HSC 2012-14 MX2 Integration Marathon (archive) (1 Viewer)

Registered User · May 9, 2013

Re: MX2 Integration Marathon

Fus Ro Dah said:
$\int \sqrt{\tan x}dx$

Method 1:

$Let\ I=\int \sqrt{\tan(x)}dx$

$Let\ u=\tan(x), du=(1+\tan^{2}(x))dx$

$I=\int \frac{\sqrt{u}}{u^{2}+1}$

$Let\ v=\sqrt{u}, dv=\frac{du}{2\sqrt{u}}$

$I=2\int \frac{v^{2}}{v^{4}+1}$

$\int_0^\infty\frac{x^2}{1+x^4}dx$

$Let\ t=\frac{1}{v} \therefore dt=\frac{-dv}{v^2}$

$\therefore I=\int \frac{\frac{1}{t^2}}{1+\frac{1}{t^4}}\times\frac{-dt}{t^2}$

$I=-\int \frac{dt}{1+t^4}$

$This was done by Sy123 as he answered my question on the 4U marathon thread so I will just copy his solution:$

$\frac{1}{t^4+1} = \frac{1}{4\sqrt{2}}\left( \frac{2t+\sqrt{2}}{t^2+\sqrt{2}t+1} - \frac{2t-\sqrt{2}}{t^2-\sqrt{2}x+1}+\frac{\sqrt{2}}{t^2+\sqrt{2}+1} + \frac{\sqrt{2}}{t^2-\sqrt{2}t+1} \right)$

$After integrating both sides and substituting$ $t=\frac{1}{\sqrt{tan(x)}}$ $we get:$

$= -\frac{1}{4\sqrt{2}}\left ( -log(\frac{1}{\sqrt{tanx}}^{2}-\sqrt{2} \frac{1}{\sqrt{tanx}}+1)+log(\frac{1}{\sqrt{tanx}}^2+\sqrt{2} \frac{1}{\sqrt{tanx}}+1)-2 tan^{-1}(1- \sqrt{2}\frac{1}{\sqrt{tanx}})+2 tan^{-1}(\sqrt{2} \frac{1}{\sqrt{tanx}}+1) \right )$

Method 2: which I think is outside the syllabus:

$First compute the following$

$\int(\sqrt{\tan x}+\sqrt{\cot x}) dx=\int\frac{\sin x +\cos x}{\sqrt{\sin x\cdot \cos x}}dx$

$=\sqrt 2\int\frac{d(\sin x - \cos x)}{\sqrt{1-(\sin x -\cos x)^2}}$

$which is same as the$ $\sqrt 2\int\frac{dz}{\sqrt{1-z^2}}=\sqrt 2\sin ^{-1}z+c$

$Now compute$

$\int(\sqrt{\tan x}-\sqrt{\cot x})dx$

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

$and this is same as$

$-\sqrt 2\int \frac {dw}{\sqrt{w^2 -1}}=-\sqrt 2\int\frac{\sec u\tan u}{\tan u} du=-\sqrt 2\ln(\sec u +\tan u)+C$

$where$ $w=\sec u.$ $Now add both the integrals to obtain the result.$

shongaponga · May 9, 2013

Re: MX2 Integration Marathon

Fus Ro Dah said:
$\int \sqrt{\tan x}dx$

$\\ $ Let $ u = \sqrt{\tan x} \\\\ \Rightarrow u^2 = \tan x \\\\ \Rightarrow 2udu = \sec^2xdx \\\\ \Rightarrow dx = \frac{2udu}{\sec^2x} \\\\ \Rightarrow dx = \frac{2udu}{1 + \tan^2x} = \frac{2udu}{1 + u^4}$

$.'. \int \sqrt{\tan x}dx = \int \frac{2u^2du}{1 + u^4}$

$= \int \frac{(u^2 +1) + (u^2 - 1)}{1+u^4}du$

$= \int \frac{(1+\frac{1}{u^2})}{(u^2+\frac{1}{u^2})}du + \int \frac{(1-\frac{1}{u^2})}{(u^2+\frac{1}{u^2})}du$

$= \int \frac{(1+\frac{1}{u^2})}{(u-\frac{1}{u})^2+2}du + \int \frac{(1-\frac{1}{u^2})}{(u+\frac{1}{u})^2-2}du $ Call this equation (*) $$

$$ Now, let $ v = (u-\frac{1}{u}) \Rightarrow dv = (1+\frac{1}{u^2}) du $ \\ and let $ t = (u+\frac{1}{u}) \Rightarrow dt = (1-\frac{1}{u^2}) du$

$$ Subbing v, t, dv and dt into (*) yields:$

$\int \sqrt{\tan x}dx = \int \frac{dv}{v^2 + 2} + \frac{dt}{t^2 -2}$

$= \frac{1}{\sqrt{2}} \tan^{-1}\frac{v}{\sqrt2} $ + $ \frac{1}{2\sqrt{2}} \ln(\frac{t-{\sqrt2}}{t+{\sqrt2}}) + C$

$= \frac{1}{\sqrt{2}}\tan^{-1}\frac{(u-\frac{1}{u})}{\sqrt{2}} + \frac{1}{2\sqrt{2}}\ln(\frac{(u+\frac{1}{u})-{\sqrt{2}}}{(u+\frac{1}{u})+\sqrt{2}}) + C$

$= \frac{1}{\sqrt{2}}\tan^{-1}\frac{(\tan x - 1)}{\sqrt{2 \tan x}} + \frac{1}{2\sqrt{2}}\ln(\frac{(\tan x + 1)-\sqrt{2\tan x}}{(\tan x + 1)+\sqrt{2\tan x}}) + C$

I really hope this is correct - took so long to Tex up.

Registered User · May 9, 2013

Re: MX2 Integration Marathon

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

$\int \frac{dx}{(x^2+a^2)^n}$

Sy123 · May 9, 2013

Re: MX2 Integration Marathon

Registered User said:
$I =\int{\frac{(1+x^{2})dx}{(1-x^{2})\sqrt{1+x^{4}}}}$

Great Question

By dividing numerator and denominator by x^2, we arrive at:

$I = \int \frac{\frac{1}{x^2} + 1}{\left(\frac{1}{x}- x \right ) \sqrt{x^2 + \frac{1}{x^2}}} \ dx$

$y = x - \frac{1}{x}$

$dy = 1 + \frac{1}{x^2} \ dx$

$I = \int \frac{dy}{-y \sqrt{y^2+2}}$

$I = -\int \frac{dy}{y\sqrt{y^2+2}}$

$y = \sqrt{2} \tan \theta$

$dy = \sqrt{2} \sec^2 \theta \ d\theta$

$I = -\int \frac{\sqrt{2} \sec^2 \theta \ d\theta}{\sqrt{2} \tan \theta \sec \theta}$

$I = \frac{1}{\sqrt{2}} \int \frac{-1}{\sin \theta} \ d\theta$

$I = \frac{1}{\sqrt{2}} \ln |\cosec \theta + \cot \theta| + c$

$I = \frac{1}{\sqrt{2}} \ln |\frac{\sqrt{2}}{y} + \sqrt{1+\frac{2}{y^2}}| + c$

$I = \frac{1}{\sqrt{2}} \ln |\frac{x\sqrt{2}}{x^2 - 1} + \frac{1}{x}\sqrt{x^2 + 2(x^2-1)^2}| + \epsilon$

Hopefully no mistakes.

Registered User · May 10, 2013

Re: MX2 Integration Marathon

Sy123 said:
Great Question

By dividing numerator and denominator by x^2, we arrive at:

$I = \int \frac{\frac{1}{x^2} + 1}{\left(\frac{1}{x}- x \right ) \sqrt{x^2 + \frac{1}{x^2}}} \ dx$

$y = x - \frac{1}{x}$

$dy = 1 + \frac{1}{x^2} \ dx$

$I = \int \frac{dy}{-y \sqrt{y^2+2}}$

$I = -\int \frac{dy}{y\sqrt{y^2+2}}$

$y = \sqrt{2} \tan \theta$

$dy = \sqrt{2} \sec^2 \theta \ d\theta$

$I = -\int \frac{\sqrt{2} \sec^2 \theta \ d\theta}{\sqrt{2} \tan \theta \sec \theta}$

$I = \frac{1}{\sqrt{2}} \int \frac{-1}{\sin \theta} \ d\theta$

$I = \frac{1}{\sqrt{2}} \ln |\cosec \theta + \cot \theta| + c$

$I = \frac{1}{\sqrt{2}} \ln |\frac{\sqrt{2}}{y} + \sqrt{1+\frac{2}{y^2}}| + c$

$I = \frac{1}{\sqrt{2}} \ln |\frac{x\sqrt{2}}{x^2 - 1} + \frac{1}{x}\sqrt{x^2 + 2(x^2-1)^2}| + \epsilon$

Hopefully no mistakes.

Nice work.
I think the other integral is very difficult (haven't done it myself though).

Sy123 · May 10, 2013

Re: MX2 Integration Marathon

Registered User said:
Nice work.
I think the other integral is very difficult (haven't done it myself though).

It can be resolved to a cosine recurrence integral with the substitution x = atan u

Or by parts we can do:

$I_n = \int (a^2+x^2)^{-n} \ dx$

$I_n = \frac{x}{(x^2+a^2)^n} + 2n \int \frac{x^2}{(x^2+a^2)^{n+1}} \ dx$

$I_n = \frac{x}{(x^2+a^2)^n} + 2n (I_n - a^2 I_{n+1})$

$a^2 I_{n+1} = (1-2n)I_{n} - \frac{x}{(x^2+a^2)^{n}}$

Which is a pain to keep recurring, its less of a pain for the cosine one I guess.

Registered User · May 10, 2013

Re: MX2 Integration Marathon

$\int \sin (\sin x) \ dx$

Registered User · May 10, 2013

Re: MX2 Integration Marathon

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

braintic · May 10, 2013

Re: MX2 Integration Marathon

Registered User said:
$\int_{0}^{\pi} \sin (\sin x) \ dx$

Pretty pointless question for practical purposes. Why would you want to find the sin of something that is not an angle?

Registered User · May 10, 2013

Re: MX2 Integration Marathon

braintic said:
Pretty pointless question for practical purposes. Why would you want to find the sin of something that is not an angle?

lol

Sy123 · May 10, 2013

Re: MX2 Integration Marathon

Registered User said:
$\int \frac{(x+1)^3 (x-1)}{(x^2+1)^2 \sqrt{x^4+x^2+1}} \, dx$

Divide by top and bottom by 1/x^3, by distributing the x's appropriately, we arrive at:

$\int \frac{(x+ 1/x +2)(1-1/x^2) \dx}{(x+1/x)^2 \sqrt{x^2 + 1 + 1/x^2}} dx$

Upon the substitution:

$y = x + 1/x$

The integral resolves into:

$\int \frac{y+2}{y^2 \sqrt{y^2-1}} \ dy$

Which upon the substiution y = sec theta

$\int 1 + 2\cos \theta \ d\theta$

$= \theta + 2\sin \theta + c$

$= \cos^{-1} (1/y) + 2\frac{\sqrt{y^2-1}}{y} + c$

$= \cos^{-1} \frac{x}{x^2+1} + \frac{2x\sqrt{x^2 + 1/x^2 +1}}{x^2+1} + c$

hopefully no mistakes, another good integral.

==========

$\int_0^1 \frac{\ln(1+x^2)}{1+x^2} \ dx$

Makematics · May 10, 2013

Re: MX2 Integration Marathon

Beautiful is the only word to describe the solutions to those awesome integrals Sy. Do questions like this ever come up in trials or in the HSC?

Sy123 · May 10, 2013

Re: MX2 Integration Marathon

Makematics said:
Beautiful is the only word to describe the solutions to those awesome integrals Sy. Do questions like this ever come up in trials or in the HSC?

Thanks.

Integrals of that difficulty won't come up unless they actually give you the substitution (I'm pretty sure). But who knows maybe they will do something different this year.

Makematics · May 10, 2013

Re: MX2 Integration Marathon

Sy123 said:
Thanks.

Integrals of that difficulty won't come up unless they actually give you the substitution (I'm pretty sure). But who knows maybe they will do something different this year.

Alright, yeah you're probably right they would usually give the substitution and say something like "by manipulating the integrand appropriately, evaluate..."

Registered User · May 10, 2013

Re: MX2 Integration Marathon

Registered User said:
$\int \sin (\sin x) \ dx$

Sorry everyone, this question can't be done using MX2 integration techniques.

You can sub u=sinx and get $\int \: \frac{\sin{u}}{\sqrt{1-u^2}}du$ but can't do anything else

Sy123 · May 11, 2013

Re: MX2 Integration Marathon

$\int \sqrt{\csc x - \sin x} \ dx$

RealiseNothing · May 11, 2013

Re: MX2 Integration Marathon

Sy123 said:
$\int \sqrt{\csc x - \sin x} \ dx$

$\int \sqrt{\frac{1-sin^2x}{sinx}}dx$

$\int \sqrt{\frac{cos^2x}{sinx}}dx$

$\int \frac{cosx}{\sqrt{sinx}}$

Let $u=sinx$

$\int \frac{du}{\sqrt{u}}$

$2\sqrt{u} + C$

$2\sqrt{sinx} + C$

Sy123 · May 11, 2013

Re: MX2 Integration Marathon

Yep, nice.

$\int_0^1 \frac{\ln(1+x^2)}{1+x^2} \ dx$

Registered User · May 11, 2013

Re: MX2 Integration Marathon

From now on I will try to post only questions that are done by very tricky substitutions, IBP or very tedious.

$\int \frac{x^{2}}{(1+x^{2})^{2}}dx$

Registered User · May 11, 2013

Re: MX2 Integration Marathon

$\int cot^{-1}(x^{2}+x+1)dx$

HSC 2012-14 MX2 Integration Marathon (archive) (1 Viewer)

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