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

Sy123 · Jul 7, 2013

Re: MX2 Integration Marathon

braintic said:
Just checking - should there be a proviso that cos alpha is not equal to +/- 1 ?

Yes you are correct, apologies.

Carrotsticks · Jul 7, 2013

Re: MX2 Integration Marathon

braintic said:
Just checking - should there be a proviso that cos alpha is not equal to +/- 1 ?

Maybe a range of alpha such as 0 < alpha < pi, which I can imagine would be a more 'standard condition' than cos alpha =/= plusminus 1.

Makematics · Jul 7, 2013

Re: MX2 Integration Marathon

lol to be fair sydney grammar leads you through it... and that is a q8 :O 7 marks for it too!

Sy123 · Jul 7, 2013

Re: MX2 Integration Marathon

Makematics said:
lol to be fair sydney grammar leads you through it... and that is a q8 :O 7 marks for it too!

'newer papers are easier' ; )

Seriously though, the tan(x/2) substitution is obvious, and the tan inverse identity is obvious, the only thing even requiring a sort of intuition is the use of double angle formulae while simplifying which in a sense is obvious when we observe what we are supposed to get, we can predict the steps needed to arrive at it!

7 marks is a bit much though..... and to think Sydney Grammar papers are usually the best out of top selective school papers.

Carrotsticks · Jul 8, 2013

Re: MX2 Integration Marathon

Sy123 said:
'newer papers are easier' ; )

Oh boy don't you start this again.

asianese · Jul 8, 2013

Re: MX2 Integration Marathon

TBH Syd grammar papers are one of the best out there...not just because they have pretty diagrams and are written in TeX...

Sy123 · Jul 8, 2013

Re: MX2 Integration Marathon

asianese said:
TBH Syd grammar papers are one of the best out there...not just because they have pretty diagrams and are written in TeX...

Yeah, relative to most other schools, definitely.

Carrotsticks · Jul 8, 2013

Re: MX2 Integration Marathon

asianese said:
TBH Syd grammar papers are one of the best out there...not just because they have pretty diagrams and are written in TeX...

2012 was a bit of a disappoint though, I must say.

asianese · Jul 8, 2013

Re: MX2 Integration Marathon

An easier question: I guess leave for someone else (nonregular..?)

$$Using the fact that $ \lim_{x\to 0^+} x\ln x = 0,$

$Evaluate$

$\int_0^1 \ln\left(\dfrac{1}{x}\right)\mathop{\mathrm{d}x}.$

Carrotsticks · Jul 8, 2013

Re: MX2 Integration Marathon

asianese said:
An easier question: I guess leave for someone else (nonregular..?)

$$Using the fact that $ \lim_{x\to 0} x\ln x = 0,$

$Evaluate$

$\int \ln\left(\dfrac{1}{x}\right)\mathop{\mathrm{d}x}.$

$\int \ln \left ( \frac{1}{x} \right ) ~dx = - \int \ln (x) ~dx = ...$

Sy123 · Jul 8, 2013

Re: MX2 Integration Marathon

$\int \sqrt{\frac{a-x}{x-b}} \ dx$

asianese · Jul 8, 2013

Re: MX2 Integration Marathon

Oops forgot limits.

Also just had a look at 2012 Syd Grammar. It resembled a random school paper. What happened?....Gosh.

Argaroth · Jul 8, 2013

Re: MX2 Integration Marathon

Sy123 said:
$\int \sqrt{\frac{a-x}{x-b}} \ dx$

Sy123 · Jul 8, 2013

Re: MX2 Integration Marathon

Argaroth said:

Well done
The idea is correct, but there is a mistake when you substitute cos(k) back in, it is not:

$\cos (k) = 1 - \frac{u^2}{(a-b)}$

But rather the square root of that expression.

An easier substitution is:

$t^2 = \frac{a-x}{x-b}$

Then re-arranging:

$x = \frac{a+b}{t^2 + 1}$

It then becomes a partial fraction problem.

Sy123 · Jul 8, 2013

Re: MX2 Integration Marathon

$$It is known that any integral of the form$ \ \ \int \frac{P(x)}{Q(x)} \ dx$

$$For some polynomials$ \ \ P \ \ $and$ \ \ Q$

$The integral can be found$

$Prove that any integral of the form$ \ \ \int_0^{\pi } x f(\sin x) \ dx \ \ $can be evaluated$

$$For some function$ \ \ f \ \ $in the form$ \ \ f(x) = \frac{P(x)}{Q(x)} \ \ $for polynomials$ \ \ P \ \ $and$ \ \ Q$

Argaroth · Jul 9, 2013

Re: MX2 Integration Marathon

Sy123 said:
Well done
The idea is correct, but there is a mistake when you substitute cos(k) back in, it is not:

$\cos (k) = 1 - \frac{u^2}{(a-b)}$

But rather the square root of that expression.

I think you misread my answer sir.

asianese · Jul 12, 2013

Re: MX2 Integration Marathon

Sy123 said:
$$It is known that any integral of the form$ \ \ \int \frac{P(x)}{Q(x)} \ dx$

$$For some polynomials$ \ \ P \ \ $and$ \ \ Q$

$The integral can be found$

$Prove that any integral of the form$ \ \ \int_0^{\pi } x f(\sin x) \ dx \ \ $can be evaluated$

$$For some function$ \ \ f \ \ $in the form$ \ \ f(x) = \frac{P(x)}{Q(x)} \ \ $for polynomials$ \ \ P \ \ $and$ \ \ Q$

Hint for anyone: use the fact that $\int_0^{\pi } x f(\sin x) \ dx = \int_0^{\pi } (\pi-x) f(\sin (\pi-x) \ dx$

Sy123 · Jul 13, 2013

Re: MX2 Integration Marathon

$\int \sqrt{a+\sqrt{b+\sqrt{x}}} \ dx$

asianese · Jul 15, 2013

Re: MX2 Integration Marathon

1. Use the substitution u^2=x and so 2u du = dx

2. Substitute into the integrand.

2. Let v^2 = b+u (so that u=v^2 - b) and 2v dv = du.

3. Substitute into the integrand.

4. Let w^2 = a+v (so that v=w^2 -a) and 2w dw = dv.

5. Substitute into the integrand.

6. Expand all the factors in terms of w.

7. Integrate wrt w.

8. Do a mass back substitution.

9. Add a constant.

$\int \dfrac{\mathop{\mathrm{d}x}}{(x+1)\sqrt{-x^2-2x}}.$

Sy123 · Jul 15, 2013

Re: MX2 Integration Marathon

Sy123 said:
$$It is known that any integral of the form$ \ \ \int \frac{P(x)}{Q(x)} \ dx$

$$For some polynomials$ \ \ P \ \ $and$ \ \ Q$

$The integral can be found$

$Prove that any integral of the form$ \ \ \int_0^{\pi } x f(\sin x) \ dx \ \ $can be evaluated$

$$For some function$ \ \ f \ \ $in the form$ \ \ f(x) = \frac{P(x)}{Q(x)} \ \ $for polynomials$ \ \ P \ \ $and$ \ \ Q$

Using the identity:

$\int_0^a f(t) \ dt = \int_0^a f(a-t) \ dt$

$I = \int_0^{\pi } xf(\sin x) \ dx = \int_0^{\pi }(\pi -x) f(\sin x) \ dx$

Therefore:

$I = \frac{\pi}{2} \int_0^{\pi} f(\sin x) \ dx$

Now, upon the generic substitution:

$t= \tan(x/2)$ , we arrive at:

$I = \frac{\pi}{2} \int_0^{\infty} f(\frac{2t}{1+t^2}) \frac{2}{1+t^2} \ dt$

The integral will always be in the form, P/Q for some polynomials P and Q since the only operations done is the division and multiplication of polynomials with polynomials, and hence the end result will be in the form P/Q for some polynomials.

Therefore by the first theorem given in the question, the integral is evaluatable.

(I do realise that there is a possibility that the integral might not converge, I apologise for this inconsistency).

asianese said:
1. Use the substitution u^2=x and so 2u du = dx

2. Substitute into the integrand.

2. Let v^2 = b+u (so that u=v^2 - b) and 2v dv = du.

3. Substitute into the integrand.

4. Let w^2 = a+v (so that v=w^2 -a) and 2w dw = dv.

5. Substitute into the integrand.

6. Expand all the factors in terms of w.

7. Integrate wrt w.

8. Do a mass back substitution.

9. Add a constant.

$\int \dfrac{\mathop{\mathrm{d}x}}{(x+1)\sqrt{-x^2-2x}}.$

Yep, alternatively you can do the substitution:

$x=((u-a)^2-b)^2$

Which essentially combines all the substitutions you have used into a nice clean one.

For your integral, upon the substitution: $(x+1) = \sin u$ , dissolves into the integral of cosecant of u
We can now do the t=tan u/2 substitution or just recognise it to be a standard integral not standard to the HSC, resulting in:

$-\ln |\frac{1}{x+1} + \sqrt{\frac{1}{(x+1)^2}-1}| + c$

=====

$\int \frac{dx}{1+ \cos^4 x}$

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

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