HSC 2015 MX2 Integration Marathon (archive) (5 Viewers)

VBN2470 · Apr 26, 2015

Re: MX2 2015 Integration Marathon

WhoStanLeee said:
Yeah, it does look a bit strange...how would I go about undergoing the substitution in one line though? To prevent having u's and x's at the same time?

You will then need to break each part of the integrand and express it in terms of u. THEN substitute in to express your final integral in terms of u ONLY and go on from there.

InteGrand · Apr 26, 2015

Re: MX2 2015 Integration Marathon

Alternatively, write $x(u)$ to remind them $x$ is a function of $u$ (but you need to restrict the domain of x to make sure u is one-to-one).

InteGrand · Apr 26, 2015

Re: MX2 2015 Integration Marathon

So it's probably best in this case to do all the manipulation separately before integrating.

Ekman · Apr 26, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
Next Question:

$\int \frac{\sin^2x}{\cos^2x+\cot^2x} dx$

Bump

Ekman · Apr 27, 2015

Re: MX2 2015 Integration Marathon

New Question:

$\int \frac{x+\sin{x}}{1+\cos{x}} dx$

Axio · Apr 27, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
New Question:

$\int \frac{x+\sin{x}}{1+\cos{x}} dx$

Can't be bothered doing latex atm...

let integral =I
let t=tanx/2, dx/dt=2/t^2+1

This simplifies down to I/2=tan^-1 + t/t^2+1

Then integration by parts on tan^-1:

I/2=t/t^2+1 + ttan^-1t -t/t^2+1
I=2ttan^-1t

subbing t back in:

I=xtanx/2 +c

dunjaaa · Apr 28, 2015

Re: MX2 2015 Integration Marathon

My personal favourite

mreditor16 · Apr 28, 2015

Re: MX2 2015 Integration Marathon

Haven't you seen in ages, dunjaaa? How's things?

Ekman · Apr 28, 2015

Re: MX2 2015 Integration Marathon

New Question

$\int \frac{1}{\sin^3{x} + cos^3{x}} dx$

dunjaaa · Apr 28, 2015

Re: MX2 2015 Integration Marathon

Hey man, things have been great haha. Just swinging by, checking out how the marathons are holding up. As for uni, it's a nice environment. Must say, I've never enjoyed so much leisure time and freedom. Currently I'm studying BCom/BSc(Adv. Maths) at UNSW atm

Wbu?

FrankXie · Apr 28, 2015

Re: MX2 2015 Integration Marathon

dunjaaa said:
My personal favourite View attachment 32038

most interesting question ever

i am getting $\frac{4}{3}+\frac{2\sqrt{3}\pi}{27}$ , kind of weird, need OP to check pls

InteGrand · Apr 28, 2015

Re: MX2 2015 Integration Marathon

FrankXie said:
most interesting question ever
i am getting $\frac{4}{3}+\frac{2\sqrt{3}\pi}{27}$ , kind of weird, need OP to check pls

You're right according to WolframAlpha: https://www.wolframalpha.com/input/?i=sum+from+0+to+oo+of+1/(2n+choose+n)

dunjaaa · Apr 28, 2015

Re: MX2 2015 Integration Marathon

Yep, that's it. The summation is probably the most difficult part of the question haha

Drsoccerball · May 2, 2015

Re: MX2 2015 Integration Marathon

$\int{x^3(log(x^2))^2}$

Drsoccerball · May 2, 2015

Re: MX2 2015 Integration Marathon

$\int{xsin^{-1}(lnx)}dx$

InteGrand · May 2, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
$\int{x^3(log(x^2))^2}$

Use integration by parts twice (each time, the function you differentiate is the one involving the log, and the power of x is the function integrated).

juantheron · May 5, 2015

Re: MX2 2015 Integration Marathon

$Let \displaystyle I = \int\frac{\sin^2 x}{\cos^2 x+\cot^2 x}dx = \int\frac{\sin^4 x}{(1+\sin^2 x)\cdot \cos^2 x}dx = \int\frac{(\sin^4 x-1)+1}{(1+\sin^2 x)\cdot \cos^2 x}dx$

$$So Integral \displaystyle I = -\int 1\cdot dx+\int\frac{1}{(1+\sin^2 x)\cdot \cos^2 x}dx = -x+\int\frac{(1+\tan^2 x)\cdot \sec^2 x}{(1+2\tan^2 x)}dx$

$Now Let $\tan x = t$\;, Then \sec^2 xdx = dt$

$So Integral \displaystyle I = -x+\frac{1}{2}\int\frac{(1+2t^2)+1}{1+2t^2}dt$

$So Integral \displaystyle I = -x+\frac{1}{2}\tan x+\frac{\sqrt{2}}{4}\tan^{-1}\left(\sqrt{2}\cdot \tan x\right)+\mathcal{C}$

$So \displaystyle I = \int\frac{\sin^2 x}{\cos^2 x+\cot^2 x}dx = -x+\frac{1}{2}\tan x+\frac{1}{2\sqrt{2}}\tan \left(\sqrt{2}\cdot \tan x\right)+\mathcal{C}.$

juantheron · May 5, 2015

Re: MX2 2015 Integration Marathon

$Let \displaystyle I = \int\frac{1}{\sin^3 x+\cos^3 x}dx = \int\frac{1}{(\sin x+\cos x)\cdot (1-\sin x\cdot \cos x)}dx$

$So we can write it as \displaystyle \frac{1}{3}\int \left[\frac{2}{\sin x+\cos x}+\frac{(\sin x+\cos x)}{1-\sin x\cdot \cos x}\right]dx$

$Now put \displaystyle $\sin x = \frac{2\tan \frac{x}{2}}{1+\tan^2 \frac{x}{2}}$ and \displaystyle \cos x = \frac{1-\tan^2\frac{x}{2}}{1+\tan^2 \frac{x}{2}}$

$And for second Integral put $(\sin x-\cos x) = t\Rightarrow 1-\sin 2x =t^2$

$And $\sin 2x = (1-t^2)$ and (\cos x+\sin x)dx = dt$

$So our Integral convert into \displaystyle \frac{1}{3}\int\frac{2\sec^2 \frac{x}{2}}{1-\tan^2\frac{x}{2}+2\tan \frac{x}{2}}dx+\frac{2}{3}\int\frac{1}{t^2+1}dt$

$Now $\tan \frac{x}{2} = u\;,$ Then $\sec^2 \frac{x}{2}dx = 2dt$

$So Integral \displaystyle I = \frac{4}{3}\int\frac{1}{1-u^2+2u}du+\frac{2}{3}\tan t+\mathcal{C}$

$Now We can Convert first integral into \displaystyle $\int\frac{1}{x^2+a^2}$

$Or in the form of \displaystyle \int\frac{1}{a^2-x^2}dx$

Drsoccerball · May 5, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
$\int{xsin^{-1}(lnx)}dx$

bump

Axio · May 5, 2015

Re: MX2 2015 Integration Marathon

$\int\frac{sin8x}{sinx}dx$

HSC 2015 MX2 Integration Marathon (archive) (5 Viewers)

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