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

RealiseNothing · May 5, 2013

Re: MX2 Integration Marathon

woops

RealiseNothing · May 5, 2013

Re: MX2 Integration Marathon

RealiseNothing said:
$\int_{\frac{1}{2}}^{1} woops dx$

wait nevermind, let me fix this question up.

RealiseNothing · May 5, 2013

Re: MX2 Integration Marathon

$\int_{0}^{1} tan^{-1}(x)~dx$

btx3 · May 5, 2013

Re: MX2 Integration Marathon

Who would of thought that sy and realise would be up late on a saturday night doing 4 unit integration questions.

Sy123 · May 5, 2013

Re: MX2 Integration Marathon

$\int_0^{\pi/2} \frac{dx}{1+ \tan^{\pi}(x)}$

HeroicPandas · May 5, 2013

Re: MX2 Integration Marathon

RealiseNothing said:
$\int_{0}^{1} tan^{-1}(x)~dx$

$$Let I = $\int_0^1 tan^{-1}(x)dx \\ $Using integration by parts: u = tan^{-1}x, $ $ and dv = dx \dots \dots \\ \therefore I = $\left [ xtan^{-1}x \right ]_0 ^1 - \frac{1}{2}\int_0^1 \frac{2x}{1+x^2}dx \\ \\ I = \frac{\pi}{4} - \frac{1}{2} \left [ ln(1+x^2) \right ]_0^1 \\ \\ I = \frac{\pi}{4} - ln(\sqrt{2})$

Question

$$Find $\int e^{cotx + 2ln(cosecx)} dx$

asianese · May 5, 2013

Re: MX2 Integration Marathon

$I=\int e^{\cot x+2\ln (\csc x)} dx\\ = \int e^{\cot x} \cdot e^{\ln(\csc^2x)}dx \\ = \int e^{\cot x} \csc^2x dx\\ $NB: $ \frac{d}{dx}\cot x = -\csc^2x \Rightarrow -d(\cot x)=\csc^2x dx \\ \Longrightarrow I= -\int e^{\cot x} \cdot d(\cot x)\\=-e^{\cot x} + \text{(Sy's face), (Sy's face) constant}$

HeroicPandas · May 5, 2013

Re: MX2 Integration Marathon

asianese said:
$I=\int e^{\cot x+2\ln (\csc x)} dx\\ = \int e^{\cot x} \cdot e^{\ln(\csc^2x)}dx \\ = \int e^{\cot x} \csc^2x dx\\ $NB: $ \frac{d}{dx}\cot x = -\csc^2x \Rightarrow -d(\cot x)=\csc^2x dx \\ \Longrightarrow I= -\int e^{\cot x} \cdot d(\cot x)\\=-e^{\cot x} + \text{(Sy's face), (Sy's face) constant}$

lol nice! but why Sy's face? xD

HeroicPandas · May 5, 2013

Re: MX2 Integration Marathon

Sy123 said:
$\int_0^{\pi/2} \frac{dx}{1+ \tan^{\pi}(x)}$

$$Let I = $\int_0^{\pi/2} \frac{dx}{1+ \tan^{\pi}(x)} \\ I= \int_0^{\pi/2} \frac{cos^{\pi}x }{cos^{\pi}x + sin^{\pi}x} dx \\ $ Use the greatest rule in integration: $\int_0^a f(x)dx = \int_0^a f(a-x)dx \\ \\ \therefore I = \int_0^{\pi/2} \frac{sin^{\pi}x}{sin^{\pi}x + cos^{\pi}x}dx \\ \\ $Add both I's \\ \\ \therefore I = $\frac{1}{2} \int_0^{\pi/2} dx \\ \\ \therefore I = \frac{\pi}{4}?$

Question:
$$Find $\int (e^{ln(\cos^{-1}(-x))} - \pi - \sin^{-1}x)dx$

Carrotsticks · May 5, 2013

Re: MX2 Integration Marathon

Sy123 said:
$\int_0^{\pi/2} \frac{dx}{1+ \tan^{\pi}(x)}$

$\\ \int_0^{\pi/2} \frac{dx}{1+ \tan^{\pi}(x)} = \int_{0}^{\frac{\pi}{2}} \frac{dx}{1+\tan^{n}(x)} \\\\ $ where $ n $ is over 9000.$$

asianese · May 5, 2013

Re: MX2 Integration Marathon

$NB: $ \arccos x + \arcsin x=\pi/2\\ \Rightarrow \arccos(-x)+\arcsin(-x)=\pi/2\\ \Rightarrow \arccos(-x)-\arcsin(x)=\pi/2 $ since arcsin is odd.$ \\ I=\int \arccos(-x)-\pi -\arcsin(x) dx\\=\int (\pi/2-\pi)dx\\ =\int -\pi/2 dx\\ = -\frac{\pi}{2}x+$pandas$

Sy123 · May 5, 2013

Re: MX2 Integration Marathon

HeroicPandas said:
$$Let I = $\int_0^{\pi/2} \frac{dx}{1+ \tan^{\pi}(x)} \\ I= \int_0^{\pi/2} \frac{cos^{\pi}x }{cos^{\pi}x + sin^{\pi}x} dx \\ $ Use the greatest rule in integration: $\int_0^a f(x)dx = \int_0^a f(a-x)dx \\ \\ \therefore I = \int_0^{\pi/2} \frac{sin^{\pi}x}{sin^{\pi}x + cos^{\pi}x}dx \\ \\ $Add both I's \\ \\ \therefore I = $\frac{1}{2} \int_0^{\pi/2} dx \\ \\ \therefore I = \frac{\pi}{4}?$

Question:
$$Find $I= \int (e^{ln(\cos^{-1}(-x))} - \pi - \sin^{-1}x)dx$

Yep, good job.

=======

By simplifying the integral and using inverse trig rules, we end up with -pi/2 x + C

======

$\int_0^a \frac{dx}{x+\sqrt{x^2-a^2}}$

RealiseNothing · May 5, 2013

Re: MX2 Integration Marathon

HeroicPandas said:
$$Let I = $\int_0^1 tan^{-1}(x)dx \\ $Using integration by parts: u = tan^{-1}x, $ $ and dv = dx \dots \dots \\ \therefore I = $\left [ xtan^{-1}x \right ]_0 ^1 - \frac{1}{2}\int_0^1 \frac{2x}{1+x^2}dx \\ \\ I = \frac{\pi}{4} - \frac{1}{2} \left [ ln(1+x^2) \right ]_0^1 \\ \\ I = \frac{\pi}{4} - ln(\sqrt{2})$

Hence integrate:

$\int_{0}^{1} cot^{-1}(x)~dx$

Registered User · May 5, 2013

Re: MX2 Integration Marathon

RealiseNothing said:
Hence integrate:

$\int_{0}^{1} cot^{-1}(x)~dx$

$Let$ $I = \int_{0}^{1} cot^{-1}(x)dx$

$Let$ $u=cot^{-1}x, du=\frac{-1}{1+x^{2}}dx, dv=dx, v=x$

$\therefore I = \frac{\pi }{4}+ln(\sqrt{2})$

seanieg89 · May 5, 2013

Re: MX2 Integration Marathon

$\int_0^{\pi/2}x\cot(x)\, dx$

its been done earlier in this thread, but if you resist the temptation to look at the solution it is a very good question.

Registered User · May 5, 2013

Re: MX2 Integration Marathon

Sy123 said:
$\int_0^a \frac{dx}{x+\sqrt{x^2-a^2}}$

$Assuming$ $a^{2}<x^{2}:$

$\int_0^a \frac{dx}{x+\sqrt{x^2-a^2}}\times \frac{x-\sqrt{x^2-a^2}}{x-\sqrt{x^2-a^2}}$

$=\frac{1}{a^{2}}\int_0^a x-\frac{1}{a^{2}}\int_0^a\sqrt{x^2-a^2}dx$

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

Registered User · May 5, 2013

Re: MX2 Integration Marathon

seanieg89 said:
$\int_0^{\pi/2}x\cot(x)\, dx$

its been done earlier in this thread, but if you resist the temptation to look at the solution it is a very good question.

Looks like a hard integral. I did IBP and got $\frac{1}{2}\int_{0}^{\frac{\pi }{2}}x^{2} cosec^{2}x dx$ which doesn't really help.

seanieg89 · May 5, 2013

Re: MX2 Integration Marathon

It is quite tricky, probably one of the hardest in this thread that doesn't require out of syllabus techniques.

RealiseNothing · May 5, 2013

Re: MX2 Integration Marathon

Registered User said:
$Let$ $I = \int_{0}^{1} cot^{-1}(x)dx$

$Let$ $u=cot^{-1}x, du=\frac{-1}{1+x^{2}}dx, dv=dx, v=x$

$\therefore I = \frac{\pi }{4}+ln(\sqrt{2})$

I wanted you to use the previous result, since:

$tan^{-1}x + cot^{-1}x = \frac{\pi}{2}$

$\int_{0}^{1}cot^{-1}xdx = \frac{\pi}{2} - \int_{0}^{1} tan^{-1}xdx$

And you know the value of the integral on the RHS already.

SpiralFlex · May 5, 2013

Re: MX2 Integration Marathon

Clue

$x=??????$

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