Challenge integral (2 Viewers)

[member 6003]

but when u divide both sides by u^2, a problem is gonna be created when u = 0 so the answers are not actually the same
View attachment 40380

no I don't think that should matter at all



You just did it wrong. man I should be doing english

 

[member 6003]

but when u divide both sides by u^2, a problem is gonna be created when u = 0 so the answers are not actually the same
View attachment 40380

besides the situation where u=0 is undefined either way since cotx is undefined

 

[unofficiallyred12]

no I don't think that should matter at all



You just did it wrong. man I should be doing english

i agree i did make a silly mistake

 

[unofficiallyred12]

besides the situation where u=0 is undefined either way since cotx is undefined

this is irrelevant since the original function being undefined at a point doesn't mean it's primitive is also undefined, eg f(x) = 1/(sqrt(1-x^2)). In this case the manipulation works (I think) and the results end up being the same due to how we define trig functions most likely, but as a general rule of thumb this isn't necessarily true e.g. for integral of (x^2+1)/(x^4+1) if we just integrate that without having done a trig sub first to obtain it, if u sketch it on desmos the thing is diff

 

[SB257426]

Good answer but there's an easier way than doing partial fraction decomposition, so I thought I might share it:



The logic is basically trying to find a way to cancel the nominator with a substitution so


then making the substitution,

which should lead to the same solution but simplified with the tan inverses combined

Thanks man i appreciate your answer, looks great

 

[Luukas.2]

but when u divide both sides by u^2, a problem is gonna be created when u = 0 so the answers are not actually the same
View attachment 40380

Your is missing a factor of 1 over the square root of 2. That is, it should be:

​

It is the same result as I obtained, just with the two terms combined into a single inverse tangent function.

Edit to add: I posted this before seeing others had made the same correction.

 

[Luukas.2]

Good answer but there's an easier way than doing partial fraction decomposition, so I thought I might share it:



The logic is basically trying to find a way to cancel the nominator with a substitution so


then making the substitution,

which should lead to the same solution but simplified with the tan inverses combined

This is making use of the same sort of approach as I used for factorising, but applied to greatly simplify the solution. Excellent!

 

[SB257426]

but when u divide both sides by u^2, a problem is gonna be created when u = 0 so the answers are not actually the same
View attachment 40380

does this mean the blue function is incorrect because of the u^2 issue ?

 

[member 6003]

does this mean the blue function is incorrect because of the u^2 issue ?

unofiiciallyred12 just made a mistake when evaluating the integral (that's why the blue function is incorrect), the u^2 issue isn't actually an issue at all

 

[member 6003]

this is irrelevant since the original function being undefined at a point doesn't mean it's primitive is also undefined, eg f(x) = 1/(sqrt(1-x^2)). In this case the manipulation works (I think) and the results end up being the same due to how we define trig functions most likely, but as a general rule of thumb this isn't necessarily true e.g. for integral of (x^2+1)/(x^4+1) if we just integrate that without having done a trig sub first to obtain it, if u sketch it on desmos the thing is diff

I don't really get what you're trying to say, I pointed out that the original integral is

so it doesn't make any sense in the first place that dividing by u on the top and bottom of the fraction would lead to a mistake since is already present (if it was 0 it's not like they can cancel), it's not being newly introduced or anything.

For more general cases: (technique being multiplying top and bottom by something)

When evaluating indefinite integrals the technique is fine, indefinite integrals are more lenient because it simply means taking the derivative of the indefinite integral gives you the integrand. That's why blob063540 's answer would pretty much always be considered correct and you can pretty much always multiply the top and bottom by something without considering anything else.

For definite integrals you need to consider the discontinuity since it can change the answer (same with the sign for blob's answer). So I believe for regular types of functions the discontinuity can either be a pointwise discontinuity or a jump discontinuity. If you do it with partial fractions (you can't always) and get something with no discontinuity then it doesn't matter, this is a better antiderivative. If you integrate over the limits a, c where the discontinuity is b, a<b<c, then the definite integral becomes idk if that's the right notation. That's mostly for jump discontinuity because the pointwise discontinuity doesn't matter unless the interval is at the discontinuity then you have to take limits for its lower/upper bound.

 

[unofficiallyred12]

I don't really get what you're trying to say, I pointed out that the original integral is

so it doesn't make any sense in the first place that dividing by u on the top and bottom of the fraction would lead to a mistake since is already present (if it was 0 it's not like they can cancel), it's not being newly introduced or anything.

For more general cases: (technique being multiplying top and bottom by something)

When evaluating indefinite integrals the technique is fine, indefinite integrals are more lenient because it simply means taking the derivative of the indefinite integral gives you the integrand. That's why blob063540 's answer would pretty much always be considered correct and you can pretty much always multiply the top and bottom by something without considering anything else.

For definite integrals you need to consider the discontinuity since it can change the answer (same with the sign for blob's answer). So I believe for regular types of functions the discontinuity can either be a pointwise discontinuity or a jump discontinuity. If you do it with partial fractions (you can't always) and get something with no discontinuity then it doesn't matter, this is a better antiderivative. If you integrate over the limits a, c where the discontinuity is b, a<b<c, then the definite integral becomes idk if that's the right notation. That's mostly for jump discontinuity because the pointwise discontinuity doesn't matter unless the interval is at the discontinuity then you have to take limits for its lower/upper bound.

What I mean is just because the original integrand is undefined doesn't mean the primitive is also undefined at that point. This is because the primitive could have a vertical tangent at the point where the original function is undefined (e.g. 1/sqrt(1-x^2) is undefined at x = +-1 yet it's primitive y = arcsinx + C is defined at these points). If the original function isn't defined at x = 0 (or u = 0) that doesn't mean u can always just divide both sides by x^2 or u^2 because in some cases we could add a point where this is undefined to the primitive. for example, if we take sqrttanx -sqrtcotx as the integrand instead, if we integrate by partial fractions, we get
If we instead do the technique u used, we get

where x = u+1/u
Now if we consider f(0) of the 2nd primitive, we get
if we consider f(0) for the first function, we get

the point where x= 0 isn't the same for the 2 functions, so they technically aren't the same. (Note I realised i should've used x = sqrt(tanx) instead here but if u graph the graphs, u'd realise the same thing occurs. )

 

[member 6003]

What I mean is just because the original integrand is undefined doesn't mean the primitive is also undefined at that point. This is because the primitive could have a vertical tangent at the point where the original function is undefined (e.g. 1/sqrt(1-x^2) is undefined at x = +-1 yet it's primitive y = arcsinx + C is defined at these points). If the original function isn't defined at x = 0 (or u = 0) that doesn't mean u can always just divide both sides by x^2 or u^2 because in some cases we could add a point where this is undefined to the primitive. for example, if we take sqrttanx -sqrtcotx as the integrand instead, if we integrate by partial fractions, we get View attachment 40405
If we instead do the technique u used, we get
View attachment 40406
where x = u+1/u
Now if we consider f(0) of the 2nd primitive, we get View attachment 40407
if we consider f(0) for the first function, we get
View attachment 40408
the point where x= 0 isn't the same for the 2 functions, so they technically aren't the same. (Note I realised i should've used x = sqrt(tanx) instead here but if u graph the graphs, u'd realise the same thing occurs. )

Ok, I don't really know how this contradicts anything I've said/explained. After simplifying there is no difference between the two methods. This integral is basically the case where there is a pointwise discontinuity if you chose preserve it and not simplify which I wouldn't. Does this mean the problem you have is not multiplying the top and bottom by 1/u but the substitution, x=u+1/u? I never tried to argue that a point can exist on a function where it has no derivative. I'm not saying you should always get the same answer for indefinite integrals (read previous post about discontinuities), the answer where there is no discontinuities is better since its easier to use it to find the definite integral.

What I said in my original post replying to you to true, because even though there can exist a point on the graph where it has no derivative, given the derivative we can technically say nothing about a point where it doesn't exist. So the point f(0) could be any value or it could even be the line x=0, where ever it doesn't have a derivative. Ofc if you take the limit it's 0. (but generally we want nice continuous functions)

This isn't a technique that I made up btw, the bos trials do something similar in 2020 and 2021, I learnt it from random math videos on youtube typically made by smart pp, also its used in the common way to integrate sqrt(tanx), + you have to use a technique like this to solve a lot of integration bee questions as well see this:

Also how do you get 2^(3/2)? I think its wrong

 

[member 6003]

Ok, I don't really know how this contradicts anything I've said/explained. After simplifying there is no difference between the two methods. This integral is basically the case where there is a pointwise discontinuity if you chose preserve it and not simplify which I wouldn't. Does this mean the problem you have is not multiplying the top and bottom by 1/u but the substitution, x=u+1/u? I never tried to argue that a point can exist on a function where it has no derivative. I'm not saying you should always get the same answer for indefinite integrals (read previous post about discontinuities), the answer where there is no discontinuities is better since its easier to use it to find the definite integral.

What I said in my original post replying to you to true, because even though there can exist a point on the graph where it has no derivative, given the derivative we can technically say nothing about a point where it doesn't exist. So the point f(0) could be any value or it could even be the line x=0, where ever it doesn't have a derivative. Ofc if you take the limit it's 0. (but generally we want nice continuous functions)

This isn't a technique that I made up btw, the bos trials do something similar in 2020 and 2021, I learnt it from random math videos on youtube typically made by smart pp, also its used in the common way to integrate sqrt(tanx), + you have to use a technique like this to solve a lot of integration bee questions as well see this:

Also how do you get 2^(3/2)? I think its wrong

On second thought thinking about a different function f(0) could be anything its just you're unable to tell what it is precisely. So like it could be continuous with a derivative at that point. I'm not so sure about this point though, thinking back to this thread if you want to read it: https://boredofstudies.org/threads/handling-questions-with-flaws.405100/#post-7496356
I don't think I mentioned it but technically you could have different parabolas (different constants) that join at the discontinuities, it would still be valid you can try it out on geogebra

 

[SB257426]

unofiiciallyred12 just made a mistake when evaluating the integral (that's why the blue function is incorrect), the u^2 issue isn't actually an issue at all

ah okay

 

[unofficiallyred12]

Ok, I don't really know how this contradicts anything I've said/explained. After simplifying there is no difference between the two methods. This integral is basically the case where there is a pointwise discontinuity if you chose preserve it and not simplify which I wouldn't. Does this mean the problem you have is not multiplying the top and bottom by 1/u but the substitution, x=u+1/u? I never tried to argue that a point can exist on a function where it has no derivative. I'm not saying you should always get the same answer for indefinite integrals (read previous post about discontinuities), the answer where there is no discontinuities is better since its easier to use it to find the definite integral.

What I said in my original post replying to you to true, because even though there can exist a point on the graph where it has no derivative, given the derivative we can technically say nothing about a point where it doesn't exist. So the point f(0) could be any value or it could even be the line x=0, where ever it doesn't have a derivative. Ofc if you take the limit it's 0. (but generally we want nice continuous functions)

This isn't a technique that I made up btw, the bos trials do something similar in 2020 and 2021, I learnt it from random math videos on youtube typically made by smart pp, also its used in the common way to integrate sqrt(tanx), + you have to use a technique like this to solve a lot of integration bee questions as well see this:

Also how do you get 2^(3/2)? I think its wrong

I was arguing that the answers u obtain using these methods aren't the same function because of the inherent restriction u put on certain values, which can cause problems such as what u pointed out in evaluating definite integrals. I am also fully aware of the trick, in fact I've used it before like in here https://boredofstudies.org/threads/integration-partial-fractions-q.403664/#post-7467072 but what I disagree with using it for questions that likely are out of the hsc syllabus because of the issues such a manipulation can create (which wouldn't apply much to this question but in the lone solving of the integrand (u^2+1)/(u^4+1) could be a problem.

 

[SB257426]

I was arguing that the answers u obtain using these methods aren't the same function because of the inherent restriction u put on certain values, which can cause problems such as what u pointed out in evaluating definite integrals. I am also fully aware of the trick, in fact I've used it before like in here https://boredofstudies.org/threads/integration-partial-fractions-q.403664/#post-7467072 but what I disagree with using it for questions that likely are out of the hsc syllabus because of the issues such a manipulation can create (which wouldn't apply much to this question but in the lone solving of the integrand (u^2+1)/(u^4+1) could be a problem.

mmhm i really want to look into this further... its quite interesting

 

[unofficiallyred12]

On second thought thinking about a different function f(0) could be anything its just you're unable to tell what it is precisely. So like it could be continuous with a derivative at that point. I'm not so sure about this point though, thinking back to this thread if you want to read it: https://boredofstudies.org/threads/handling-questions-with-flaws.405100/#post-7496356
I don't think I mentioned it but technically you could have different parabolas (different constants) that join at the discontinuities, it would still be valid you can try it out on geogebra

not too sure abt this either, but i think it probably just isn't properly defined yet?

 

[unofficiallyred12]

I was arguing that the answers u obtain using these methods aren't the same function because of the inherent restriction u put on certain values, which can cause problems such as what u pointed out in evaluating definite integrals. I am also fully aware of the trick, in fact I've used it before like in here https://boredofstudies.org/threads/integration-partial-fractions-q.403664/#post-7467072 but what I disagree with using it for questions that likely are out of the hsc syllabus because of the issues such a manipulation can create (which wouldn't apply much to this question but in the lone solving of the integrand (u^2+1)/(u^4+1) could be a problem.

expanding on this, would just say that such a method just isn't rigourous but if there are applicable qs in the hsc u would use it cos they don't rlly care abt this stuff.

 

[member 6003]

I was arguing that the answers u obtain using these methods aren't the same function because of the inherent restriction u put on certain values, which can cause problems such as what u pointed out in evaluating definite integrals. I am also fully aware of the trick, in fact I've used it before like in here https://boredofstudies.org/threads/integration-partial-fractions-q.403664/#post-7467072 but what I disagree with using it for questions that likely are out of the hsc syllabus because of the issues such a manipulation can create (which wouldn't apply much to this question but in the lone solving of the integrand (u^2+1)/(u^4+1) could be a problem.

I agree in part, tbh I dont think they're going to ask a question where you could use this method in the hsc anyway. I also agree that partial fractions leads to a better antiderivative, but for speed its x10 worse and i hate partial fractions algebra.

 

[unofficiallyred12]

eh i have seen such questions in trials (killara iirc) and have been called out for using it by other ppl LOL