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

pluto · Sep 5, 2012

Re: MX2 Integration Marathon

Hey Asianese, just notice your HSC subjects including Janpanese Continuers. Do you know how Jap Cont scale compare with Physic and Chemistry.
My planed for HSC subjects very much the same like yours. Just worry if Jap Continuers good in scale...Thanks

Carrotsticks · Sep 5, 2012

Re: MX2 Integration Marathon

Pluto, you can send him a Private Message, or put something on his Visitors Message on his profile. Please don't use this thread to ask him unrelated questions.

cutemouse · Sep 5, 2012

Re: MX2 Integration Marathon

Shadowdude said:
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

Haha, complex analysis is really really nice. How's Peter Brown? Gotta love how he personifies maths

Shadowdude · Sep 5, 2012

Re: MX2 Integration Marathon

Peter Brown? Yeah, great! I'm never bored in his lectures.

Also helps that, as you said, he teaches a really really nice subject.

asianese · Sep 5, 2012

Re: MX2 Integration Marathon

Nice little one

$\int e^{\cos ^2x-\sin ^2 x}\cos x\sin x \;dx$

Sindivyn · Sep 5, 2012

Re: MX2 Integration Marathon

$\int \frac{dx}{\sqrt{\sin x}} \\ $let$ u = {sin x}^{1/2} \\ {du} = \frac{cosx}{sin x}^{1/2}/2 \\ $now$ u^{2} = {sin x} \\ $rearranging we get$ {cos x} = {\sqrt{1-u^{4}} \\ $subbing in$ \int \ frac{\sqrt{1-u^{4}}{du}/2 \\ = frac{2{(1-u^{4})^{3/2}/12u^{3}$

asianese · Sep 5, 2012

Re: MX2 Integration Marathon

No....

Sindivyn · Sep 5, 2012

Re: MX2 Integration Marathon

Yeah, just noticed a lot of mistakes, hate working in latex (a lot of powers where they shouldn't be)

asianese · Sep 5, 2012

Re: MX2 Integration Marathon

That is definitely not your main concern. You have done the old non-linear-treat-as-linear bracket thing!!

Sindivyn · Sep 5, 2012

Re: MX2 Integration Marathon

I know, fixing that part now

, you should be able to work with what I had before hand (obviously with correct powers) to get the answer

asianese · Sep 5, 2012

Re: MX2 Integration Marathon

I will let you continue.

Sindivyn · Sep 5, 2012

Re: MX2 Integration Marathon

Done, took 4 pages of working so I don't think I can Latex it, will post directions in a sec though

Sindivyn · Sep 5, 2012

Re: MX2 Integration Marathon

After finding that the integral is equal to 1/2 * (1-u^4) ^1/2
Let u = sinA
Use difference of two squares, one of the terms will equal 1-sin^2A = 1
Now you have the integration of 1/2* cosA * (1+sin^2 A)^1/2
Let A = sinx
=1/2 * root (1+k^2)
Make a substitution of tanC
= 1/2 * integration of sec^3 x
Integration by parts and done ( don't forget to sub EVERYTHING back in

)

nightweaver066 · Sep 5, 2012

Re: MX2 Integration Marathon

asianese said:
Nice little one

$\int e^{\cos ^2x-\sin ^2 x}\cos x\sin x \;dx$

$\int e^{cos2x} \times \frac{1}{2}sin2x dx$

$= \frac{-1}{4} \int e^{cos2x} d(cos2x)$

$= -\frac{1}{4}e^{cos2x} + c$

asianese · Sep 5, 2012

Re: MX2 Integration Marathon

I'm not sure about this bit

Let A = sinx
=1/2 * root (1+k^2)
Make a substitution of tanC
= 1/2 * integration of sec^3 x
Integration by parts and done ( don't forget to sub EVERYTHING back in

)

is that the same x as before? If so you went in a huge circle. Where did a k come from?? (I'm actually confused - trying to work it out)

Trebla · Sep 5, 2012

Re: MX2 Integration Marathon

$\displaystyle\int e^{x+e^x}\,dx$

Carrotsticks · Sep 5, 2012

Re: MX2 Integration Marathon

Trebla said:
$\displaystyle\int e^{x+e^x}\,dx$

I think a similar one was posted earlier.

Rather sneaky integral.

TheProfessional · Sep 5, 2012

Re: MX2 Integration Marathon

Trebla said:
$\displaystyle\int e^{x+e^x}\,dx$

$\int e^{e^{x}+x}dx=\int e^{e^{x}}\times e^{x}dx$
Let $u=e^{x}$
$\int e^{u}\times e^{x}\frac{du}{e^{x}}=e^{e^{x}}+c$

barbernator · Sep 5, 2012

Re: MX2 Integration Marathon

e^e^x

beeeeeten

nightweaver066 · Sep 5, 2012

Re: MX2 Integration Marathon

Trebla said:
$\displaystyle\int e^{x+e^x}\,dx$

$\int e^{x + e^x} dx$

$= \int e^x \times e^{e^x} dx$

$= \int e^{e^x} d(e^x)$

$= e^{e^x} + c$

lol 3 answers

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