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

RealiseNothing · May 24, 2013

Re: MX2 Integration Marathon

Makematics said:
I think something might be wrong, i got some weird shit with the limits too :/

The integral doesn't converge lol, although if you were to do an indefinite integral I'm fairly sure what I got is right.

anomalousdecay · May 24, 2013

Re: MX2 Integration Marathon

RealiseNothing said:
edit: nope, my answer is definitely right lol.

http://www.wolframalpha.com/input/?...^2+-+1/9)/(x^6(square+root+of+x^2+-+1/9)))+dx

http://integrals.wolfram.com/index.jsp?expr=5/(x^6(sqrt[x^2-(1/9)]))+++69&random=false

This site proves that you get a tendency to infinity in the answer.

Makematics · May 24, 2013

Re: MX2 Integration Marathon

RealiseNothing said:
The integral doesn't converge lol, although if you were to do an indefinite integral I'm fairly sure what I got is right.

hmmm i got something nice for the indefinite integral too

anomalousdecay · May 24, 2013

Re: MX2 Integration Marathon

Makematics said:
1/40200 using the property of a definite integral: integral of f(x)= integral of f(a-x)

That is correct. Well done.

RealiseNothing · May 24, 2013

Re: MX2 Integration Marathon

Posted this in 3U thread but why not:

Show that:

$\frac{\pi}{2} > \lim_{\alpha \to 1^-} \int_{0}^{\alpha} \sqrt{1+x^2+x^4+...+x^{2n}}~dx > 1$

Carrotsticks · May 25, 2013

Re: MX2 Integration Marathon

Move along...

asianese · May 25, 2013

Re: MX2 Integration Marathon

$Find$

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

Makematics · May 25, 2013

Re: MX2 Integration Marathon

Carrotsticks said:
Move along...

cheers

Sy123 · May 25, 2013

Re: MX2 Integration Marathon

asianese said:
$Find$

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

Using the substitution:

$u^2=1+x^2$

Resolves the integral to:

$\int \sqrt{2-u^2} \ du$

Upon the subtitution:

$u=\sqrt{2} \sin \theta$

The integral resolves to:

$2\int \cos^2 \theta \ d\theta$

Upon computation:

$= \sin \theta \cos \theta + \theta + c_1$

Substituting u back in:

$= \sin^{-1} \left(\frac{u}{\sqrt{2}} \right ) + \frac{u}{2} \sqrt{2-u^2} + c_2$

Substituting back in x:

$= \sin^{-1} \sqrt{\frac{1+x^2}{2}} + \sqrt{1-x^4} + \epsilon$

asianese · May 25, 2013

Re: MX2 Integration Marathon

Alternatively you could have rationalised the numerator and put x^2=sin@.

Trebla · May 25, 2013

Re: MX2 Integration Marathon

$\displaystyle\int \dfrac{\sin^{2n}x}{\cos x}\,dx$

Where n is some positive integer

Capt Rifle · May 25, 2013

Re: MX2 Integration Marathon

Makematics said:
Not 4u difficulty but w/e

people utilised integration by parts at my school to solve it so yea..

Sy123 · May 25, 2013

Re: MX2 Integration Marathon

Trebla said:
$\displaystyle\int \dfrac{\sin^{2n}x}{\cos x}\,dx$

Where n is some positive integer

$I_n = \int \sin^{2n}x \sec x \ dx$

$I_{n-1}-I_n = \int \sin^{2n-2}x \sec x (1-\sin^2 x) \ dx$

$I_{n-1}-I_n = \int \sin^{2n-2}x \cos x \ dx = \frac{1}{2n-2} \sin^{2n-1} x$

$\sum_{k=1}^{n} I_{k-1}-I_k = \sum_{k=1}^{n} \frac{1}{2k-1} \sin^{2k-1}x$

$I_n = I_0 - \sum_{k=1}^{n} \frac{1}{2k-1} \sin^{2k-1} x$

$\fbox{\int \frac{\sin^{2n}x}{\cos x}\ dx = \ln |\sec x + \tan x| - \sum_{k=1}^{n} \frac{1}{2k-1} \sin^{2k-1}x + c}$

Is this the answer you expected or is it possible to find that sum? Because I guess you could find that sum in integral form by integrating a geometric series of common ratio sin^2 x
But I think that just leads to the same thing as when you integrate by parts.

Sy123 · May 25, 2013

Re: MX2 Integration Marathon

$\int \frac{1}{\sqrt{x} + \sqrt[3]{x}} \ dx$

Trebla · May 25, 2013

Re: MX2 Integration Marathon

Sy123 said:
$I_n = \int \sin^{2n}x \sec x \ dx$

$I_{n-1}-I_n = \int \sin^{2n-2}x \sec x (1-\sin^2 x) \ dx$

$I_{n-1}-I_n = \int \sin^{2n-2}x \cos x \ dx = \frac{1}{2n-2} \sin^{2n-1} x$

$\sum_{k=1}^{n} I_{k-1}-I_k = \sum_{k=1}^{n} \frac{1}{2k-1} \sin^{2k-1}x$

$I_n = I_0 - \sum_{k=1}^{n} \frac{1}{2k-1} \sin^{2k-1} x$

$\fbox{\int \frac{\sin^{2n}x}{\cos x}\ dx = \ln |\sec x + \tan x| - \sum_{k=1}^{n} \frac{1}{2k-1} \sin^{2k-1}x + c}$

Is this the answer you expected or is it possible to find that sum? Because I guess you could find that sum in integral form by integrating a geometric series of common ratio sin^2 x
But I think that just leads to the same thing as when you integrate by parts.

Pretty much the answer I was after. I also had in mind a bit of manipulation to generate the series as an alternative approach.

$\displaystyle\int \dfrac{\sin^{2n}x}{\cos x}\,dx\\\\ = \displaystyle\int \left(\dfrac{1}{\cos x} - \dfrac{1 - \sin^{2n}x}{\cos x}\right)\,dx\\\\ = \ln |\sec x + \tan x| - \displaystyle\int \dfrac{\cos x(1 - \sin^{2n}x)}{1-\sin^2x}\,dx\\\\=\ln |\sec x + \tan x| - \displaystyle\int \displaystyle\sum_{k=0}^{n-1} \sin^{2k}x\cos x\,dx \\\\=\ln |\sec x + \tan x| - \displaystyle\sum_{k=0}^{n-1} \dfrac{1}{2k+1}\sin^{2k+1}x + c$

jmk123 · May 25, 2013

Re: MX2 Integration Marathon

Sy123 said:
$\int \frac{1}{\sqrt{x} + \sqrt[3]{x}} \ dx$

I got a final answer of x^3-3x+(1/x)+c, where x=sqrtu.
Is that right?

Sy123 · May 25, 2013

Re: MX2 Integration Marathon

jmk123 said:
I got a final answer of x^3-3x+(1/x)+c, where x=sqrtu.
Is that right?

I don't think so, what was your method briefly?

jmk123 · May 25, 2013

Re: MX2 Integration Marathon

Sy123 said:
I don't think so, what was your method briefly?

let u^3=x, 3u^2du=dx

So integral is 3u^2/(u^3/2+u).du
take out factor of u on bottom, it cancels with one u on top so u have

3u/(sqrtu+1)

let x=sqrtu and divide 3x^2 by (x+1), get -3/3x^2 as remainder, so -1/x^2, and the rest expands to 3x^2-3

So just integrated each bit separately.

Sy123 · May 25, 2013

Re: MX2 Integration Marathon

jmk123 said:
let u^3=x, 3u^2du=dx

So integral is 3u^2/(u^3/2+u).du
take out factor of u on bottom, it cancels with one u on top so u have

3u/(sqrtu+1)

let x=sqrtu and divide 3x^2 by (x+1), get -3/3x^2 as remainder, so -1/x^2, and the rest expands to 3x^2-3

So just integrated each bit separately.

Upon your substitution, $v=\sqrt{u}$ , you didn't find du.

jmk123 · May 25, 2013

Re: MX2 Integration Marathon

Sy123 said:
Upon your substitution, $v=\sqrt{u}$ , you didn't find du.

K I did that and got int(6x^3+6-(1/((x-0.5)^2+0.75)))

Final result is (3x^4)/2+6x-(2/sqrt3).tan^-1((2x-1)/sqrt3)
That right?

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