HSC 2012-14 MX2 Integration Marathon (archive) (5 Viewers)

RealiseNothing · Aug 29, 2012

Re: MX2 Integration Marathon

My question anyway:

$\int \sqrt{\sin 4x + \sin 2x}~dx$

Just put in a random function, sorry if it's too easy or impossible etc.

nightweaver066 · Aug 29, 2012

Re: MX2 Integration Marathon

RealiseNothing said:
$\int \frac{dx}{\sqrt{\sin x}}$

$Let~u = \sqrt{sinx}$

$dx = \frac{2\sqrt{sinx}du}{cosx}$

$dx = \frac{2udu}{cosx}$

$\int \frac{1}{u} \frac{2udu}{cosx}$

$\int \frac{2}{cosx}du$

$\frac{2u}{cosx} + C$

$\frac{2\sqrt{sinx}}{cosx} + C$

I did not know you were allowed or could do that..

bleakarcher · Aug 29, 2012

Re: MX2 Integration Marathon

nightweaver066 said:
I did not know you were allowed or could do that..

This was precisely my reaction. Holy shit.

Carrotsticks · Aug 29, 2012

Re: MX2 Integration Marathon

Wait I actually didn't read it properly. Give me a few moments.

RealiseNothing · Aug 29, 2012

Re: MX2 Integration Marathon

lol yer I wasn't exactly sure if you could do that, hence my first post with a question mark.

seanieg89 · Aug 29, 2012

Re: MX2 Integration Marathon

It's wrong.

bleakarcher · Aug 29, 2012

Re: MX2 Integration Marathon

RealiseNothing said:
lol yer I wasn't exactly sure if you could do that, hence my first post with a question mark.

lol I wish that was legit. Unfortunately it's not.

RealiseNothing · Aug 29, 2012

Re: MX2 Integration Marathon

seanieg89 said:
It's wrong.

lol damn.

Carrotsticks · Aug 29, 2012

Re: MX2 Integration Marathon

You integrated 1/cos(x) with respect to U, then did U/cos(x).

You can't do this because U is dependent on X. You can only do this if it is independent.

Shadowdude · Aug 29, 2012

Re: MX2 Integration Marathon

it's wrong because wolfram alpha says it's wrong

trollface.jpg

asianese · Aug 30, 2012

Re: MX2 Integration Marathon

$\int \frac{dx}{\sqrt{\sin x}} \\ $Let $ u^2=\sin x\\ 2udu=\cos x dx\\$Draw a triangle with x as an acute angle such that $ \cos x=\sqrt{1-u^2}\\ \therefore dx=\frac{2u}{\sqrt{1-u^2}}du\\ I=\int \frac{2u}{u\sqrt{1-u^2}}du\\=2\int\frac{du}{\sqrt{1-u^2}}\\=2\sin^{-1}u\\ = 2\sin^{-1}(\sqrt{\sinx})+C$

double post

asianese · Aug 30, 2012

Re: MX2 Integration Marathon

$\int \frac{dx}{\sqrt{\sin x}} \\ $Let $ u^2=\sin x\\ 2udu=\cos x dx\\$Draw a right triangle with x as an acute angle such that $ \cos x=\sqrt{1-u^4}\\ \therefore dx=\frac{2u}{\sqrt{1-u^4}}du\\ I=\int \frac{2u}{u\sqrt{1-u^4}}du\\=2\int\frac{du}{\sqrt{1-u^4}}\\=$WTFBBQ$+C$

Hm.

Sy123 · Aug 30, 2012

Re: MX2 Integration Marathon

asianese said:
$\int \frac{dx}{\sqrt{\sin x}} \\ $Let $ u^2=\sin x\\ 2udu=\cos x dx\\$Draw a right triangle with x as an acute angle such that $ \cos x=\sqrt{1-u^2}\\ \therefore dx=\frac{2u}{\sqrt{1-u^2}}du\\ I=\int \frac{2u}{u\sqrt{1-u^2}}du\\=2\int\frac{du}{\sqrt{1-u^2}}\\=2\sin^{-1}u\\ = 2\sin^{-1}(\sqrt{\sin x})+C$

Why doesn't wolfram have a solution with simpler functions? (Mebs I'm wrong)

Differentiating your answer leads to a result which isnt the integral (and subbing in 1 radian to both the differential of your answer, and the function that you integrated has different results)

asianese · Aug 30, 2012

Re: MX2 Integration Marathon

Hmm where is the algebraic flaw?

Carrotsticks · Aug 30, 2012

Re: MX2 Integration Marathon

asianese said:
Hmm where is the algebraic flaw?

When you did the triangle.

D94 · Aug 30, 2012

Re: MX2 Integration Marathon

asianese said:
Hmm where is the algebraic flaw?

cos(x) ≠ √(1-u²)

barbernator · Sep 3, 2012

Re: MX2 Integration Marathon

spent 20min on this before i went to wolfram.

http://www.wolframalpha.com/input/?i=integrate:+(sinx)^(-1/2)dx.....

asianese · Sep 3, 2012

Re: MX2 Integration Marathon

I don't think it is integratable?? (Idk)

Shadowdude · Sep 3, 2012

Re: MX2 Integration Marathon

ohh that integral

yeah, i dont even know how to solve that (yet)

it's in the next section of my complex analysis course

will get back to you in about three weeks after i covered the material necessary

Carrotsticks · Sep 5, 2012

Re: MX2 Integration Marathon

Haha I saw that 4025808.

HSC 2012-14 MX2 Integration Marathon (archive) (5 Viewers)

what is that?It is Cowpea

Well-Known Member

Active Member

Retired

what is that?It is Cowpea

Well-Known Member

Active Member

what is that?It is Cowpea

Retired

Cult of Personality

Σ

Σ

This too shall pass

Σ

Retired

New Member

Active Member

Σ

Cult of Personality

Retired

Users Who Are Viewing This Thread (Users: 0, Guests: 5)