HSC 2012-14 MX2 Integration Marathon (archive) (1 Viewer)

Carrotsticks · May 13, 2013

Re: MX2 Integration Marathon

HeroicPandas said:
did u get part of this from a sydney boys trial paper?

It's a classic proof for the fact that the common Year 7-10 approximation of pi actually exceeds it.

Registered User · May 14, 2013

Re: MX2 Integration Marathon

Makematics said:
Ahhh right very nice I got stuck on the 6th last line in my solution:L and still dont get it... How does the 1/2 change to a 1 and the limits change from pi to pi/2

Make the substitution 2 theta = theta

Sy123 · May 14, 2013

Re: MX2 Integration Marathon

$\int \sin^{-1} \sqrt{\frac{x}{x+a}} \ dx$

Registered User · May 16, 2013

Re: MX2 Integration Marathon

Sy123 said:
$\int \sin^{-1} \sqrt{\frac{x}{x+a}} \ dx$

I will skip the IBP step:

$\int \sqrt{\frac{1}{a-x}}\sqrt{\frac{1}{x-a}}dx=\int \frac{1}{\sqrt{-(x-a)^{2}}}dx$

$Let$ $u=x-a, du=dx$

$\therefore -i\int \frac{1}{u}du=-ilogu=-ilog(x-a)+c$

$\therefore \int \arcsin(\frac{x}{x-a})=x \arcsin(\frac{x}{x-a})+-ilog(x-a)+c$

Sy123 · May 18, 2013

Re: MX2 Integration Marathon

Registered User said:
I will skip the IBP step:

$\int \sqrt{\frac{1}{a-x}}\sqrt{\frac{1}{x-a}}dx=\int \frac{1}{\sqrt{-(x-a)^{2}}}dx$

$Let$ $u=x-a, du=dx$

$\therefore -i\int \frac{1}{u}du=-ilogu=-ilog(x-a)+c$

$\therefore \int \arcsin(\frac{x}{x-a})=x \arcsin(\frac{x}{x-a})+-ilog(x-a)+c$

You have done it for when there is no square root inside the asine function.

===============

$\int_0^{\infty} \frac{1}{(x^2+x+1)^3} \ dx$

Remember if you can't be bothered to write a solution, a summary at each major step is sufficient.

(i.e. Do IBP to get (text), then substitute u = f(x) to get (text), etc.)

Sy123 · May 18, 2013

Re: MX2 Integration Marathon

Sy123 said:
$\int \sin^{-1} \sqrt{\frac{x}{x+a}} \ dx$

$u^2 = \frac{x}{x+a}$

$x = -a\frac{u^2}{u^2-1}$

$dx = \frac{2a u}{(u^2-1)^2} \ du$

$I = 2a \int \frac{u \sin^{-1}u}{(u^2-1)^2} \ du$

$u = \sin \theta$

$I = 2a\int \frac{\theta \sin \theta}{\cos^3 \theta} \ d\theta$

Integrating by parts

$u = \theta \ \ dv = \tan \theta \sec \theta \sec \theta$

$du = 1 \ \ v = \frac{1}{2} \sec^2 \theta$

$I = 2a \left(\frac{1}{2 }\sec^2 \theta - \frac{1}{2} \int \sec^2 \theta \ d\theta \right )$

$I = a\sec^2 \theta - a \tan \theta + c$

$Then substitute in for$ \ u \ $then substitute in for$ \ \ x \ \ $to yield the answer$

Sy123 · May 21, 2013

Re: MX2 Integration Marathon

$\int \sin(\ln x) + \cos(\ln x) \ dx$

Drongoski · May 22, 2013

Re: MX2 Integration Marathon

Is answer = xsin(lnx) + C ?

$I = \int sin(lnx) + cos(lnx) dx$

$= xsin(lnx) -\int x \frac {cos(lnx)}{x} dx + x cos(lnx) - \int x \frac {-sin(lnx)}{x} dx$

$= xsin(lnx) + xcos(lnx) - \int cos(lnx)dx + \int sin(lnx)dx$

$= [ \dots] - xcos(lnx) - \int sin(lnx)dx + xsin(lnx) - \int cos(lnx)dx$

$= 2sin(lnx) - I$

$\therefore I = sin(lnx) + C$

braintic · May 22, 2013

Re: MX2 Integration Marathon

Drongoski said:
Is answer = xsin(lnx) + C ?

$I = \int sin(lnx) + cos(lnx) dx$

$= xsin(lnx) -\int x \frac {cos(lnx)}{x} dx + x cos(lnx) - \int x \frac {-sin(lnx)}{x} dx$

$= xsin(lnx) + xcos(lnx) - \int cos(lnx)dx + \int sin(lnx)dx$

$= [ \dots] - xcos(lnx) - \int sin(lnx)dx + xsin(lnx) - \int cos(lnx)dx$

$= 2sin(lnx) - I$

$\therefore I = sin(lnx) + C$

The derivative of your answer is not the question.

Try: ∫sin(lnx)dx = xsin(lnx) - ∫cos(lnx) dx [by parts]
Rearranging: ∫[sin(lnx) + cos(lnx)] dx = xsin(lnx) + c

Drongoski · May 22, 2013

Re: MX2 Integration Marathon

braintic said:
The derivative of your answer is not the question.

Try: ∫sin(lnx)dx = xsin(lnx) - ∫cos(lnx) dx [by parts]
Rearranging: ∫[sin(lnx) + cos(lnx)] dx = xsin(lnx) + c

That's so much simpler. Why didn't I see that?

braintic · May 22, 2013

Re: MX2 Integration Marathon

Drongoski said:
That's so much simpler. Why didn't I see that?

I only saw this after I chose to do the two integrals separately to avoid the mess.
I doubt many people would see to do it this way other than through experimentation.

Drongoski · May 22, 2013

Re: MX2 Integration Marathon

You make me feel better now, braintic. Haha.

Sy123 · May 22, 2013

Re: MX2 Integration Marathon

$I = \int_0^{\pi /2} \frac{f(x) }{1+x^2} \ dx = \frac{\pi^2}{8}$

$Find an example of a function$ \ \ f(x) \ \ $ such that the integral holds$

Carrotsticks · May 22, 2013

Re: MX2 Integration Marathon

Sy123 said:
$I = \int_0^{\pi /2} \frac{f(x) }{1+x^2} \ dx = \frac{\pi^2}{8}$

$Find an example of a function$ \ \ f(x) \ \ $ such that the integral holds$

$\\ f(x) = \frac{\pi}{4} (1+x^2)$

=)

(is this cheating)

seanieg89 · May 22, 2013

Re: MX2 Integration Marathon

Sy123 said:
$I = \int_0^{\pi /2} \frac{f(x) }{1+x^2} \ dx = \frac{\pi^2}{8}$

$Find an example of a function$ \ \ f(x) \ \ $ such that the integral holds$

$f(x)=\frac{\pi^2}{8\tan^{-1}(\pi/2)}$

Or did you want any additional conditions on this function?

seanieg89 · May 22, 2013

Re: MX2 Integration Marathon

Carrotsticks said:
$\\ f(x) = \frac{\pi}{4} (1+x^2)$

=)

(is this cheating)

Haha started my post before I saw yours.

Sy123 · May 22, 2013

Re: MX2 Integration Marathon

Carrotsticks said:
$\\ f(x) = \frac{\pi}{4} (1+x^2)$

=)

(is this cheating)

lol nope hehe

A harder one:

$\int \sqrt{1+\sqrt{x}} \ \ dx$

Sy123 · May 22, 2013

Re: MX2 Integration Marathon

seanieg89 said:
$f(x)=\frac{\pi^2}{8\tan^{-1}(\pi/2)}$

Or did you want any additional conditions on this function?

Well my intent for the question was for someone to say f(x) = atan(x), but I guess I just posed this without really thinking about it :s

braintic · May 23, 2013

Re: MX2 Integration Marathon

Sy123 said:
$\int \sqrt{1+\sqrt{x}} \ \ dx$

(4/15) × (1+√x)^(3/2) × (3√x - 2)
[by letting x=(u^2 - 1)^2
Y/N?

Sy123 · May 23, 2013

Re: MX2 Integration Marathon

braintic said:
(4/15) × (1+√x)^(3/2) × (3√x - 2)
[by letting x=(u^2 - 1)^2
Y/N?

The substitution (u^2-1)^2 can be modified slightly to make the integral much more easier. Its close though.

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