A level integration help

OP

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

Reply 4

OP

here is the corrected (i think) working

Reply 5

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 2 months ago)

Reply 6

OP

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

Reply 7

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

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.

Reply 8

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

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 2 months ago)

Reply 9

OP

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

Reply 10