HSC 2016 MX2 Integration Marathon (archive) (1 Viewer)

InteGrand · Jan 13, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
http://www.wolframalpha.com/input/?i=integrate+sqrt(arcsin(sqrtx))

What.

KingOfActing said:
Are you sure it has a nice closed form? My first substitution was $x = \sin^2{(u^2)}$ which made the integral to to $\frac{-u}{2}\cos(2u^2)+\frac{1}{2}\int\cos(2u^2)\,du$ by integration by parts, and I'm pretty sure that second integral doesn't have a closed form solution.

Yeah that second integral can't be found using elementary functions (requires a Fresnel Integral). The C in the WolframAlpha output refers to a Fresnel Integral (a non-elementary function). More about Fresnel Integrals: https://en.wikipedia.org/wiki/Fresnel_integral

Drsoccerball · Jan 13, 2016

Re: MX2 2016 Integration Marathon

$I= \int{\sqrt{\sin^{-1}\sqrt{x}}}dx$

$$ Let $ u^2=x$

$I= \int{2u{\sqrt{\sin^{-1}u}}}du$

$$ Let $ \theta ^2 = \sin^{-1}u$

$I= \int{2\theta ^2 sin(2\theta ^2)}d\theta$

$$ Now IBP letting $ u= 2\theta $ and $ dv=\theta sin(2\theta ^2)$

$-\frac{\theta}{2}\cos{2\theta ^2} +\frac{1}{2} \int{\cos{2\theta ^2 }d\theta}$

$Well here's where I dropped the pen and thought the integral was easy :P sorry guys...$

Paradoxica · Jan 13, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
$I= \int{\sqrt{\sin^{-1}\sqrt{x}}}dx$

$$ Let $ u^2=x$

$I= \int{2u{\sqrt{\sin^{-1}u}}}du$

$$ Let $ \theta ^2 = \sin^{-1}u$

$I= \int{2\theta ^2 sin(2\theta ^2)}d\theta$

$$ Now IBP letting $ u= 2\theta $ and $ dv=\theta sin(2\theta ^2)$

$-\frac{\theta}{2}\cos{2\theta ^2} +\frac{1}{2} \int{\cos{2\theta ^2 }d\theta}$

$Well here's where I dropped the pen and thought the integral was easy :P sorry guys...$

$$\noindent You monster. Luckily, when I looked at the integral, and saw the square root of an inverse trig, I instantly knew there was no elementary solution, due to my awareness of Fresnel Integrals$\\ $\mathbf{Next Question}$ \\ \int \cot^{-1}{\sqrt[3]{x}} $d$x$

kawaiipotato · Jan 13, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$$\noindent You monster. Luckily, when I looked at the integral, and saw the square root of an inverse trig, I instantly knew there was no elementary solution, due to my awareness of Fresnel Integrals$\\ $\mathbf{Next Question}$ \\ \int \cot^{-1}{\sqrt[3]{x}} $d$x$

Not a fan of IBP so I think there would be an alternate method

$$ let $ y^3 = x$

$3y^2 dy = dx$

$$I$ = \int \cot ^{-1} y * 3y^2 dy$

$$Using IBP, letting u $ = \cot ^{-1} y $and$ dv = 3y^2 dy$

$$I$ = y^3 \cot ^{-1}y + \int \frac{y^3}{y^2 + 1} dy$

$= y^3 \cot ^{-1}y + \int \frac{y^3 + y - y}{y^2 + 1} dy$

$= y^3 \cot ^{-1}y + \int \frac{y(y^2 + 1) - y}{y^2 + 1} dy$

$= y^3 \cot ^{-1}y + \int y - \frac{y}{y^2 + 1} dy$

$= y^3 \cot ^{-1}y + \frac{y^2}{2} - \frac{1}{2} \ln (y^2 + 1) + \mathcal{C}$

$= x \cot ^{-1}{\sqrt[3]{x}} + \frac{1}{2}\left({\sqrt[3]{x^2}} - \ln ({\sqrt[3]{x^2}} + 1)\right) + \mathcal{C}$

omegadot · Jan 13, 2016

Re: MX2 2016 Integration Marathon

$\noindent \textbf{New Question}$

$$\noindent Here is something a little easier. Find $\int \frac{x^4}{(2 - x)^3} \, dx$ without using polynomial long division and partial fractions.$

Paradoxica · Jan 13, 2016

Re: MX2 2016 Integration Marathon

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

$$\noindent Here is something a little easier. Find $\int \frac{x^4}{(2 - x)^3} \, dx$ with using polynomial long division and partial fractions.$

Hint: Subtract and add a certain perfect fourth power to reduce the denominator by one power to make decomposition less tedious.

omegadot · Jan 13, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
Hint: Subtract and add a certain perfect fourth power to reduce the denominator by one power to make decomposition less tedious.

$$\noindent The question should read \textit{without} using polynomial long division and partial fractions.$

leehuan · Jan 13, 2016

Re: MX2 2016 Integration Marathon

seanieg89 said:
Bump.

Any hint on how to get the solution? Haven't seen these enough to know a starting point

Paradoxica · Jan 13, 2016

Re: MX2 2016 Integration Marathon

omegadot said:
$$\noindent The question should read \textit{without} using polynomial long division and partial fractions.$

Yeesh, such specificity.

Substitute in order to translate the binomial term to the top. Expand using the binomial theorem and bring down all powers. Integrate accordingly.

seanieg89 · Jan 13, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
Any hint on how to get the solution? Haven't seen these enough to know a starting point

Key ingredients (in my solution at least):

1. $\int_0^\infty \frac{\sin{x}}{x}\, dx=\frac{\pi}{2}$

You can assume this if you don't know how to prove it, but it has been done before in this and similar threads, and is within MX2 scope although slightly tricky (you would be guided through it slowly in an exam question).

2. Basic trig manipulation, and trig limits.

seanieg89 · Jan 13, 2016

Re: MX2 2016 Integration Marathon

So I guess a natural followup question then if it hasn't been done recently is to prove that:

$\int_0^\infty \frac{\sin{x}}{x}\, dx = \frac{\pi}{2}$

which I ask people to attempt separately (with MX2 appropriate methods).

Paradoxica · Jan 13, 2016

Re: MX2 2016 Integration Marathon

seanieg89 said:
On the easy side, but:

$\lim_{R\rightarrow\infty}\int_{-R}^R \frac{\sin(ax)\sin(bx)}{x^2}\, dx \qquad (a,b\in\mathbb{R}).$

I remember correctly now, I did not post this, I posted the degenerate integral where a=b.

omegadot · Jan 13, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
Yeesh, such specificity.

Substitute in order to translate the binomial term to the top. Expand using the binomial theorem and bring down all powers. Integrate accordingly.

$\noindent Just trying to introduce a neat little method which avoids partial fractions to students who may not have seen such a thing before$

Drsoccerball · Jan 14, 2016

Re: MX2 2016 Integration Marathon

Wolfram says that the integral doesn't converge... Also if this isn't true is the answer 0 ?

Paradoxica · Jan 14, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
Wolfram says that the integral doesn't converge... Also if this isn't true is the answer 0 ?

Dude.

DUUUUUDE

THE INTEGRAND IS EVEN YOU ********

Drsoccerball · Jan 14, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
Dude.

DUUUUUDE

THE INTEGRAND IS EVEN YOU ********

Paradoxica · Jan 14, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:

And this isn't the first time. You have made this exact assumption before. Stop repeating history.

InteGrand · Jan 14, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
Wolfram says that the integral doesn't converge... Also if this isn't true is the answer 0 ?

$\noindent In reality, it does converge. Maybe try actual values for $a$ and $b$ if you use WolframAlpha, because maybe it can't do it for generic $a,b$.$

seanieg89 · Jan 14, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
Wolfram says that the integral doesn't converge... Also if this isn't true is the answer 0 ?

The integral actually does converge because it decays as x^(-2) and does not blow up anywhere finite.

Notice that sometimes we work with integrals that aren't convergent in the traditional sense though, like the sin(x)/x integral. when we talk about integrating this thing over the real line, we are really just talking about taking the Cauchy principal value integral of this quantity. (In the case of the sin(x)/x integrand, this is the limit of the integral over the interval [-R,R] as R->inf.)

For convergent integrals over R, it wouldn't matter at what "rate" we sent the two limits to the respective ends of the real line, but for things that are not integrable (because of not decaying fast enough at the boundary, not because of regularity) it does, and we need to specify something like the Cauchy principal value to unambiguously make sense of "the integral of f over the reals".

The answer is not 0 either.

omegadot · Jan 15, 2016

Re: MX2 2016 Integration Marathon

$\noindent I think we need a \textbf{New Question}$

$$\noindent Evaluate $\int^{\sqrt{3}}_0 \sin^{-1} \left (\frac{2x}{1 + x^2} \right ) \, dx$

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