions-derivatives

https://www.quantamagazine.org/how-isaac-newton-discovered-the-binomial-power-series-20220831/

https://www.youtube.com/watch?v=gMlf1ELvRzc&ab_channel=Veritasium

You should immediately think of circles/pythagoras/trig when you see similar calculus problems.

Both methods are substitution, its just method 2 is based on a trig substitution. Doing a pythagorean substitution as you seem to be trying to do so

y^2 + x^2 = 1

wont really change the integration problem (by symmetry). As 1/sqrt(1-x^2) is the derivative of arcsin(x) then in order to integrate it up youd have to do a trig substitution or a power series, but I cant see how a substitution similar to what youre proposing will work out.

Edit - your modified post has a similar problem as youre not being clear that the 1/x is inside the integral. So in both cases, write the new integral down clearly as in the previous reply.

(edited 4 months ago)

Ok thank you for the additional info.

I don't see how my 'method 1' is a substitution as i have simply integrated by reversing the chain rule.

And as for your edit what do you mean by the 1/x is not clearly inside the integral?

Thanks for the help

I don't see how my 'method 1' is a substitution as i have simply integrated by reversing the chain rule.

And as for your edit what do you mean by the 1/x is not clearly inside the integral?

Thanks for the help

Original post by firestudent

Ok thank you for the additional info.

I don't see how my 'method 1' is a substitution as i have simply integrated by reversing the chain rule.

And as for your edit what do you mean by the 1/x is not clearly inside the integral?

Thanks for the help

I don't see how my 'method 1' is a substitution as i have simply integrated by reversing the chain rule.

And as for your edit what do you mean by the 1/x is not clearly inside the integral?

Thanks for the help

As in the previous post you have

dy/dx = x/sqrt(1-x^2)

so seperation of variables, substitute for x and integrating

Int 1/x dy = Int 1/sqrt(1-y^2) dy = Int 1/sqrt(1-x^2) dx

So as you should expect with the symmetry of a circle/pythagoras x^2+y^2=1, the integration problem is unchanged.

So you cant write down

I = -(1-x^2)^(1/2) / x + c

Youve got to be clear about what happens to the extra "x". So expand the lines between dy/dx = ... and I =... and be clear about what youre integrating.

Original post by firestudent

Ok thank you for the additional info.

I don't see how my 'method 1' is a substitution as i have simply integrated by reversing the chain rule.

And as for your edit what do you mean by the 1/x is not clearly inside the integral?

Thanks for the help

I don't see how my 'method 1' is a substitution as i have simply integrated by reversing the chain rule.

And as for your edit what do you mean by the 1/x is not clearly inside the integral?

Thanks for the help

Just an aside, if you plot the integrand and the two solutions

https://www.desmos.com/calculator/8xzihirim2

you see that arcsin() has about the right form, so infinite gradient as x->+/-1 and has gradient 1 when x=0 (based on the form of the integrand). Its also increasing as x increases because the integrand is always positive.

Your solution does the opposite so is 0 when x=+/-1 (negative gradient/decreasing to begin with) and tends to +/-inf when x->0. Just thinking about the area of thin strips under the integrand would show this cant be correct.

(edited 4 months ago)

Ok thanks, I haven't yet learnt about differential equations which I believe may have confused me slightly but this seems to make more sense now knowing that it requires the separation of variables

Original post by firestudent

Ok thanks, I haven't yet learnt about differential equations which I believe may have confused me slightly but this seems to make more sense now knowing that it requires the separation of variables

Its just not the right way to do this type of integal. Seperation of variables is really just based on the reverse chain rule so its what youre trying to do. Having said that, you need to make sure youre clear about the mistake youre making between the two steps dy/dx = ... and I = ...

Similarly, have a read of the previous links and get a clear idea of why the trig subs are the way to go for this type of pythagoras integrand so (1-x^2)^(+/-1/2)

(edited 4 months ago)

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