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

Paradoxica · Jul 22, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
I got something like (assuming no mistakes i'm pretty sure my method was correct). :

$I = \frac{4(b-a)}{(e^b+e^a) \cdot \sqrt{(e^b+e^a)^2 -(b-a)^2}} + C$

That's not even correct for $a=\log{2}, b=\log{3}$ , which is $\frac{\pi}{\sqrt{6}}$

If I recall correctly, the derivation contains inverse trigonometry.

Drsoccerball · Jul 22, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
That's not even correct for $a=\log{2}, b=\log{3}$ , which is $\frac{\pi}{\sqrt{6}}$

If I recall correctly, the answer contains inverse trigonometry.

Well what I did was let : $u = e^x - \frac{e^b + e^a}{2}$

$$ Then $ u = (\frac{e^b+e^a}{2}) \sin{\theta}$

$$ My integral was an easy $ \frac{1}{1 + \sin{\theta}}$

InteGrand · Jul 29, 2016

Re: MX2 2016 Integration Marathon

$\noindent Find $\int \left(a^2 -x^2\right)^{\frac{3}{2}}\, \mathrm{d}x$.$

Paradoxica · Jul 29, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
Well what I did was let : $u = e^x - \frac{e^b + e^a}{2}$

$$ Then $ u = (\frac{e^b+e^a}{2}) \sin{\theta}$

$$ My integral was an easy $ \frac{1}{1 + \sin{\theta}}$

The answer is $\frac{\pi}{\sqrt{e^{a+b}}}$ , which Mathematica correctly verifies.

leehuan · Aug 2, 2016

Re: MX2 2016 Integration Marathon

From class (lol)

$\int \frac { x^{ \frac { 2 }{ 3 } } }{ \left( 1+x^{ \frac { 3 }{ 5 } } \right) ^{ 2 } } dx$

Paradoxica · Aug 2, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
From class (lol)

$\int \frac { x^{ \frac { 2 }{ 3 } } }{ \left( 1+x^{ \frac { 3 }{ 5 } } \right) ^{ 2 } } dx$

You're evil.

Paradoxica · Aug 2, 2016

Re: MX2 2016 Integration Marathon

For anyone concerned with the above integral, don't waste your time.

It is, at minimum, expressible using the real and imaginary parts of the 9^th, 18^th, and 36^th roots of unity, and is extremely horrible, even if it does have a closed form primitive.

Paradoxica · Aug 2, 2016

Re: MX2 2016 Integration Marathon

However, on the off note that Leehuan did make a typo...

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

Which can easily be done by inspection.

leehuan · Aug 3, 2016

Re: MX2 2016 Integration Marathon

Haha lol no typo. My lecturer just went through it over 5min

Yeah. Don't try to attempt that question unless you have a whole day to spend on it.

Paradoxica · Aug 3, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
Haha lol no typo. My lecturer just went through it over 5min

Yeah. Don't try to attempt that question unless you have a whole day to spend on it.

Or a CAS to assist with PFE

InteGrand · Aug 3, 2016

Re: MX2 2016 Integration Marathon

$\noindent Find $\int \frac{\sqrt{2+x^2}}{x}\, \mathrm{d}x$.$

jathu123 · Aug 3, 2016

Re: MX2 2016 Integration Marathon

InteGrand said:
$\noindent Find $\int \frac{\sqrt{2+x^2}}{x}\, \mathrm{d}x$.$

$I= \int \frac{\sqrt{2+x^2}}{x}$ d$x \\ =\int \frac{2+x^2}{x\sqrt{2+x^2}}$ d$x \\ $ let $ u=\sqrt{2+x^2} \Rightarrow $ d$u=\frac{x}{\sqrt{2+x^2}}$ d$x\\\therefore $ d$x=\frac{\sqrt{2+x^2}}{x}\\I=\int \frac{u^2}{u^2-2}$ d$u = \int 1+\frac{2}{u^2-2}$ d$u\\ = u+\int \frac{1}{\sqrt{2}}\left ( \frac{1}{u-\sqrt{2}}-\frac{1}{u+\sqrt{2}} \right )$ d$u \\ =u+\frac{1}{\sqrt{2}}(\ln|u-\sqrt{2}|-\ln|u-\sqrt{2}|)+C \\\\ = \sqrt{2+x^2}+\frac{1}{\sqrt{2}}\ln\left | \frac{\sqrt{2+x^2}-\sqrt{2}}{\sqrt{2+x^2}+\sqrt{2}} \right |+C$

juantheron · Aug 4, 2016

Re: MX2 2016 Integration Marathon

$Evaluation of $\int \frac{1-\cot^{9}x}{\tan x+\cot^{10}x}dx$

$Evaluation of $\int\frac{1}{1+x^6}dx$

$Evaluation of $\int\frac{\sqrt{\cot x}-\sqrt{\tan x}}{1+3\sin 2x}dx$

Paradoxica · Aug 4, 2016

Re: MX2 2016 Integration Marathon

juantheron said:
$Evaluation of $\int \frac{1-\cot^{9}x}{\tan x+\cot^{10}x}dx$

$Evaluation of $\int\frac{1}{1+x^6}dx$

$Evaluation of $\int\frac{\sqrt{\cot x}-\sqrt{\tan x}}{1+3\sin 2x}dx$

1: Clear denominators and reverse the chain rule to yield 1/11 × log(sin¹¹x + cos¹¹x) + C

2: Tedious and trivial by partial fractional decomposition

3: The hopeful substitutions u² = tanx, v = u + 1/u, will yield the answer of arccot((√tanx + √cotx)/2) + C

juantheron · Aug 6, 2016

Re: MX2 2016 Integration Marathon

$Evaluation of $\int\frac{\sin (x+\alpha)}{\cos^3 x}\sqrt{\frac{\csc x+\sec x}{\csc x-\sec x}}dx$

Paradoxica · Aug 7, 2016

Re: MX2 2016 Integration Marathon

juantheron said:
$Evaluation of $\int\frac{\sin (x+\alpha)}{\cos^3 x}\sqrt{\frac{\csc x+\sec x}{\csc x-\sec x}}\text{d}x$

Can't deal with α so expand the integral into two different integrals.

Integral 1:

$$\noindent$ \cos{\alpha} \int \frac{\tan{x}}{\cos^2{x}} \sqrt{\tan{\left(x+\frac{\pi}{4}\right)}} \text{d}x \quad \quad \quad u=x+\frac{\pi}{4} \\ \cos{\alpha} \int \frac{\tan{u} -1}{2(\tan{u} +1)^3} \sec^2{u} \sqrt{\tan{u}} \text{d}x \quad t^2 = \tan{u} \\ \cos{\alpha} \int \frac{(t^2-1)t^2}{(t^2+1)^3} \text{d}t \quad \quad \quad \quad \quad \quad \quad \quad \quad t = \tan{\theta} \\ \cos{\alpha} \int -\sin^2{\theta} \cos{2\theta} \, \text{d}\theta \quad \quad \quad \quad \quad \quad $double angles$ \\ \frac{\cos{\alpha}}{2} \int \cos^2{2\theta} - \cos{2\theta} \, \text{d}\theta \quad \quad \quad \quad \quad $and again...$ \\ \frac{\cos{\alpha}}{4} \int 1 + \cos{4\theta} - 2\cos{2\theta}\, \text{d}\theta \\ \frac{\cos{\alpha}}{4} \left( \theta + \frac{\sin{4\theta}}{4} - \sin{2\theta} \right) = \frac{\cos{\alpha}}{4}\left( \tan^{-1}{t} + \frac{t(1-t^2)}{(1+t^2)^2} - \frac{2t}{1+t^2}\right)$

$$\noindent$ \frac{\cos{\alpha}}{4} \left( \tan^{-1}{\sqrt{\tan{u}}} - \frac{1+3\tan{u}}{(1+\tan{u})^2}\sqrt{\tan{u}} \right) \\ \frac{\cos{\alpha}}{4} \left( \tan^{-1}{\sqrt{\frac{1+\tan{x}}{1-\tan{x}}}} - \dfrac{1+3\frac{1+\tan{x}}{1-\tan{x}}}{(1+\frac{1+\tan{x}}{1-\tan{x}})^2} \right) \\ \frac{\cos{\alpha}}{8} \left( \tan^{-1}{\sqrt{\cot^2{x}-1}} + (\tan{x}-1)(2+\tan{x}) \right)$

Paradoxica · Aug 7, 2016

Re: MX2 2016 Integration Marathon

juantheron said:
$Evaluation of $\int\frac{\sin (x+\alpha)}{\cos^3 x}\sqrt{\frac{\csc x+\sec x}{\csc x-\sec x}}dx$

Integral 2:

$$\noindent$ \sin{\alpha} \int \frac{1}{\cos^2{x}} \sqrt{\tan{\left( x+\frac{\pi}{4} \right)} } \text{d}x \quad u = x+\frac{\pi}{4} \\ \sin{\alpha} \int \frac{\sec^2{u}}{2(\tan{u}+1)^2}\sqrt{\tan{u}} \text{d}u \quad t^2 = \tan{u} \\ \sin{\alpha} \int \frac{t^2}{(t^2+1)^2} \text{d}t \quad \quad \quad \quad \quad \quad \quad t = \tan{\theta}\\ \sin{\alpha} \int \sin^2{\theta} \text{d}\theta = \frac{\sin{\alpha}}{2} \int 1-\cos{2\theta} \, \text{d}\theta \\ \frac{\sin{\alpha}}{2} \left( \theta - \sin{\theta}\cos{\theta} \right) = \frac{\sin{\alpha}}{2}\left( \tan^{-1}{\sqrt{\tan{u}}} - \frac{\sqrt{\tan{u}}}{1+\tan{u}} \right) \\ \frac{\sin{\alpha}}{4} \left(\tan^{-1}{\sqrt{\cot^2{x}-1}} - \sqrt{1-\tan^2{x}} \right)$

Then the complete integral follows as the sum of the two prior integrals.

Paradoxica · Aug 7, 2016

Re: MX2 2016 Integration Marathon

$$\noindent Evaluate:$ \\ \int \frac{2\cos{\theta}}{25+16\sin{2\theta}} \text{d}\theta$

I STRONGLY DISADVISE THE TANGENT HALF ANGLE SUBSTITUTION

juantheron · Aug 8, 2016

Re: MX2 2016 Integration Marathon

$Let $I = \int\frac{2\cos \theta}{25+16\sin 2\theta}d\theta = \int\frac{\cos \theta+\sin\theta}{25+16\sin 2\theta}d\theta + \int\frac{\cos \theta-\sin \theta}{25+16\sin 2\theta}d\theta$

$So $I = \int\frac{\cos \theta+\sin \theta}{25+16\left[1-(\sin \theta-\cos \theta)^2\right]}d\theta+\int\frac{\cos \theta-\sin \theta}{25+16\left[(\sin \theta+\cos \theta)^2-1\right]}d\theta$

$Put $\sin \theta -\cos \theta = t$\;, Then $(\sin \theta+\cos \theta)d\theta = dt$

$and $(\sin \theta+\cos \theta) = u\;,$ Then $(\cos \theta-\sin \theta)d\theta = du$

$So $I = \int\frac{1}{41-16t^2}dt+\int\frac{1}{9+16u^2}du$

$= \frac{1}{16}\int\frac{1}{\left(\frac{ \sqrt{41}}{4}\right)^2-t^2}dt+\frac{1}{16}\int\frac{1}{\left(\frac{3}{4}\right)^2+u^2}du$

$So $I = \frac{1}{32\sqrt{41}}\ln\left|\frac{\sqrt{41}+t}{ \sqrt{41}-t}\right|+\frac{1}{12}\tan^{-1}\left(\frac{4u}{3}\right)+\mathcal{C}$

$So $I = \frac{1}{32\sqrt{41}}\ln \left|\frac{\sqrt{41}+\sin \theta-\cos \theta}{\sqrt{41}-\sin \theta+\cos \theta}\right|+\frac{1}{12}\tan^{-1}\left(\frac{4(\sin \theta+\cos \theta)}{3}\right)+\mathcal{C}$

juantheron · Aug 8, 2016

Re: MX2 2016 Integration Marathon

$Evaluation of $\int_{0}^{1}\frac{\sin \theta \left(\cos^2 -\cos^2 \frac{\pi}{5}\right)\cdot \left(\cos^2 -\cos^2 \frac{2\pi}{5}\right)}{\sin 5\theta}d\theta$

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