HSC 2016 MX2 Integration Marathon (archive) (5 Viewers)

omegadot · Feb 18, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$\int \frac{\text{d}x}{\sqrt{x}(1+\sqrt[3]{x})}$

$$\noindent Let $x = \tan^6 \theta, dx = 6 \tan^5 \theta \sec^2 \theta \, d\theta$. So\\\begin{align*}\int \frac{dx}{\sqrt{x}(1 + \sqrt[3]{x})} &= \int \frac{6 \tan^5 \theta \sec^2 \theta}{\tan^3 \theta (1 + \tan^2 \theta)} \, d\theta\\ &= 6 \int \tan^2 \theta \, d\theta\\ &= 6 \int (\sec^2 \theta - 1) \, d\theta\\ &= 6 \tan \theta - 6 \theta + \cal{C}\\ &= 6 \sqrt[6]{x} - 6 \tan^{-1} (\sqrt[6]{x}) + \cal{C}\end{align*}$

omegadot · Feb 18, 2016

Re: MX2 2016 Integration Marathon

$\noindent \textbf{Next Question}$

$$\noindent Briefly explain what is wrong with the following ``proof''.$

$$\noindent If $f(x) \neq 0$, from integration by parts we have$\\\begin{align*}\int \frac{f'(x)}{f(x)} \, dx = \frac{f(x)}{f(x)} + \int \frac{f(x) \cdot f'(x)}{[f(x)]^2} \, dx = 1 + \int \frac{f'(x)}{f(x)} \, dx\end{align*}\\$Hence $1 = 0.$

Paradoxica · Feb 18, 2016

Re: MX2 2016 Integration Marathon

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

$$\noindent Briefly explain what is wrong with the following ``proof''.$

$$\noindent If $f(x) \neq 0$, from integration by parts we have$\\\begin{align*}\int \frac{f'(x)}{f(x)} \, dx = \frac{f(x)}{f(x)} + \int \frac{f(x) \cdot f'(x)}{[f(x)]^2} \, dx = 1 + \int \frac{f'(x)}{f(x)} \, dx\end{align*}\\$Hence $1 = 0.$

The two sides are only equal down to an arbitrary constant. Hence, the conclusion that the two sides are identically equivalent is false. The error is revealed when you perform a definite integral, as opposed to an indefinite one.

Paradoxica · Feb 18, 2016

Re: MX2 2016 Integration Marathon

$\int \frac{x}{(1-x^2)^n \sqrt{1-x^2}} \, $d$x$

leehuan · Feb 18, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$\int \frac{x}{(1-x^2)^n \sqrt{1-x^2}} \, $d$x$

Is this meant to be way easier than it looks?

$\frac { { \left( 1-{ x }^{ 2 } \right) }^{ \frac { 1-2n }{ 2 } } }{ 2n-1 } +C$

Paradoxica · Feb 18, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
Is this meant to be way easier than it looks?

$\frac { { \left( 1-{ x }^{ 2 } \right) }^{ \frac { 1-2n }{ 2 } } }{ 2n-1 } +C$

Yes.

Paradoxica · Feb 18, 2016

Re: MX2 2016 Integration Marathon

$\int_0^\infty \log{\left(\frac{x}{x^2+1}\right)}\frac{\text{d}x}{(1+x^2)^2}$

omegadot · Feb 18, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
Is this meant to be way easier than it looks?

$\frac { { \left( 1-{ x }^{ 2 } \right) }^{ \frac { 1-2n }{ 2 } } }{ 2n-1 } +C$

$$\noindent Provided $n \neq \frac{1}{2}$. When $n = \frac{1}{2}$ one has:$\\-\frac{1}{2} \ln |1 - x^2| + \cal{C}.$

leehuan · Feb 18, 2016

Re: MX2 2016 Integration Marathon

omegadot said:
$$\noindent Provided $n \neq \frac{1}{2}$. When $n = \frac{1}{2}$ one has:$\\-\frac{1}{2} \ln |1 - x^2| + \cal{C}.$

True, forgot about that case.

leehuan · Feb 18, 2016

Re: MX2 2016 Integration Marathon

I don't know what I'm doing anymore

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Paradoxica · Feb 18, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
I don't know what I'm doing anymore

You're getting there. Trust yourself. What you have can definitely be done using high school techniques. (That last one has been posted before by me)

Ironically, if you plug in my original question into Wolfram Mathematica, it can't find the closed form solution. But humans can do it.

leehuan · Feb 18, 2016

Re: MX2 2016 Integration Marathon

Thinking about the symmetry about x=pi/4 but I've completely forgotten how to show it.

Besides that I might've figured it out.

EDIT: OH CRUD. f(x)=f(pi/2-x) isn't it.....

leehuan · Feb 18, 2016

Re: MX2 2016 Integration Marathon

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Paradoxica · Feb 18, 2016

Re: MX2 2016 Integration Marathon

$\int \frac{\text{d}x}{x\sqrt{x^{n}+1}}$

KingOfActing · Feb 18, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$\int \frac{\text{d}x}{x\sqrt{x^{n}+1}}$

$\begin{align*} &\text{First, let }x^n = \tan^2{\theta},\,nx^{n-1}\,dx=2\tan\theta\sec^2\theta\,d\theta\\&\int \frac{1}{x\sqrt{x^n+1}}\,dx = \int \frac{x^{n-1}}{x^n\sqrt{x^n+1}}\,dx = \frac{2}{n}\int\frac{\tan\theta\sec^2\theta}{\tan^2\theta\sec\theta}\,d\theta\\&= \frac{2}{n} \int \csc\theta\,d\theta = \frac{-2}{n}\ln\left |\csc\theta+\cot\theta\right | + C = \frac{-2}{n}\ln\left |\frac{\sqrt{x^n+1}}{\sqrt{x^n}} + \frac{1}{\sqrt{x^n}}\right | + C \\&=\frac{2}{n}\ln\left | \frac{\sqrt{x^n}}{\sqrt{x^n+1}+1}\right | + C = \ln|x| - \frac{2}{n}\ln|1+\sqrt{x^n+1}| + C \end{align*}$

Paradoxica · Feb 18, 2016

Re: MX2 2016 Integration Marathon

$\noindent$ \int \frac{\sqrt{1-x}}{1-\sqrt{x}}\,$d$x \\ \int \frac{1-\sqrt{x}}{\sqrt{1-x}} \, $d$x\\$You need only evaluate one of them, then the other one can be obtained rather easily.$

omegadot · Feb 19, 2016

Re: MX2 2016 Integration Marathon

KingOfActing said:
$\begin{align*} &\text{First, let }x^n = \tan^2{\theta},\,nx^{n-1}\,dx=2\tan\theta\sec^2\theta\,d\theta\\&\int \frac{1}{x\sqrt{x^n+1}}\,dx = \int \frac{x^{n-1}}{x^n\sqrt{x^n+1}}\,dx = \frac{2}{n}\int\frac{\tan\theta\sec^2\theta}{\tan^2\theta\sec\theta}\,d\theta\\&= \frac{2}{n} \int \csc\theta\,d\theta = \frac{-2}{n}\ln\left |\csc\theta+\cot\theta\right | + C = \frac{-2}{n}\ln\left |\frac{\sqrt{x^n+1}}{\sqrt{x^n}} + \frac{1}{\sqrt{x^n}}\right | + C \\&=\frac{2}{n}\ln\left | \frac{\sqrt{x^n}}{\sqrt{x^n+1}+1}\right | + C = \ln|x| - \frac{2}{n}\ln|1+\sqrt{x^n+1}| + C \end{align*}$

$$\noindent Provided $n \neq 0$. When $n = 0$ we just have $\frac{1}{\sqrt{2}} \ln |x| + \cal{C}.$

Paradoxica · Feb 19, 2016

Re: MX2 2016 Integration Marathon

omegadot said:
$$\noindent Provided $n \neq 0$. When $n = 0$ we just have $\ln |x| + \cal{C}.$

$$\noindent Actually, we have $\frac{\log{x}}{\sqrt{2}}+ \cal{C}$

InteGrand · Feb 19, 2016

Re: MX2 2016 Integration Marathon

$\textbf{NEW QUESTION}$

$\noindent Find $\int \frac{1}{\left(x^2 -1 \right)^2}\text{ d}x$.$

omegadot · Feb 19, 2016

Re: MX2 2016 Integration Marathon

Oh yes. Silly me

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