HSC 2015 MX2 Integration Marathon (archive) (1 Viewer)

swagswagyoloswag · Mar 30, 2015

Re: MX2 2015 Integration Marathon

wtf can someone edit my post or something that was my first attempt at latex

FrankXie · Mar 30, 2015

Re: MX2 2015 Integration Marathon

swagswagyoloswag said:
$\int_{0}^{100\pi} \sqrt{cos^{2}(x) + sin^{2}(x) - cos^{2}(x) + sin^{2}(x)} dx \\ \int_{0}^{100\pi} \sqrt{{2}} \sqrt{sin^{2}(x)} dx \\ int_{0}^{100\pi} \sqrt{{2}} |sin(x)| \\ \text{Now, the graph of \|sin(x)| will repeat every pi/2 interval as it is a reflection of the negative y values across the x-axis. For every interval of the area under \|sinx|, there are 100pi/{pi/2} = 200 pieces.} \\ \\text {therefore, the integral can be represented as 200*\\sqrt{{2}} \int_{0}^{\pi/2} \sin(x) dx \\ = 200\sqrt{{2}} * (1) = 200\sqrt{{2}}}$

swagswagyoloswag said:
wtf can someone edit my post or something that was my first attempt at latex

$\int_{0}^{100\pi} \sqrt{cos^{2}(x) + sin^{2}(x) - cos^{2}(x) + sin^{2}(x)} dx=\int_{0}^{100\pi} \sqrt{{2}} \sqrt{sin^{2}(x)} dx=\int_{0}^{100\pi} \sqrt{{2}} |sin(x)|dx$
$$ Now, the graph of $ |sin(x)| $ will repeat every pi/2 interval as it is a reflection of the negative y values across the x-axis. For every interval of the area under $ |sinx|, $ there are 100pi/{pi/2} = 200 pieces. therefore, the integral can be represented as $ 200\sqrt{{2}} \int_{0}^{\pi/2} \sin(x) dx = 200\sqrt{{2}} * (1) = 200\sqrt{{2}}}$

Ekman · Mar 30, 2015

Re: MX2 2015 Integration Marathon

Next Question:

$\int \sec^3(x) dx$

VBN2470 · Mar 30, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
Next Question:

$\int \sec^3(x) dx$

Split it into sec(x)*sec^2(x) and use integration by parts. Then compare it with the integral of sec(x)*(1+tan^2(x)) so that you can evaluate the messy integral of sec(x)*tan^2(x) and you will end up getting $$\frac{1}{2}\sec{x}\tan{x} + \frac{1}{2}\ln|\sec{x}+\tan{x}| + C$$

VBN2470 · Mar 30, 2015

Re: MX2 2015 Integration Marathon

NEXT QUESTION:

$Evaluate \\ $\int_{0}^{1} \sin({\cos^{-1}{x}) dx$

swagswagyoloswag · Mar 30, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
NEXT QUESTION:

$Evaluate \\ $\int_{0}^{1} \sin({\cos^{-1}{x}) dx$

$$ let $ \cos^{-1}(x) = \alpha \\ cos(\alpha) = x \\ $Drawing a triangle, we can get all three sides relating to $ \alpha \\ $ Therefore, $ \int_{0}^{1} \sin({\cos^{-1}{x}) \\ = \int_{0}^{1} \sin(\alpha) \\ = \int_{0}^{1} \sqrt{1-x^{2}} \\ $ The integral can then be evaluated as the area of a quarter of a circle of radius 1. $ \\ = \frac{\pi *(1^{2})}{4} = \frac{\pi}{4}$

swagswagyoloswag · Mar 31, 2015

Re: MX2 2015 Integration Marathon

New q:
$\int_{0}^{\pi} \cos^{2}(tan^{-1}(sin(cot^{-1}x)))$

Ekman · Mar 31, 2015

Re: MX2 2015 Integration Marathon

swagswagyoloswag said:
New q:
$\int_{0}^{\pi} \cos^{2}(tan^{-1}(sin(cot^{-1}x)))$

Don't forget about the dx at the end, its really important that you mention it for this question.
$\int_{0}^{\pi} \cos^{2}(tan^{-1}(sin(cot^{-1}x)))dx$

VBN2470 · Apr 1, 2015

Re: MX2 2015 Integration Marathon

$\int_{0}^{\pi} \cos^{2}(\tan^{-1}(\sin(\cot^{-1}x)))dx = \int_{0}^{\pi} \frac{1+x^2}{2+x^2}dx = \pi - \frac{\tan^{-1}\frac{\pi}{\sqrt2}}{\sqrt2}$

nightweaver066 · Apr 2, 2015

Re: MX2 2015 Integration Marathon

A fun integral from 2015 UNSW Actuarial Society Integration Bee,

$\int \frac{\sqrt{x^2+1}\left[\ln\left(x^2+1\right) - 2\ln x\right]}{x^4} \, dx$

VBN2470 · Apr 2, 2015

Re: MX2 2015 Integration Marathon

nightweaver066 said:
A fun integral from 2015 UNSW Actuarial Society Integration Bee,

$\int \frac{\sqrt{x^2+1}\left[\ln\left(x^2+1\right) - 2\ln x\right]}{x^4} \, dx$

Nice question!

Essentially you want to rearrange the integrand through log laws and basic factorisation/algebraic manipulation of expressions and use the substitution $$u = 1 + \frac{1}{x^2}$$ . So you would have
$\int \frac{\sqrt{x^2+1}\left[\ln\left(x^2+1\right) - 2\ln x\right]}{x^4} \, dx \\ = -\frac{1}{2}\int \sqrt\frac{x^2+1}{x^2}\ln\left(\frac{x^2+1}{x^2} \right)\frac{-2}{x^3} \, dx \\ = -\frac{1}{2}\int \sqrt{u}\ln(u) \, du = ...$

Then use integration by parts to evaluate the rest. Final expression will probably look a bit messy though..

nightweaver066 · Apr 2, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
Nice question!

Essentially you want to rearrange the integrand through log laws and basic factorisation/algebraic manipulation of expressions and use the substitution $$u = 1 + \frac{1}{x^2}$$ . So you would have
$\int \frac{\sqrt{x^2+1}\left[\ln\left(x^2+1\right) - 2\ln x\right]}{x^4} \, dx \\ = -\frac{1}{2}\int \sqrt\frac{x^2+1}{x^2}\ln\left(\frac{x^2+1}{x^2} \right)\frac{-2}{x^3} \, dx \\ = -\frac{1}{2}\int \sqrt{u}\ln(u) \, du = ...$

Then use integration by parts to evaluate the rest. Final expression will probably look a bit messy though..

Good job.

I used a trig substitution and I think it worked out, though I couldn't be bothered resubbing in for $x$ haha.

InteGrand · Apr 2, 2015

Re: MX2 2015 Integration Marathon

NEW QUESTION:

$$Find $ I=\int \frac{1}{\sqrt{1+\mathrm{e}^x}}\text{ d}x.$

Ekman · Apr 2, 2015

Re: MX2 2015 Integration Marathon

InteGrand said:
NEW QUESTION:

$$Find $ I=\int \frac{1}{\sqrt{1+\mathrm{e}^x}}\text{ d}x.$

Im sure there is a quicker method:
$$ Let u$=\sqrt{1+e^x}$

$\therefore dx= \frac{2du \cdot u}{u^2-1}$

$$Hence the integral is: $ \int \frac{2 du}{u^2-1} = \int \frac{1}{u-1} - \frac{1}{u+1} du$

$=ln(\frac{u-1}{u+1}) + c \Rightarrow ln(\frac{\sqrt{1+e^x}-1}{\sqrt{1+e^x}+1}) +c$

Drsoccerball · Apr 2, 2015

Re: MX2 2015 Integration Marathon

InteGrand said:
NEW QUESTION:

$$Find $ I=\int \frac{1}{\sqrt{1+\mathrm{e}^x}}\text{ d}x.$

$$ Make a triangle with side length $ 1,e^{\frac{x}{2}}, \sqrt{(1+e^x)}$

$Cos\theta = \frac{1}{\sqrt{1+e^x}}$

$tan\theta =e^{\frac{x}{2}}$

$sin\theta = \frac{e^{ \frac{x}{2}}}{ \sqrt{1+e^x}}$

$$ After evaluating you get: $ ln(\frac{(\sqrt{1+e^x}-1)^2}{e^x}) +C$

FrankXie · Apr 3, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
$$ Make a triangle with side length $ 1,e^{\frac{x}{2}}, \sqrt{(1+e^x)}$

$Cos\theta = \frac{1}{\sqrt{1+e^x}}$

$tan\theta =e^{\frac{x}{2}}$

$sin\theta = \frac{e^{ \frac{x}{2}}}{ \sqrt{1+e^x}}$

what is dx then ?

Drsoccerball · Apr 3, 2015

Re: MX2 2015 Integration Marathon

FrankXie said:
what is dx then ?

Differenciate tan equation and you get sec^2theta dtheta=1/2 e^[x/2]dx where e^(x/2)=tantheta

Drsoccerball · Apr 3, 2015

Re: MX2 2015 Integration Marathon

New question:
$\int_0^1{\frac{x^7-1}{logx}}$

braintic · Apr 3, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
New question:
$\int_0^1{\frac{x^7-1}{logx}}$

Umm ...... log 0 and log 1 are a bit of a problem ....

studynerd · Apr 3, 2015

Re: MX2 2015 Integration Marathon

braintic said:
Umm ...... log 0 and log 1 are a bit of a problem ....

Ahahahahaha [emoji23] [emoji23] [emoji23]

But.. ignoring the boundaries.. is it possible to find the primitive with only mx2 knowledge? I'm stuck :/

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