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

seanieg89 · Jul 1, 2016

Re: MX2 2016 Integration Marathon

InteGrand said:
Are there any useful sufficient conditions that'd make the arc length of the limit curve equal the limit of the arc lengths?

For sure if you just want sufficient. Prepping for a meeting right now so will work one out later, but basically you just want to use a stronger norm than the uniform norm (to avoid wiggly approximations that are uniformly close but have massive arc length), such as looking at the C^1 norm. (Such a criterion would not quite be directly applicable here because the limit curve is not everywhere smooth and the obvious parametrisation of the curves have velocity blowing up near the x=1 endpoint, but you could most likely get a limit result with some trickery.)

Working with general rectifiable curves is pretty gross though, there would be a lot of old school papers on things like this.

seanieg89 · Jul 1, 2016

Re: MX2 2016 Integration Marathon

Such as http://www.ams.org/journals/bull/1948-54-06/S0002-9904-1948-09034-6/S0002-9904-1948-09034-6.pdf

Paradoxica · Jul 3, 2016

Re: MX2 2016 Integration Marathon

$\int\frac{2x^{12}+5x^{9}}{(x^5+x^3+1)^3}$d$x$

Paradoxica · Jul 3, 2016

Re: MX2 2016 Integration Marathon

$\int\frac{\sin x-\cos x}{(\sin x+\cos x)\sqrt{\sin x\cos x+\sin^2 x\cos^2 x}}$d$x$

Paradoxica · Jul 4, 2016

Re: MX2 2016 Integration Marathon

$\int \frac{1+\cos{4x}}{\cot{x} - \tan{x}}$d$x$

KingOfActing · Jul 4, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$\int \frac{1+\cos{4x}}{\cot{x} - \tan{x}}$d$x$

$\begin{align*}\int \frac{1+\cos 4x}{\cot x - \tan x}\,dx &= \int \frac{2\cos^2{2x}}{\frac{\cos^2 x - \sin ^2 x}{\cos x \sin x}}\,dx\\&= \int \frac{\cos^2 2x \sin 2x}{\cos 2x}\,dx\\&= \int \frac12\sin 4x\,dx\\&= -\frac18\cos{4x} + C\end{align*}$

jathu123 · Jul 4, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$\int\frac{\sin x-\cos x}{(\sin x+\cos x)\sqrt{\sin x\cos x+\sin^2 x\cos^2 x}}$d$x$

since no one did this, I gave it a try;
$I=\int\frac{\sin x-\cos x}{(\sin x+\cos x)\sqrt{\sin x\cos x+\sin^2 x\cos^2 x}}$ d$x \\ $ let $u=\sin x\cos x \Rightarrow $ d$u=(\cos ^2x-\sin ^2x)$ d$x\\ \therefore $ d$x=\frac{ du}{(\cos x+\sin x)(\cos x-\sin x)} \\$ when substituted, the numberator cancels out leaving -1.$ \\ I =-\int \frac{1}{(1+2u)\sqrt{u+u^2}}$ d$u \\ $ let $ v=\sqrt{u+u^2}\Rightarrow $ d$v=\frac{1+2u}{2\sqrt{u+u^2}}$ d$u \\\therefore $ d$u=\frac{2\sqrt{u+u^2}}{1+2u}$ d$v \\$ when substituted, the square roots cancels out, leaving the denominator as $(1+2u)^{2} = 1+4u+4u^2 = 1+4v^2 \\ I=-\int \frac{2}{1+4v^2}$ d$v \\$ let $ w = 2v\Rightarrow $ d$v=\frac{dw}{2} \\ I=-\int \frac{1}{1+w^2}$ d$w =-\tan ^{-1}w+ C \\ $ reversing all the substitutions, we get$\\ \\I=-\tan ^{-1}(2\sqrt{\sin x\cos x+\sin ^2x\cos ^2x})+C$

Paradoxica · Jul 4, 2016

Re: MX2 2016 Integration Marathon

jathu123 said:
since no one did this, I gave it a try;
$I=\int\frac{\sin x-\cos x}{(\sin x+\cos x)\sqrt{\sin x\cos x+\sin^2 x\cos^2 x}}$ d$x \\ $ let $u=\sin x\cos x \Rightarrow $ d$u=(\cos ^2x-\sin ^2x)$ d$x\\ \therefore $ d$x=\frac{ du}{(\cos x+\sin x)(\cos x-\sin x)} \\$ when substituted, the numberator cancels out leaving -1.$ \\ I =-\int \frac{1}{(1+2u)\sqrt{u+u^2}}$ d$u \\ $ let $ v=\sqrt{u+u^2}\Rightarrow $ d$v=\frac{1+2u}{2\sqrt{u+u^2}}$ d$u \\\therefore $ d$u=\frac{2\sqrt{u+u^2}}{1+2u}$ d$v \\$ when substituted, the square roots cancels out, leaving the denominator as $(1+2u)^{2} = 1+4u+4u^2 = 1+4v^2 \\ I=-\int \frac{2}{1+4v^2}$ d$v \\$ let $ w = 2v\Rightarrow $ d$v=\frac{dw}{2} \\ I=-\int \frac{1}{1+w^2}$ d$w =-\tan ^{-1}w+ C \\ $ reversing all the substitutions, we get$\\ \\I=-\tan ^{-1}(2\sqrt{\sin x\cos x+\sin ^2x\cos ^2x})+C$

This question generally goes with the product of the trig ratios or the sum of the trig ratios as the substitution. The product is more elegant than the sum, as deceptive as it may be.

Drsoccerball · Jul 6, 2016

Re: MX2 2016 Integration Marathon

This forums so dead...

leehuan · Jul 6, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
This forums so dead...

I came to realise that learning new maths > doing more integrals (at least for now)

I have other excuses for this holiday period though

Paradoxica · Jul 6, 2016

Re: MX2 2016 Integration Marathon

Have fun

$\noindent$\int \frac{\sin^2x+\sin x}{\sin x + \cos x +1}$d$x \\ \int \frac{\cos^2x+\cos x}{\sin x + \cos x +1}$d$x \\ \int \frac{\cos 7x- \cos 8x}{1+ 2\cos 5x}$d$x \\ \int_0^\pi \ln(1 - 2 \alpha \cos x + \alpha^{2}) $d$x \quad \forall |\alpha| < 1\\ \int_0^\pi \ln(1 - 2 \alpha \cos x + \alpha^{2}) $d$x \quad \forall |\alpha| = 1\\ \int_0^\pi \ln(1 - 2 \alpha \cos x + \alpha^{2}) $d$x \quad \forall |\alpha| > 1 \\ \int_0^\pi \frac{\text{d}x}{(5 - 3 \cos x)^{3}} \\ \int^b_a \dfrac{dx}{\sqrt{(e^x-e^a)(e^b-e^x)}} \quad 0<a<b \\ \int\frac{1 + a x^{t}}{x+bx^{2t+1}} $d$x \quad \forall \, t \in \mathbb{N} \\ \int \frac{\cos{2x}}{\cot^4{x}}$d$x\\ \int \frac{x^4+1}{x^{2}\sqrt{x^4-1}}$d$x \\ \int \frac{\sqrt{1-x^2}}{2x^4-x^2} $d$x \\ \int \frac{\sqrt{1+x^2}}{x} $d$x \\ \int_0^a \frac{\log(1+ax)}{1+x^2} $d$x$

$$\noindent$\int \frac{\csc^2{x} - 2005}{\cos^{2005}{x}} $d$x \\ \int \left(\frac{\tan^{-1}{x}}{x-\tan^{-1}{x}}\right)^2 $d$x \\ \int \frac{(\sin^3 \theta-\cos^3 \theta-\cos^2 \theta)(\sin \theta+\cos \theta+\cos^2 \theta)^{2010}}{(\sin \theta \cdot \cos \theta)^{2012}}$d$\theta$

juantheron · Jul 7, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$\int\frac{2x^{12}+5x^{9}}{(x^5+x^3+1)^3}$d$x$

$Let $I = \int\frac{2x^{12}+5x^9}{(x^5+x^3+1)^3}dx = \int\frac{2x^{12}+5x^9}{x^{15}\cdot \left(1+x^{-2}+x^{-5}\right)^3}$

$So $I = \int\frac{2x^{-3}+5x^{-6}}{\left(1+x^{-2}+x^{-5}\right)^3}dx$

$Now put $\left(1+x^{-2}+x^{-5}\right) = t\;,$ Then $\left(2x^{-3}+5x^{-6}\right)dx =-dt$

$So $I = -\int\frac{1}{t^3}dt = \frac{1}{2t^2}+\mathcal{C} = \frac{x^{10}}{2\left(x^5+x^3+1\right)^2}+\mathcal{C}$

juantheron · Jul 7, 2016

Re: MX2 2016 Integration Marathon

$Let $I = \int\frac{\cos 7x-\cos 8x}{1+2\cos 5x}dx = \int\frac{2\sin 5x \left(\cos 7x-\cos 8x\right)}{\cos 5x+\sin 10x}dx$

$Let $I = \int\frac{2\sin 5x \cdot \sin \frac{15x}{2}\sin \frac{x}{2}}{2\cos \frac{15x}{2}\cos \frac{5x}{2}}dx = \int 2\sin \frac{5x}{2}\sin \frac{x}{2}dx$

$So $I = \int \left(\cos 2x-\cos 3x\right)dx = \frac{\sin 2x}{2}-\frac{\sin 3x}{3}+\mathcal{C}$

juantheron · Jul 7, 2016

Re: MX2 2016 Integration Marathon

$Evaluation of $\int\sqrt{\tan^2 x-2016}\;d\mathrn{x}$

Paradoxica · Jul 7, 2016

Re: MX2 2016 Integration Marathon

juantheron said:
$Evaluation of $\int\sqrt{\tan^2 x-2016}\;d\mathrn{x}$

This one is not nice, but it does have a closed form.

Drsoccerball · Jul 9, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
Have fun

$$\noindent$ I = \int \frac{\sin^2x+\sin x}{\sin x + \cos x +1}$d$x$

$J = \int \frac{\cos^2x+\cos x}{\sin x + \cos x +1}$d$x \\$

Excuse the mistakes if any.

$$ Multiply top and bottom by $ (\sin{x} + 1) - \cos{x}.$

$I = \frac{1}{2} \int{\frac{(\sin{x}+1)^2 - \cos{x}(1+\sin{x})}{\sin{x} + 1}}dx$

$I = \frac{1}{2} \int{\sin{x} - \cos{x} + 1}dx$

$I = \frac{x}{2} - \frac{\sin{x} + \cos{x}}{2} + C$

$$ Notice that $ I + J = x \\ \\ \therefore J = \frac{x}{2} + \frac{\sin{x} + \cos{x}}{2} + D$

Paradoxica · Jul 9, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
Excuse the mistakes if any.

$$ Multiply top and bottom by $ (\sin{x} + 1) - \cos{x}.$

$I = \frac{1}{2} \int{\frac{(\sin{x}+1)^2 - \cos{x}(1+\sin{x})}{\sin{x} + 1}}dx$

$I = \frac{1}{2} \int{\sin{x} - \cos{x} + 1}dx$

$I = \frac{x}{2} - \frac{\sin{x} + \cos{x}}{2} + C$

$$ Notice that $ I + J = x \\ \\ \therefore J = I = \frac{x}{2} + \frac{\sin{x} + \cos{x}}{2} + D$

dat last line tho

Drsoccerball · Jul 9, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$\int_0^\pi \ln(1 - 2 \alpha \cos x + \alpha^{2}) $d$x \quad \forall |\alpha| = 1\\$

$Not too sure about the other values of $ \alpha $ but will attempt soon.$

$\int_0^{\pi}{\ln{(1- 2\alpha \cos{x} + \alpha ^ 2)}}dx$

$$ Take note that $ |\alpha| = 1 $ means that $ \alpha ^2 = 1$

$I = \int_0^{\pi}{\ln{(2- 2\alpha \cos{x})}}dx$

$$ Using the property $ \int_0^b{f(x)}dx = \int_0^b{f(b-x)}dx$

$I = \int_0^{\pi}{\ln{(4- 4\alpha^2 \cos^2{x})}}dx$

$2I = \int_0^{\pi}{\ln{4} + \ln{(\sin{x})}}dx$

$I = 0 $ (From evaluating the integral of ln(sin(x)) we realise it is just the negative of $ \frac{\pi \ln{4}}{2}$). $$

Paradoxica · Jul 9, 2016

MX2 2016 Integration Marathon

Drsoccerball said:
$Not too sure about the other values of $ \alpha $ but will attempt soon.$

$\int_0^{\pi}{\ln{(1- 2\alpha \cos{x} + \alpha ^ 2)}}dx$

$$ Take note that $ |\alpha| = 1 $ means that $ \alpha ^2 = 1$

$I = \int_0^{\pi}{\ln{(2- 2\alpha \cos{x})}}dx$

$$ Using the property $ \int_0^b{f(x)}dx = \int_0^b{f(b-x)}dx$

$I = \int_0^{\pi}{\ln{(4- 4\alpha^2 \cos^2{x})}}dx$

$2I = \int_0^{\pi}{\ln{4} + \ln{(\sin{x})}}dx$

$I = 0 $ (From evaluating the integral of ln(sin(x)) we realise it is just the negative of $ \frac{\pi \ln{4}}{2}$). $$

There exists a method of proving the integral equals 0 without evaluating it.
(check our facebook conversation)

leehuan · Jul 9, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
$$ Multiply top and bottom by $ (\sin{x} + 1) - \cos{x}.$

Was there any hint towards this step?

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