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

Thread starter Sy123
Start date Dec 7, 2014

Status: Not open for further replies.

leehuan

Well-Known Member

#901

Re: MX2 2015 Integration Marathon

$I=-2\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \ln { \left( \cos { \theta } \right) } d\theta } =-2\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \ln { \left( \sin { \theta } \right) } d\theta } \\ Adding:\\ 2I=-2\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \ln { \left( \sin { \theta } \cos { \theta } \right) } d\theta } \\ =-2\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \ln { \left( \sin { 2\theta } \right) } -\ln { 2 } d\theta }$

Sure. I wouldn't have made it past there though. Periodicity of trigonometric functions in integrals meant I would've had to draw a graph first.

Ekman

Well-Known Member

#902

Re: MX2 2015 Integration Marathon

I was able to manipulate the question down to this integral:

$-2 \int_{0}^{1} \frac{\ln{\left(\frac{4x(1-x^2)}{(1+x^2)^2}\right)}}{1+x^2} dx$

If someone can find a way to make it equal to $\frac{\pi}{2}\ln{2}$ then the integral can be solved

Carrotsticks

Retired

#903

Re: MX2 2015 Integration Marathon

leehuan said:
$\\ \theta =\arctan { x } \Rightarrow d\theta =\frac { dx }{ 1+{ x }^{ 2 } } \\ Let\quad I=\int _{ 0 }^{ \infty }{ \frac { \ln { \left( 1+{ x }^{ 2 } \right) } }{ 1+{ x }^{ 2 } } dx } \\ I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \ln { \left( \sec ^{ 2 }{ \theta } \right) } d\theta } \\ =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \ln { \left( \sec ^{ 2 }{ \left( \frac { \pi }{ 2 } -\theta \right) } \right) } d\theta } \\ =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \ln { \left( \csc ^{ 2 }{ \theta } \right) } d\theta } \\$

My bank of integral tricks isn't really that large, but if I try adding those two integrals up I don't get anywhere either..

And wolfram does give an answer. pi log2.

Continuing on from what you have, that integral becomes

$-2 \int_{0}^{\frac{\pi}{2}} \ln \left ( \sin \theta \right ) d\theta$

And this integral (ignoring the -2 at the front) evaluates to $-\frac{\pi}{2} \ln 2$ , which gives the answer. But the only way I know of evaluating this is the classic method taught in complex analysis (analytically continuing the integral of a power of sine and differentiating the resultant gamma expression). There may be a highschool friendly way of doing this, but none that I know of at the moment.

D

Drsoccerball

Well-Known Member

#904

Re: MX2 2015 Integration Marathon

Carrotsticks said:
Continuing on from what you have, that integral becomes

$-2 \int_{0}^{\frac{\pi}{2}} \ln \left ( \sin \theta \right ) d\theta$

And this integral (ignoring the -2 at the front) evaluates to $-\frac{\pi}{2} \ln 2$ , which gives the answer. But the only way I know of evaluating this is the classic method taught in complex analysis (analytically continuing the integral of a power of sine and differentiating the resultant gamma expression). There may be a highschool friendly way of doing this, but none that I know of at the moment.

Someone solved using elementary methods on one of the previous pages.

seanieg89

Well-Known Member

#905

Re: MX2 2015 Integration Marathon

$I=\int_0^{\pi/2}\log(\sin x)\, dx\\ \\ = \int_0^{\pi/2}\log(\cos x)\, dx\\ \\\Rightarrow 2I=\int_0^{\pi/2}\log(\sin x\cos x)\, dx\\ \\ = \int_0^{\pi/2}\log(\sin 2x)\, dx-\frac{\pi}{2}\log(2)\\ \\ = I-\frac{\pi}{2}\log(2)\\ \\ \Rightarrow I=-\frac{\pi}{2}\log(2).$

Last edited: Sep 27, 2015

kawaiipotato

Well-Known Member

#906

Re: MX2 2015 Integration Marathon

Carrotsticks said:
Continuing on from what you have, that integral becomes

$-2 \int_{0}^{\frac{\pi}{2}} \ln \left ( \sin \theta \right ) d\theta$

And this integral (ignoring the -2 at the front) evaluates to $-\frac{\pi}{2} \ln 2$ , which gives the answer. But the only way I know of evaluating this is the classic method taught in complex analysis (analytically continuing the integral of a power of sine and differentiating the resultant gamma expression). There may be a highschool friendly way of doing this, but none that I know of at the moment.

I've seen people do it by using int f(pi/2 - x) = int f(x)
giving I = int 0 to pi/2 of ln(costheta) dtheta
Then adding, giving 2I = int 0 to pi/2 of ln(sin2theta) - ln2 dtheta
Then using u = 2theta, 2I = int 0 to pi of ln(sinu) - ln2 du
And then they evaluate the int 0 to pi of ln(sinu) to be zero because it's an odd around theta = pi/2 or something

seanieg89

Well-Known Member

#907

Re: MX2 2015 Integration Marathon

kawaiipotato said:
I've seen people do it by using int f(pi/2 - x) = int f(x)
giving I = int 0 to pi/2 of ln(costheta) dtheta
Then adding, giving 2I = int 0 to pi/2 of ln(sin2theta) - ln2 dtheta
Then using u = 2theta, 2I = int 0 to pi of ln(sinu) - ln2 du
And then they evaluate the int 0 to pi of ln(sinu) to be zero because it's an odd around theta = pi/2 or something

You are mostly right, not the last line though. Will fix up my eqn which isn't displaying for some reason.

Carrotsticks

Retired

#908

Re: MX2 2015 Integration Marathon

Drsoccerball said:
Someone solved using elementary methods on one of the previous pages.

Actually I think I know what you are referring to. I remember seeing something like that a while ago, and iirc braintic made a pdf about it too.

kawaiipotato

Well-Known Member

#909

Re: MX2 2015 Integration Marathon

seanieg89 said:
$I=\int_0^{\pi/2}\log(\sin x)\, dx\\ \\ = \int_0^{\pi/2}\log(\cos x)\, dx\\ \\\Rightarrow 2I=\int_0^{\pi/2}\log(\sin x\cos x)\, dx\\ \\ = \int_0^{\pi/2}\log(\sin 2x)\, dx-\frac{\pi}{2}\log(2)\\ \\ = I-\frac{\pi}{2}\log(2)\\ \\ \Rightarrow I=-\frac{\pi}{2}\log(2).$

Why is I = int 0 to pi/2 ln(sinx)dx = int 0 to pi/2 ln(sin2x) dx ?

D

Drsoccerball

Well-Known Member

#910

Re: MX2 2015 Integration Marathon

kawaiipotato said:
Why is I = int 0 to pi/2 ln(sinx)dx = int 0 to pi/2 ln(sin2x) dx ?

You use dummy variabales by letting u=2x

D

Drsoccerball

Well-Known Member

#911

Re: MX2 2015 Integration Marathon

Carrotsticks said:
Actually I think I know what you are referring to. I remember seeing something like that a while ago, and iirc braintic made a pdf about it too.

Sean did it just up there^

seanieg89

Well-Known Member

#912

Re: MX2 2015 Integration Marathon

kawaiipotato said:
Why is I = int 0 to pi/2 ln(sinx)dx = int 0 to pi/2 ln(sin2x) dx ?

$\int_0^{\pi/2}\log(\sin 2x)\, dx\\ \\ = \frac{1}{2}\int_0^{\pi}\log(\sin x)\, dx \\ \\ = \frac{1}{2}\int_0^{\pi/2}\log(\sin x)\,dx+\frac{1}{2}\int_{\pi/2}^{\pi}\log(\sin x)\, dx\\ \\ = \frac{1}{2}(I+\int_0^{\pi/2}\log(\sin(\pi-x))\, dx)\\ \\ =\frac{1}{2}(I+I)\\ \\ =I.$

You should of course be painstakingly slow in the jumps in your working for your HSC exams, I am just lazy / students trying to fill in blanks is actually decent exercise.

Last edited: Sep 27, 2015

Carrotsticks

Retired

#913

Re: MX2 2015 Integration Marathon

Drsoccerball said:
Sean did it just up there^

I remember that proof now. A bit ashamed that it didn't stick with me first time I read it. It's neat proof.

kawaiipotato

Well-Known Member

#914

Re: MX2 2015 Integration Marathon

seanieg89 said:
$\int_0^{\pi/2}\log(\sin 2x)\, dx\\ \\ = \frac{1}{2}\int_0^{\pi}\log(\sin x)\, dx \\ \\ = \frac{1}{2}\int_0^{\pi/2}\log(\sin x)\,dx+\frac{1}{2}\int_{\pi/2}^{\pi}\log(\sin x)\, dx\\ \\ = \frac{1}{2}(I+\int_0^{\pi/2}\log(\sin(\pi-x))\, dx)\\ \\ =\frac{1}{2}(I+I)\\ \\ =I.$

You should of course be painstakingly slow in the jumps in your working for your HSC exams.

Ah that's cool
Thanks

seanieg89

Well-Known Member

#915

Re: MX2 2015 Integration Marathon

kawaiipotato said:
Ah that's cool
Thanks

No worries. Remember the pattern, because of log laws you can often exploit the x -> a-x substitution in ways like this.

D

Drsoccerball

Well-Known Member

#916

Re: MX2 2015 Integration Marathon

seanieg89 said:
$\int_0^{\pi/2}\log(\sin 2x)\, dx\\ \\ = \frac{1}{2}\int_0^{\pi}\log(\sin x)\, dx \\ \\ = \frac{1}{2}\int_0^{\pi/2}\log(\sin x)\,dx+\frac{1}{2}\int_{\pi/2}^{\pi}\log(\sin x)\, dx\\ \\ = \frac{1}{2}(I+\int_0^{\pi/2}\log(\sin(\pi-x))\, dx)\\ \\ =\frac{1}{2}(I+I)\\ \\ =I.$

You should of course be painstakingly slow in the jumps in your working for your HSC exams, I am just lazy / students trying to fill in blanks is actually decent exercise.

Instead of (a-x) couldn't you just say that the area under sinx from 0 to pi/2 is the same as pi/2 to pi ?

kawaiipotato

Well-Known Member

#917

Re: MX2 2015 Integration Marathon

Drsoccerball said:
Instead of (a-x) couldn't you just say that the area under sinx from 0 to pi/2 is the same as pi/2 to pi ?

How wouldyou prove that it's the same besides using substitution: u = pi/2 - x?

D

Drsoccerball

Well-Known Member

#918

Re: MX2 2015 Integration Marathon

kawaiipotato said:
How wouldyou prove that it's the same besides using substitution: u = pi/2 - x?

Bruh its a sin graph we know that its the same area...

kawaiipotato

Well-Known Member

#919

Re: MX2 2015 Integration Marathon

Drsoccerball said:
Bruh its a sin graph we know that its the same area...

lol that's a very articulate proof

seanieg89

Well-Known Member

#920

Re: MX2 2015 Integration Marathon

Drsoccerball said:
Instead of (a-x) couldn't you just say that the area under sinx from 0 to pi/2 is the same as pi/2 to pi ?

But why is the area under sin(x) between 0 and pi/2 the same as that between pi/2 and pi?

This transformation is exactly how you would prove this rigorously (*).

(*) At least as rigorously as possible in high school. A more subtle point here is that area is not really rigorously defined in high school. Before getting to more advanced notions of volume in measure theory, we typically define the area only for certain regions. (Such as those bounded by the x axis, two vertical lines, and the graph of a positive continuous function f(x). For such regions we would literally DEFINE area as being the definite integral of this function between the two vertical lines.)

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