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

leehuan · Mar 29, 2016

Re: MX2 2016 Integration Marathon

Is In = (2n-1)/(2n) . In-1?

Paradoxica · Apr 2, 2016

Re: MX2 2016 Integration Marathon

$\int_{-1}^1 \cos^{-1}{(2x^2-1)}\,$d$x$

Carrotsticks · Apr 2, 2016

Re: MX2 2016 Integration Marathon

Sigh ignore please.

http://i.imgur.com/I446dGA.jpg

leehuan · Apr 6, 2016

Re: MX2 2016 Integration Marathon

Instructive question for 2016ers

$\int_{0}^{\frac{\pi}{2}}{\ln{\left(\tan{x}\right)}dx}$

KingOfActing · Apr 6, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
Instructive question for 2016ers

$\int_{0}^{\frac{\pi}{2}}{\ln{\left(\tan{x}\right)}dx}$

$I = \int_{0}^{\frac{\pi}{2}}{\ln{\left(\tan{x}\right)}\,dx} \stackrel{x \to \frac{\pi}{2} - x}{=} \int_{0}^{\frac{\pi}{2}}{\ln{\left(\cot{x}\right)}\,dx}\\2I = \int_{0}^{\frac{\pi}{2}}{\ln{\left(\tan{x}\right)} +\ln{\left(\cot{x}\right) }\,dx} =\int_{0}^{\frac{\pi}{2}}{\ln{1}\,dx} = 0\\ \therefore \int_{0}^{\frac{\pi}{2}}{\ln{\left(\tan{x}\right)}dx} = 0$

leehuan · Apr 6, 2016

Re: MX2 2016 Integration Marathon

Above ordinary level questions for 2016ers

$\int_{0}^{1}{4\ln{\left(\sqrt{1-x}+\sqrt{1+x}\right)}dx}$

$\int{15\sqrt{1+\sqrt{x}}dx}$

$\int_{1}^{64}{\frac{dx}{{x}^{\frac{1}{2}}+{x}^{\frac{1}{3}}}}$

A cliche

$\int{12\sqrt{x+\sqrt{x+\sqrt{x+\dots}}}dx}$

KingOfActing · Apr 6, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
Above ordinary level questions for 2016ers

$\int_{0}^{1}{4\ln{\left(\sqrt{1-x}+\sqrt{1+x}\right)}dx}$

Expand 2 into the logarithm to get

$\begin{align*} \int_{0}^{1} 2\ln{(2+2\sqrt{1-x^2})\,dx} &= 2\ln2 + \int_0^1 \ln(1 + \sqrt{1-x^2})\,dx \\&= 2\ln2 + 2\int_0^1 \frac{x^2}{(\sqrt{1-x^2})(1+\sqrt{1-x^2})}\,dx\\&= 2\ln{2} + 2\int_0^{\frac{\pi}{2}} \frac{\sin^2{\theta}\cos{\theta}}{\cos{\theta}(1+\cos{\theta})}\,d\theta \\&= 2\ln2 + 2\int_0^{\frac{\pi}{2}} 1 - \cos{\theta}\,d\theta \\ &= 2\ln{2} + \pi - 2\end{align*}$

The second line is IBP

Paradoxica · Apr 6, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
A cliche

$\int{12\sqrt{x+\sqrt{x+\sqrt{x+\dots}}}dx}$

$6x + (1+4x)^{\frac{3}{2}} + \mathcal{C}$

KingOfActing · Apr 7, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
$\int{15\sqrt{1+\sqrt{x}}dx}$

Let x = u^2
Let u + 1 = y^2

A simple integral and some back substitutions later:

$\int 15\sqrt{1+\sqrt{x}}\,dx = 12x\sqrt{1+\sqrt{x}} + 4\sqrt{x}\sqrt{1+\sqrt{x}} - 8\sqrt{1+\sqrt{x}} + C$

leehuan · Apr 7, 2016

Re: MX2 2016 Integration Marathon

A cliche

$\int{\sqrt{\cot{x}}\, dx}$

Instructive question for 2016ers

$$\noindent$ \int_1^{64} \frac{\text{d}x}{\sqrt{x}+\sqrt[3]{x}}$

Bringing back hell for 2016ers (no cheating)

$\int \frac{(x-1)\sqrt{x^4+2x^3+4x^2+2x+1}}{x^2(x+1)}$d$x$

Paradoxica · Apr 7, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
A cliche

$\int{\sqrt{\cot{x}}\, dx}$

Instructive question for 2016ers

$$\noindent$ \int_1^{64} \frac{\text{d}x}{\sqrt{x}+\sqrt[3]{x}}$

Bringing back hell for 2016ers (no cheating)

$\int \frac{(x-1)\sqrt{x^4+2x^3+4x^2+2x+1}}{x^2(x+1)}$d$x$

By inspection....

jks

InteGrand · Apr 7, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
By inspection....

jks

I was just about to post "I think they are all within the realms of inspection for Paradoxica".

Paradoxica · Apr 8, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$\int \frac{2(x+2)(x^4+1)}{x^8-2x^4+1}\,$d$x$

Paradoxica said:
For those who aren't bothered doing the above integral, here is an intriguing generalised integral.
$\int_{-a}^a \frac{\text{d}x}{b+x^{2 n-1}+\sqrt{b^2+x^{4 n-2}}}$
Remember to check special cases.

Paradoxica said:
$$\noindent Given $n$ is a non-negative integer, find the recurrence relation for $\\ I_n = \int_{-\infty}^\infty \frac{\text{d}x}{(x^2+1)^{n+1}} \\ $Hence, prove the following statement using induction:$ \\ \int_{-\infty}^\infty \frac{\text{d}x}{(x^2+1)^{n+1}} = \frac{(2n)!\,\pi}{2^{2n}(n!)^2}$

Paradoxica said:
$\int_{-1}^1 \cos^{-1}{(2x^2-1)}\,$d$x$

.

juantheron · Apr 8, 2016

Re: MX2 2016 Integration Marathon

$Given $\displaystyle \int\frac{(x-1)\cdot \sqrt{x^4+2x^3+4x^2+2x+1}}{x^2(x+1)}dx = \int\frac{(x^2-1)\cdot \sqrt{x^4+2x^3+4x^2+2x+1}}{x^2(x^2+2x+1)}dx$

$Above we multiply both $\bf{N_{r}}$ and $\bf{D_{r}}$ by $(x+1).$

$So $\displaystyle = \int\frac{\left(1-\frac {1}{x^2}\right)\cdot \sqrt{ \left(x^2+2x+4+\frac{2}{x}+\frac{1}{x^2}\right)}}{ \left(x+2+\frac{1}{x}\right)}dx$

$Now Let $ \left(x+\frac{1}{x}\right) = t\;,$ Then $\displaystyle \left(1-\frac{1}{x^2}\right)dx = dt$

$So Integral $\displaystyle = \int\frac{\sqrt{t^2+2t+2}}{t+2}dt = \frac{t^2+2t+2}{(t+2)\sqrt{t^2+2t+2}}dt = \int\frac{t(t+2)+2}{(t+2)\sqrt{t^2+2t+2}}dt$

$So Integral $\displaystyle = \underbrace{\int\frac{t}{\sqrt{t^2+2t+2}}dt}_{I} - \underbrace{\int\frac{3}{(t+2)}\cdot \frac{1}{\sqrt{t^2+2t+2}}dt}_{J}$

$Now after that we can solve $I$ and $J$ Seperately$

Paradoxica · Apr 8, 2016

Re: MX2 2016 Integration Marathon

juantheron said:
$Given $\displaystyle \int\frac{(x-1)\cdot \sqrt{x^4+2x^3+4x^2+2x+1}}{x^2(x+1)}dx = \int\frac{(x^2-1)\cdot \sqrt{x^4+2x^3+4x^2+2x+1}}{x^2(x^2+2x+1)}dx$

$Above we multiply both $\bf{N_{r}}$ and $\bf{D_{r}}$ by $(x+1).$

$So $\displaystyle = \int\frac{\left(1-\frac {1}{x^2}\right)\cdot \sqrt{ \left(x^2+2x+4+\frac{2}{x}+\frac{1}{x^2}\right)}}{ \left(x+2+\frac{1}{x}\right)}dx$

$Now Let $ \left(x+\frac{1}{x}\right) = t\;,$ Then $\displaystyle \left(1-\frac{1}{x^2}\right)dx = dt$

$So Integral $\displaystyle = \int\frac{\sqrt{t^2+2t+2}}{t+2}dt = \frac{t^2+2t+2}{(t+2)\sqrt{t^2+2t+2}}dt = \int\frac{t(t+2)+2}{(t+2)\sqrt{t^2+2t+2}}dt$

$So Integral $\displaystyle = \underbrace{\int\frac{t}{\sqrt{t^2+2t+2}}dt}_{I} - \underbrace{\int\frac{3}{(t+2)}\cdot \frac{1}{\sqrt{t^2+2t+2}}dt}_{J}$

$Now after that we can solve $I$ and $J$ Seperately$

It would be more effective to use the substitution

$x+\frac{1}{x} +1= t$

as it wipes clean the entire integral in one step, making it look infinitely nicer.

leehuan · Apr 9, 2016

Re: MX2 2016 Integration Marathon

Just looking at random integrals now.

$\int{\sqrt{1-\sin{2x}}$ d$x}$

Paradoxica · Apr 9, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
Just looking at random integrals now.

$\int{\sqrt{1-\sin{2x}}$ d$x}$

The only existing form that is within the realms of elementary functions is

$\text{sgn}(\sin{x}-\cos{x})(\sin{x}+\cos{x}) + \mathcal{C}$

leehuan · Apr 9, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
The only existing form that is within the realms of elementary functions is

$\text{sgn}(\sin{x}-\cos{x})(\sin{x}+\cos{x}) + \mathcal{C}$

Supposedly the answer can be expressed as this though

$\frac{\sin{x}+\cos{x}}{\cos{x}-\sin{x}}\sqrt{1-\sin{\left(2x\right)}}$

InteGrand · Apr 9, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
Supposedly the answer can be expressed as this though

$\frac{\sin{x}+\cos{x}}{\cos{x}-\sin{x}}\sqrt{1-\sin{\left(2x\right)}}$

$\noindent Yes, this is essentially equivalent to Paradoxica's answer. This is because $\sqrt{1-\sin{\left(2x\right)}} = \sqrt{\left(\sin x - \cos x \right)^2}=|\sin x - \cos x|$, so $\frac{\sqrt{1-\sin{\left(2x\right)}}}{\cos{x}-\sin{x}}=\frac{|\sin x - \cos x|}{\cos{x}-\sin{x}}=\mathrm{sgn} \left(\sin x - \cos x\right)$, since $\mathrm{sgn} \left(\xi\right)=\frac{|\xi|}{\xi}$ for real $\xi \neq 0$. The only difference is that Paradoxica's answer will say the derivative is 0 at the `corners' of the graph (see link below), whereas the one you typed will be undefined there (and the derivative is indeed undefined at these places). The reason Paradoxica's will yield 0 is that $\mathrm{sgn} \left(\xi\right)$ is defined to be $0$ when $\xi=0$, as the `corners' of the graph occur at the places where the argument of his sign function is 0.$

• Graph linked here: http://www.wolframalpha.com/input/?i=sketch+y+=+sqrt(1-sin(2x))
• Further info on the sign function: https://en.wikipedia.org/wiki/Sign_function

Paradoxica · Apr 9, 2016

Re: MX2 2016 Integration Marathon

InteGrand said:
$\noindent Yes, this is essentially equivalent to Paradoxica's answer. This is because $\sqrt{1-\sin{\left(2x\right)}} = \sqrt{\left(\sin x - \cos x \right)^2}=|\sin x - \cos x|$, so $\frac{\sqrt{1-\sin{\left(2x\right)}}}{\cos{x}-\sin{x}}=\frac{|\sin x - \cos x|}{\cos{x}-\sin{x}}=\mathrm{sgn} \left(\sin x - \cos x\right)$, since $\mathrm{sgn} \left(\xi\right)=\frac{|\xi|}{\xi}$ for real $\xi \neq 0$. The only difference is that Paradoxica's answer will say the derivative is 0 at the `corners' of the graph (see link below), whereas the one you typed will be undefined there (and the derivative is indeed undefined at these places). The reason Paradoxica's will yield 0 is that $\mathrm{sgn} \left(\xi\right)$ is defined to be $0$ when $\xi=0$, as the `corners' of the graph occur at the places where the argument of his sign function is 0.$

• Graph linked here: http://www.wolframalpha.com/input/?i=sketch+y+=+sqrt(1-sin(2x))
• Further info on the sign function: https://en.wikipedia.org/wiki/Sign_function

I don't want to be nitpicky, but isn't the technical term "cusp"?

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