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

asianese · Sep 21, 2013

Re: MX2 Integration Marathon

Are you sure?

My working got me down to

$(n+1)a^{n+1}\ln a=a^{n+1}$

and so a can't be 0 since $\ln 0$ is undefined.

Sy123 · Sep 21, 2013

Re: MX2 Integration Marathon

asianese said:
Are you sure?

My working got me down to

$(n+1)a^{n+1}\ln a=a^{n+1}$

and so a can't be 0 since $\ln 0$ is undefined.

Ah yes of course, that a little silly of me

seanieg89 · Sep 21, 2013

Re: MX2 Integration Marathon

But in the initial integration, a=0 means that you are integrating over an interval of length 0, of course this will give you 0.

asianese · Sep 21, 2013

Re: MX2 Integration Marathon

True.

Sy123 · Sep 21, 2013

Re: MX2 Integration Marathon

$\int \frac{x+2}{(x^2+3x+3) \sqrt{x+1}} \ dx$

chris_power_96 · Sep 22, 2013

Re: MX2 Integration Marathon

Sy123 said:
$\int \frac{x+2}{(x^2+3x+3) \sqrt{x+1}} \ dx$

Use the substitution u^2=x+1 to give you ln(x+1)+c

Sy123 · Sep 22, 2013

Re: MX2 Integration Marathon

chris_power_96 said:
Use the substitution u^2=x+1 to give you ln(x+1)+c

I don't think that is correct

Sy123 · Sep 27, 2013

Re: MX2 Integration Marathon

Sy123 said:
$\int \frac{x+2}{(x^2+3x+3) \sqrt{x+1}} \ dx$

$u = \sqrt{x+1}$

$2u du = dx$

$\int \frac{2u(u^2+1)}{u (u^4 + u^2 + 1)} \ du$

$2\int \frac{1 + \frac{1}{u^2}}{u^2 + 1 + \frac{1}{u^2} } \ du$

$t = u - \frac{1}{u}$

.
.
.
and so on

=========

$\int \frac{dx}{ax^2 + bx + c}$

HeroicPandas · Sep 27, 2013

Re: MX2 Integration Marathon

Sy123 said:
$\int \frac{dx}{ax^2 + bx + c}$

$I = \frac{2}{\sqrt{4ac-b^2}} \left [ tan^{-1}\frac{2ax+b}{\sqrt{4ac-b^2}}\right ]$ ?

Makematics · Sep 27, 2013

Re: MX2 Integration Marathon

constant brah! but yeh i got the same

Sy123 · Sep 27, 2013

Re: MX2 Integration Marathon

HeroicPandas said:
$I = \frac{2}{\sqrt{4ac-b^2}} \left [ tan^{-1}\frac{2ax+b}{\sqrt{4ac-b^2}}\right ]$ ?

That is partially correct (even with the constant)

Carrotsticks · Sep 27, 2013

Re: MX2 Integration Marathon

HeroicPandas said:
$I = \frac{2}{\sqrt{4ac-b^2}} \left [ tan^{-1}\frac{2ax+b}{\sqrt{4ac-b^2}}\right ]$ ?

The solution depends on what the values of a,b and c are. More specifically, whether the discriminant is positive, zero, or negative.

Makematics · Sep 27, 2013

Re: MX2 Integration Marathon

Sy123 said:
That is partially correct (even with the constant)

the other solution is so painfullll

Sy123 · Sep 27, 2013

Re: MX2 Integration Marathon

Makematics said:
the other solution is so painfullll

Replace the roots with alpha and beta if you want, and anyway, I don't really want a solution I know people can do partial fractions.

The aim of the question was more getting people to recognize the different cases

Sy123 · Sep 29, 2013

Re: MX2 Integration Marathon

$\int_0^{\pi} \frac{\sin(nx)}{\sin x} \ dx$

JJ345 · Sep 30, 2013

Re: MX2 Integration Marathon

Sy123 said:
$\int_0^{\pi} \frac{\sin(nx)}{\sin x} \ dx$

I derived the reduction formula using trig identities etc. But got a final answer of 2^(n/2) I(0)--> So what would my final answer be? I(0) is just a constant isn't it? Probably did something wrong.

Sy123 · Sep 30, 2013

Re: MX2 Integration Marathon

JJ345 said:
I derived the reduction formula using trig identities etc. But got a final answer of 2^(n/2) I(0)--> So what would my final answer be? I(0) is just a constant isn't it? Probably did something wrong.

The answer is:

$I_n=\int_0^{\pi} \frac{\sin(nx)}{\sin x} \ dx = \begin{cases} 0 , \ \ $for even$ \ \ n \\ \pi, \ \ $for odd$ \ \ n \end{cases}$

This is because, $I_{n} = I_{n-2}$ (using sum to product formula)

So if n is even, it becomes I(0), and if n is odd, it becomes I(1)

JJ345 · Sep 30, 2013

Re: MX2 Integration Marathon

Sy123 said:
The answer is:

$I_n=\int_0^{\pi} \frac{\sin(nx)}{\sin x} \ dx = \begin{cases} 0 , \ \ $for even$ \ \ n \\ \pi, \ \ $for odd$ \ \ n \end{cases}$

This is because, $I_{n} = I_{n-2}$ (using sum to product formula)

So if n is even, it becomes I(0), and if n is odd, it becomes I(1)

Yeah, I didn't recognise the cases. But my final reduction formula was I(n)=2I(n-2) ???

Sy123 · Sep 30, 2013

Re: MX2 Integration Marathon

JJ345 said:
Yeah, I didn't recognise the cases. But my final reduction formula was I(n)=2I(n-2) ???

You may have made a silly mistake, because:

$I_n - I_{n-2} = \int_0^{\pi} \frac{\sin(nx) - \sin(n-2)x}{\sin x} \ dx = \int_0^{\pi} \frac{2\sin \left(\frac{nx - (n-2)x}{2} \right ) \cos \left(\frac{nx + (n-2)x}{2} \right)}{\sin x} \ dx$

$I_n -I_{n-2} = 2\int_0^{\pi} \cos(n-1) x \ dx = \frac{2\sin(n-1) x}{n-1} |_0^{\pi} = 0$

$\therefore \ \ I_n = I_{n-2}$

JJ345 · Sep 30, 2013

Re: MX2 Integration Marathon

Sy123 said:
You may have made a silly mistake, because:

$I_n - I_{n-2} = \int_0^{\pi} \frac{\sin(nx) - \sin(n-2)x}{\sin x} \ dx = \int_0^{\pi} \frac{2\sin \left(\frac{nx - (n-2)x}{2} \right ) \cos \left(\frac{nx + (n-2)x}{2} \right)}{\sin x} \ dx$

$I_n -I_{n-2} = 2\int_0^{\pi} \cos(n-1) x \ dx = \frac{2\sin(n-1) x}{n-1} |_0^{\pi} = 0$

$\therefore \ \ I_n = I_{n-2}$

Your right, but for some reason I still can't find the mistake in mine :/ Its wayyy longer than your method so probably did make a mistake somewhere along the line.

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

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