HSC 2016 MX2 Integration Marathon (archive) (2 Viewers)

Paradoxica · Feb 8, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
Now that I think about it.

The original function has no discontinuity at x=0 yet the primitive does. Does this matter?

when integrand divided down x^2 he implicitly assumed that x can not be zero. to resolve this matter, take the x back to the top by multiplying.

the value can then be verified with l'hopital's rule.

InteGrand · Feb 8, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
Now that I think about it.

The original function has no discontinuity at x=0 yet the primitive does. Does this matter?

$\noindent Technically it does matter. The primitive I gave has the integrand as a derivative for all $x\neq 0$, but the function has a (jump) discontinuity at 0 and so is not differentiable there.$

$\noindent Let $F(x) = \frac{1}{2\sqrt{2}}\tan^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{2}}\right)+\frac{1}{4\sqrt{2}} \ln \left(\frac{x+\frac{1}{x}-\sqrt{2}}{x+\frac{1}{x}+\sqrt{2}}\right)$ for real $x>0$ (this was our primitive restricted to the positive reals, ignore the $+c$ for now). (Also, we don't actually need absolute values for the logarithm's argument, since that is positive for all real non-zero $x$. This can be proved using the fact that $x+\frac{1}{x}\geq 2$ whenever $x>0$ and $x+\frac{1}{x}\leq -2$ whenever $x<0$.) (Also, if $x$ is allowed to be any real number that's not 0, i.e. unrestricted $F$, we would have $F$ be an odd function (easily, proved by replacing $x$ with $-x$ in $F(x)$ and using log laws and other things).) Note that as $x \to 0^{+}$, $F(x) \to -\frac{1}{2 \sqrt{2}}\cdot \frac{\pi}{2}+\ln (1) = -\frac{\pi} {4\sqrt{2}}$. So let's define $F(0)=-\frac{\pi}{4\sqrt{2}}$. Then $F$ becomes a function that is continuous on the non-negative reals.$

$\noindent (The unrestricted primitive would have a jump discontinuity at $x=0$ due to the fact that the expression would approach $+\frac{\pi} {4\sqrt{2}}$ as $x\to 0^{-}$.)$

$\noindent Now, for all positive $x$, we have $F^\prime (x) = \frac{x^2}{1+x^4}$, where $F(x)=\begin{cases} \frac{1}{2\sqrt{2}}\tan^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{2}}\right)+\frac{1}{4\sqrt{2}} \ln \left(\frac{x+\frac{1}{x}-\sqrt{2}}{x+\frac{1}{x}+\sqrt{2}}\right), &x>0 \\-\frac{\pi} {4\sqrt{2}}, & x= 0\end{cases}$. We have a continuous function defined on the non-negative reals whose derivative matches that of the original integrand for all $x>0$. If we can show that $F^\prime (0)$ exists and equals 0, then we can extend this function $F$ to a continuous function on all $\mathbb{R}$ with the desired derivative by having the primitive be: $F_2 (x)= \begin{cases}F(x) - F(0), &x\geq0 \\ -F_2(-x), & x<0 \end{cases}$.$

$\noindent (Basically, we're taking the function for the positive part, moving it so it goes through the origin, and flipping it about the origin so that the entire function is odd (and thus has even derivative as we want) and continuous everywhere, including at the origin. This will also match the original integrand at $0$ if $F^\prime (0)=0$, since the original integrand equals 0 when $x=0$.)$

$\noindent So our final claim to prove is that $F^\prime (0) = 0$. Note that this is a one-sided limit due to the definition of $F$, i.e. $F^\prime (0) = \lim _{x\to 0^{+}} \frac{F(x)-F(0)}{x-0}$. (So whenever we've said $F^\prime (0)$ here, we've been talking about a \textit{one-sided derivative}, namely a \textit{right derivative}. Actually, the proper notation for this to use is: $\partial_{+}F(0)$, which we will use from now on. I didn't use it before because it would probably confuse most HSC students when seeing it first. A link to the relevant Wikipedia page is provided at the end for anyone curious.) Since we defined $F(0)$ in such a way that $F$ is continuous at $x=0$, the numerator has limit$

$\begin{align*} \lim _{x\to 0^{+}} \left(F(x)-F(0)\right) &= \lim _{x\to 0^{+}} \left(F(x)\right) - \lim_{x\to 0^{+}} \left(F(0)\right) \\ &= F(0)-F(0) \quad (\text{by continuity of }F) \\ &= 0 \end{align*}$

$\noindent and the denominator also goes to 0 as $x\to 0^{+}$. So we can use L'H$\hat{\text{o}}$pital's rule.$

$\noindent Note that for $x> 0$, we always have $F^\prime (x) = \frac{x^2}{1+x^4}$, and the limit $\lim _{x\to 0^{+}} \frac{F^\prime (x)}{(x)^\prime}$ exists, as $\lim _{x\to 0^{+}} \frac{F^\prime (x)}{(x)^\prime} = \lim _{x\to 0^{+}} \frac{\frac{x^2}{1+x^4}}{1} = 0$. It follows from L'H$\hat{\text{o}}$pital's rule that $\lim _{x\to 0^{+}} \frac{F(x)-F(0)}{x-0} = 0$, i.e. $\partial_{+} F(0) = 0$, as desired.$

$\noindent So in summary, the antiderivative is $F_2 (x) +C$, where $F_2 = \begin{cases}F(x)-F(0), &x\geq0 \\ -F_2(-x), & x<0 \end{cases}$, where $F(x)= \begin{cases} \frac{1}{2\sqrt{2}}\tan^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{2}}\right)+\frac{1}{4\sqrt{2}} \ln \left(\frac{x+\frac{1}{x}-\sqrt{2}}{x+\frac{1}{x}+\sqrt{2}}\right), &x>0 \\-\frac{\pi} {4\sqrt{2}}, & x= 0\end{cases}$,$

$\noindent and $C$ is an arbitrary constant.$
_____
• Left and right derivatives: https://en.wikipedia.org/wiki/Left_and_right_derivative
• L'Hôpital's rule: https://en.wikipedia.org/wiki/L'Hôpital's_rule

Paradoxica · Feb 8, 2016

Re: MX2 2016 Integration Marathon

$$\noindent $ \int_{-a}^{a} \frac{\text{d}x}{1+x^5+\sqrt{1+x^{10}}}$

omegadot · Feb 9, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$$\noindent $ \int_{-a}^{a} \frac{\text{d}x}{1+x^5+\sqrt{1+x^{10}}}$

$$\noindent Let $I = \int_{-a}^{a} \frac{\text{d}x}{1+x^5+\sqrt{1+x^{10}}}.$

$\noindent Using the result $\int^b_a f(x) \, dx = \int^b_a f(b + a - x) \, dx$, which for symmetric limits is the same thing as making a substitution of $x = -u$, we have$

$I = \int^a_{-a} \frac{dx}{1 - x^5 + \sqrt{1 + x^{10}}}$

$$\noindent Adding the two results for $ I$ together gives$\\\begin{align*}I + I &= \int^a_{-a} \left [\frac{1}{1 + x^5 + \sqrt{1 + x^{10}}} + \frac{1}{1 - x^5 + \sqrt{1 + x^{10}}} \right ] \, dx\\ \Rightarrow 2I &= \int^a_{-a} dx\\\Rightarrow I &= a\end{align*}$

omegadot · Feb 9, 2016

Re: MX2 2016 Integration Marathon

$\noindent \textbf{Next Question}$

$$\noindent (a) [easy] Evaluate $\int^\pi_0 \frac{\cos (2x) - \cos (2y)}{\cos x - \cos y} \, dx$

$$\noindent (b) [hard] Evaluate $\int^\pi_0 \frac{\cos (nx) - \cos (ny)}{\cos x - \cos y} \, dx,$ where $n = 0,1,2,\ldots$

Paradoxica · Feb 9, 2016

Re: MX2 2016 Integration Marathon

omegadot said:
$\noindent \textbf{Next Question}$

$$\noindent (a) [easy] Evaluate $\int^\pi_0 \frac{\cos (2x) - \cos (2y)}{\cos x - \cos y} \, dx$

$$\noindent (b) [hard] Evaluate $\int^\pi_0 \frac{\cos (nx) - \cos (ny)}{\cos x - \cos y} \, dx,$ where $n = 0,1,2,\ldots$

$$\noindent By inspection, the first integral is equal to $2\pi\cos{y} \\ $Define $I_n = \int_0^\pi \frac{\cos{nx}-\cos{ny}}{\cos{x}-\cos{y}}$d$x \\ I_{n+1}+I_{n-1}-2I_n\cos{y} = \int_0^\pi \frac{\cos{(n+1)x} + \cos{(n-1)x} - \cos{(n+1)y} -\cos{(n-1)y} -2\cos{nx}\cos{y}+2\cos{ny}\cos{y}} {\cos{x}-\cos{y}}$d$x \\ I_{n+1}+I_{n-1}-2I_n\cos{y} = \int_0^\pi \frac{2\cos{nx}\cos{x}-2\cos{ny}\cos{y}-2\cos{nx}\cos{y}+2\cos{ny}\cos{y}}{\cos{x}-\cos{y}}$d$x \\ I_{n+1}+I_{n-1}-2I_n\cos{y}=\int_0^\pi \frac{2\cos{nx}(\cos{x}-\cos{y})}{\cos{x}-\cos{y}}$d$x = \int_0^\pi 2\cos{nx} $d$x = 0 \\ $By inspection, $ I_1 = \pi , \, I_2 = 2\pi\cos{y}$

$$\noindent Claim: $I_n = \frac{\sin{ny}}{\sin{y}} \pi \, \, \forall n \in \mathbb{N} \\ $The claim is clearly true for $n = 1,\, 2 \\ $Assume the claim is true for $\{n,\,n-1\} \in \mathbb{N} \\ $From the reduction recurrence relation, we have $I_{n+1} = 2\cos{y}I_n -I_{n-1} = \frac{2\pi\sin{ny}\cos{y} - \pi\sin{(n-1)y}}{\sin{y}}=\frac{\sin{(n+1)y} + \sin{(n-1)y} - \sin{(n-1)y}}{\sin{y}}\pi \\ I_{n+1} = \frac{\sin{(n+1)y}}{\sin{y}}\pi \\ $Thus, our claim is true for $n+1$ if it is true for $n,\,n-1$. Since it is true for $n=1,\,2,$ by the Axiom of Mathematical Induction, our claim is true $\forall n \in \mathbb{N} \\ \therefore \int_0^\pi \frac{\cos{nx}-\cos{ny}}{\cos{x}-\cos{y}}$d$x = \frac{\sin{ny}}{\sin{y}}\pi$

omegadot · Feb 9, 2016

Re: MX2 2016 Integration Marathon

$$\noindent$\ldots$ and as a \textit{bonus}, find $\int \frac{\cos nx - \cos ny}{\cos x - \cos y} \, dx$ for $n = 2,3,\ldots$

Paradoxica · Feb 9, 2016

Re: MX2 2016 Integration Marathon

omegadot said:
$$\noindent$\ldots$ and as a \textit{bonus}, find $\int \frac{\cos nx - \cos ny}{\cos x - \cos y} \, dx$ for $n = 2,3,\ldots$

Mathematica could not find a closed form. Are you certain one exists? I know the reduction recurrence relation can be used to determine a given value of n, it doesn't imply the existence of a closed form. Even the case of y = 0 requires the hypergeometric function in order to express. The fourth argument is complex. This does not look nice.

omegadot · Feb 9, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
Mathematica could not find a closed form. Are you certain one exists? I know the reduction recurrence relation can be used to determine a given value of n, it doesn't imply the existence of a closed form. Even the case of y = 0 requires the hypergeometric function in order to express. The fourth argument is complex. This does not look nice.

$\noindent One does. What one needs to do is try and express the integrand in a form which can be suitably integrated in elementary terms. It's not easy, and would require significant hand holding if it ever appeared in an HSC exam (which I doubt), but it is doable!$

omegadot · Feb 9, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
Also attempt anyone except paradoxica his insight is to good for such a crap question:

$\int \frac{(x+1)^4(x-1)^2\sqrt{x^4+x^2+1}}{(x^2+1)^2(x^6-1)} dx$

Paradoxica said:
$$By inspection, the integral is $\sec^{-1}\left(x+\frac{1}{x} \right)+2\sqrt{\frac{x^4+x^2+1}{(x^2+1)^2}}+C$

$$\noindent For those who lack Paradoxica's insight into this one (or though I must admit his comment \textit{by inspection} did make me laugh) I will aim to find the integral using a symmetric substitution of the form $u = x + \frac{1}{x}.$

$$\noindent Calling the integral $I$ and working on the integrand for a bit, since$\\x^6 - 1 = (x^3 - 1)(x^3 + 1) = (x - 1)(x + 1)(x^4 + x^2 + 1)\\$we have$\\\begin{align*}I &= \int \frac{(x + 1)^3(x - 1) \sqrt{x^4 + x^2 + 1}}{(x^2 + 1)^2 (x^4 + x^2 + 1)} \, dx\\&= \int \frac{(x + 1)^3 (x - 1)}{(x^2 + 1)^2 \sqrt{x^4 + x^2 + 1}} \, dx\\&= \int \frac{(x + 1)^2 (x + 1)(x - 1)}{(x^2 + 1)^2 \sqrt{x^4 + x^2 + 1}} \, dx\\&= \int \frac{(x + 1)^2 (x^2 - 1)}{(x^2 + 1)^2 \sqrt{x^4 + x^2 + 1}}\\ &= \int \frac{(x^2 + 2x + 1) x^2 (1-1/x^2)}{x^2 (x + 1/x) x \sqrt{x^2 + 1 + 1/x^2}} \, dx\\&= \int \frac{\left (x + 2 + \frac{1}{x} \right ) \left (1 - \frac{1}{x^2} \right )}{\left (x + \frac{1}{x} \right )^2 \sqrt{\left (x + \frac{1}{x} \right )^2 - 1}} \, dx\\&= \int \frac{\left [\left (x + \frac{1}{x} \right ) + 2 \right ] \left (1 - \frac{1}{x^2} \right )}{\left (x + \frac{1}{x} \right )^2 \sqrt{\left (x + \frac{1}{x} \right )^2 - 1}} \, dx\end{align*}$

$$\noindent We now make the promised symmetric substitution of $\\u = x + \frac{1}{x}, du = \left (1 - \frac{1}{x^2} \right ) \, dx.$\\Thus$\\I = \int \frac{u + 2}{u^2 \sqrt{u^2 - 1}} \, du.$

$$\noindent The resulting integral can be found by making a substitution of $u = \sec \theta, du = \sec \theta \tan \theta \, d\theta.$ After substituting in and simplifying we have$\\\begin{align*}I &= \int (1 + 2\cos \theta) \, d\theta\\&= \theta + 2 \sin \theta + \cal{C}\\&= \sec^{-1} u + \frac{2}{u} \sqrt{u^2 - 1} + \cal{C}\\&= \sec^{-1} \left (x + \frac{1}{x} \right ) + 2 \sqrt{\frac{x^4 + x^2 + 1}{(x^2 + 1)^2}} + \cal{C}\end{align*}\\$as expected.$

Paradoxica · Feb 11, 2016

Re: MX2 2016 Integration Marathon

omegadot said:
$$\noindent$\ldots$ and as a \textit{bonus}, find $\int \frac{\cos nx - \cos ny}{\cos x - \cos y} \, dx$ for $n = 2,3,\ldots$

$$\noindent Using the reduction recurrence relation from the definite integral, we have $\\I_{n+1} - 2\cos{y}I_n + I_{n-1} = \int 2\cos{nx}dx \\ $Differentiating both sides and multiplying throughout by $\sin{y}\\\sin{y}I'_{n+1} = 2\cos{y}\sin{y}I'_n - \sin{y}I'_{n-1} +2\sin{y}\cos{nx} \\ $Claim: $\sin{y}I'_n = 2\left(\sum_{r=1}^{n-1} \sin{(n-r)y}\cos{rx}\right) + \sin{ny} \\ $By inspection, $\sin{y}I'_2 = 2\sin{y}\cos{x} + \sin{2y} \\ \sin{y}I'_3 = 2\sin{2y}\cos{x} + 2\sin{y}\cos{2x} + \sin{3y} \\ $The formula is clearly true for $n=2,\,3 \\ $Assuming the formula is true for $n,\,{n-1},$ using the reduction recurrence relation, we have $\sin{y}I'_{n+1} = 4\cos{y}\left(\sum_{r=1}^{n-1} \sin{(n-r)y}\cos{rx}\right) - 2\left(\sum_{r=1}^{n-2} \sin{(n-1-r)y}\cos{rx}\right) + 2\cos{y}\sin{ny} - \sin{(n-1)y} +2\sin{y}\cos{nx}$

$\noindent Extract the $(n-1)^{\text{th}}$ term from the first sum, merge the two sums$\\$together, and use products to sums on the third term, yielding$ \\ 2\left(\sum_{r=1}^{n-2} 2\cos{y}\sin{(n-r)y}\cos{rx}-\sin{(n-1-r)y}\cos{rx}\right) + 4\cos{y}\sin{y}\cos{(n-1)x}+2\sin{y}\cos{nx}-\sin{(n-1)y}+\sin{(n+1)y}+\sin{(n-1)y} \\ $Use products to sums again on the general term inside the sum, and reverse double angle on the second term, yielding $\\2\left(\sum_{r=1}^{n-2} \sin{(n+1-r)y}\cos{rx}+\sin{(n-1-r)y}\cos{rx}-\sin{(n-1-r)y}\cos{rx}\right) + 2\sin{2y}\cos{(n-1)x}+2\sin{y}\cos{nx}+\sin{(n+1)y} \\ \Rightarrow 2\left(\sum_{r=1}^{n-2} \sin{(n+1-r)y}\cos{rx}\right) = 2\sin{2y}\cos{(n-1)x}+2\sin{y}\cos{nx}+\sin{(n+1)y} \\ $Notice that the middle two terms match the correct indexes for the $(n-1)^{\text{th}}$ and $n^{\text{th}}$ general term, so we can absorb it into the sum. Finally, we have$

$\noindent$ \sin{y} I'_{n+1} = 2\left(\sum_{r=1}^n \sin{(n+1-r)y}\cos{rx}\right) + \sin{(n+1)y} \\ $Thus, the claim is true for $n$ only if it is true for $n,\,n-1$. As the claim is true for $n = 2,\,3$, by the Axiom of Mathematical Induction, our claim is true $\forall n \in \mathbb{N} \land n \neq 1 \\ \therefore I_n = \frac{\sin{ny}}{\sin{y}}x + 2\sum_{r=1}^{n-1} \frac{\sin{(n-r)y}\sin{rx}}{r\sin{y}} + C \\ $Remark: I do not consider this a closed form, for me, a closed form cannot require compact sigma notation, and must be expressible in a finite number of terms independent of the integral's external parameter.$

aoc · Feb 11, 2016

Re: MX2 2016 Integration Marathon

How do you know this stuff if youre doing youre hsc this year wtf? I have to compete with you...

Paradoxica · Feb 11, 2016

Re: MX2 2016 Integration Marathon

aoc said:
How do you know this stuff if youre doing youre hsc this year wtf? I have to compete with you...

I have the patience, tenacity and perseverance to spend over one day solving a single symbolic integral. I do NOT have the determinism to be competitive and speedy at examination level questions. These questions are beyond examination level.

Drsoccerball · Feb 11, 2016

Re: MX2 2016 Integration Marathon

aoc said:
How do you know this stuff if youre doing youre hsc this year wtf? I have to compete with you...

LOL he knows uni stuff... Dw the only way you can beat him is by him making silly mistakes

Since HSC is easy its possible that both of you could get 100

Drsoccerball · Feb 11, 2016

Re: MX2 2016 Integration Marathon

The reduction formula for these integrals appear the same is there a general rule to memorise these or are they only done through rigorous expansion and +- techniques?

Paradoxica · Feb 11, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
The reduction formula for these integrals appear the same is there a general rule to memorise these or are they only done through rigorous expansion and +- techniques?

Combine the n-1 th term and the n+1 th term, use sums to products and then guess the coefficient of the nth term, which is usually negative, but can definitely be positive. (Will write one soon)

You want to be able to obtain the denominator in the numerator so you can cancel and then easily integrate.

Also, go back to sleep. You don't have morning classes like I do.

Drsoccerball · Feb 11, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
Combine the n-1 th term and the n+1 th term, use sums to products and then guess the coefficient of the nth term, which is usually negative, but can definitely be positive. (Will write one soon)

You want to be able to obtain the denominator in the numerator so you can cancel and then easily integrate.

Also, go back to sleep. You don't have morning classes like I do.

We pray 5 times a day

And I havn't sleept yet

Drsoccerball · Feb 11, 2016

Re: MX2 2016 Integration Marathon

$Here's a little easy one I thought of just now :$

$\int{(tan^{-1}x)^n}dx$

leehuan · Feb 11, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
$Here's a little easy one I thought of just now :$

$\int{(tan^{-1}x)^n}dx$

I think I asked this one before but Paradoxica found a flaw in it.

Paradoxica · Feb 11, 2016

Re: MX2 2016 Integration Marathon

$$\noindent \textbf{NEW (LEGIT) QUESTION}$ \\ \int \tan^{-1}{\left(\frac{1}{\sqrt{1-x^2}}\right)}$d$x$

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