HSC 2012-14 MX2 Integration Marathon (archive) (2 Viewers)

RealiseNothing · May 5, 2013

Re: MX2 Integration Marathon

SpiralFlex said:
Clue

$x=??????[/QUOTE]$

Lets just give it away.

SpiralFlex · May 5, 2013

Re: MX2 Integration Marathon

RealiseNothing said:
Lets just give it away.

I will remove my post.

Registered User · May 5, 2013

Re: MX2 Integration Marathon

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

$I=\int_0^{\pi/2}x\cot xdx$

$=\int_0^{\pi/2}x(\ln\sin x)^\prime dx=-\int_0^{\pi/2}\ln\sin xdx$

$Let$ $x=\frac{\pi }{2}-u,$ $we have$

$I=-\int_0^{\pi/2}\ln\cos udu$

$Therefore$

$2I=-\int_0^{\pi/2}(\ln\sin x+\ln\cos x)dx$

$=\frac\pi2\ln2-\int_0^{\pi/2}\ln\sin2xdx$

$I just need to prove that:$ $\int_0^{\pi/2}\ln\sin2xdx=I$

Can you give me a hint on how I can do that?

$If I prove that, then:$ $I=\frac{\pi }{2}ln(2)$

Registered User · May 5, 2013

Re: MX2 Integration Marathon

$Evaluate the following integral:$

$I(r)=\int_0^\pi\ln(1-2r\cos x+r^2)dx$ $where$ $\lvert r\rvert\neq1$

seanieg89 · May 5, 2013

Re: MX2 Integration Marathon

Registered User said:
$I=\int_0^{\pi/2}x\cot xdx$

$=\int_0^{\pi/2}x(\ln\sin x)^\prime dx=-\int_0^{\pi/2}\ln\sin xdx$

$Let$ $x=\frac{\pi }{2}-u,$ $we have$

$I=-\int_0^{\pi/2}\ln\cos udu$

$Therefore$

$2I=-\int_0^{\pi/2}(\ln\sin x+\ln\cos x)dx$

$=\frac\pi2\ln2-\int_0^{\pi/2}\ln\sin2xdx$

$I just need to prove that:$ $\int_0^{\pi/2}\ln\sin2xdx=I$

Can you give me a hint on how I can do that?

$If I prove that, then:$ $I=\frac{\pi }{2}ln(2)$

To me that final integral seems harder than the first...may be wrong though.

Registered User · May 5, 2013

Re: MX2 Integration Marathon

A way to come up with really complicated integrals that can be done using elementary functions: make up a complicated function and differentiate it

asianese · May 5, 2013

Re: MX2 Integration Marathon

Registered User said:
A way to come up with really complicated integrals that can be done using elementary functions: make up a complicated function and differentiate it

That's usually how integrals are made up lols. But you should be careful because not all functions are integratable, but nearly all are differentiatable.

Sy123 · May 5, 2013

Re: MX2 Integration Marathon

seanieg89 said:
To me that final integral seems harder than the first...may be wrong though.

For the cot one, is it a really obscure substitution of the difficulty of like x = (1-u)/(1+u)?

SpiralFlex · May 5, 2013

Re: MX2 Integration Marathon

No substitution req.

seanieg89 · May 5, 2013

Re: MX2 Integration Marathon

Sy123 said:
For the cot one, is it a really obscure substitution of the difficulty of like x = (1-u)/(1+u)?

Not the way I did it.

Sy123 · May 5, 2013

Re: MX2 Integration Marathon

seanieg89 said:
Not the way I did it.

I am guessing I need to multiply 1 in a clever way, no?

I need to find a way to prove that:

$\int_0^{\pi/2} \ln(\sin(2x)) = \int_0^{\pi /2} \ln(\sin(x)) \ dx$

I am considering some sort of periodicity argument but I don't think that might work.

SpiralFlex · May 5, 2013

Re: MX2 Integration Marathon

Well I did it by manipulating the x to something.

Sy123 · May 5, 2013

Re: MX2 Integration Marathon

Sy123 said:
I am guessing I need to multiply 1 in a clever way, no?

I need to find a way to prove that:

$\int_0^{\pi/2} \ln(\sin(2x)) = \int_0^{\pi /2} \ln(\sin(x)) \ dx$

I am considering some sort of periodicity argument but I don't think that might work.

Ok here is my attempt at proving the above equality.

We can establish that:

$\int_0^{\pi /4} \ln(\sin(2x)) \ dx = \frac{1}{2} \int_{0}^{\pi /2} \ln(\sin(x)) \ dx$

By simply using the substitution 2x = u

And since sin(2x) is symmetrical about pi/4 from x=0 to x=pi/2. Hence (by taking the integral of ln(sin(2x)) from pi/4 to pi/2)

$\int_0^{\pi/2} \ln(\sin(2x)) = \int_0^{\pi /2} \ln(\sin(x)) \ dx$

===========

This relates to the problem, because by integrating xcot x by parts:

$I = - \int_0^{\pi /2} \ln(\sin x) \ dx$

Taking that specific integral on the RHS, then applying that definite integrals property (x) -> (a-x), then adding and log properties, we arrive at:

$2\int_{0}^{\pi/2} \ln(\sin x) \ dx = \int_{0}^{\pi /2}\ln(\sin(2x)) \ dx - \frac{\pi}{2} \ln(2)$

Since the integrals are equal, then $I = \frac{\pi}{2} \ln(2)$

====

I think that is valid, the only fishy thing maybe is the symmetry arguments but I think that's fine. EDIT: Actually a simple substitution pi - x = u for ln(sin(x)) is sufficient to prove the symmetry for ln(sin(x)) and x = pi/2 - u for ln(sin(2x)).
So I am absolutely sure its all correct.

Now what's the elegant method!?

seanieg89 · May 5, 2013

Re: MX2 Integration Marathon

Good solution Sy, pretty much the same speed as mine really:

seanieg89 said:
$I=\int_0^{\pi/2}x\cot(x)\,dx=\int_0^{\pi/2}(\pi/2-x)\tan(x)\,dx\\ \Rightarrow 2I=\int_0^{\pi/2}\frac{2x\cos^2(x)+2(\pi/2-x)\sin^2(x)}{2\sin(x)\cos(x)}\, dx\\=\int_0^{\pi/2}\frac{2x\cos(2x)+\frac{\pi}{2}(1-\cos(2x))}{\sin(2x)}\,dx\\\Rightarrow 4I=\int_0^{\pi}\frac{x\cos(x)+\frac{\pi}{2}(1-\cos(x))}{\sin(x)}\, dx\\=\int_0^{\pi/2}x\cot(x)\,dx+\frac{\pi}{2}\int_0^{\pi/2}\csc(x)-\cot(x)\,dx+\int_{\pi/2}^\pi (x-\pi)\cot(x)\,dx+\frac{\pi}{2}\int_{\pi/2}^\pi\csc(x)+\cot(x)\,dx\\=2I+\pi\int_0^{\pi/2}\csc(x)-\cot(x)\,dx\\=2I-\pi\log|1+\cos(x)|]^{\pi/2}_0\\\Rightarrow I=\frac{\pi}{2}\log(2).$

Sy123 · May 5, 2013

Re: MX2 Integration Marathon

seanieg89 said:
Good solution Sy, pretty much the same speed as mine really:

Alright nice.

asianese · May 5, 2013

Re: MX2 Integration Marathon

We should have an integration party.

RealiseNothing · May 6, 2013

Re: MX2 Integration Marathon

Sy123 said:
Now what's the elegant method!?

Spiral's method was to let $x = tan^{-1}(tan(x))$

seanieg89 · May 6, 2013

Re: MX2 Integration Marathon

RealiseNothing said:
Spiral's method was to let $x = tan^{-1}(tan(x))$

How does that lead to something faster than the previous solutions?

RealiseNothing · May 6, 2013

Re: MX2 Integration Marathon

seanieg89 said:
How does that lead to something faster than the previous solutions?

idk I didn't actually try the method myself lol.

seanieg89 · May 6, 2013

Re: MX2 Integration Marathon

RealiseNothing said:
idk I didn't actually try the method myself lol.

Ah well, am interested to know.

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