HSC 2012-14 MX2 Integration Marathon (archive) (4 Viewers)

seanieg89 · Sep 16, 2012

Re: MX2 Integration Marathon

Alternatively alternatively,

$I=\int \frac{2}{(\sqrt{2}\cos(x+\pi/4))^2}\, dx=\int \sec^2(x+\pi/4)\,dx=\tan(x+\pi/4)+C$

bleakarcher · Sep 16, 2012

Re: MX2 Integration Marathon

seanieg89 said:
Alternatively alternatively,

$I=\int \frac{2}{(\sqrt{2}\cos(x+\pi/4))^2}\, dx=\int \sec^2(x+\pi/4)\,dx=\tan(x+\pi/4)+C$

oh right, I didn't see this. nice.

seanieg89 · Sep 16, 2012

Re: MX2 Integration Marathon

Keep the trick in mind, it is a handy way to bypass some t-substitution bashes.

bleakarcher · Sep 16, 2012

Re: MX2 Integration Marathon

seanieg89 said:
Keep the trick in mind, it is a handy way to bypass some t-substitution bashes.

Sure will.

asianese · Sep 16, 2012

Re: MX2 Integration Marathon

OO aux angle..

Carrotsticks · Sep 18, 2012

Re: MX2 Integration Marathon

$\int \frac{dx}{1-e^{kx}}$

asianese · Sep 18, 2012

Re: MX2 Integration Marathon

$\int \frac{dx}{1-e^{kx}} \\ u=1-e^{kx} \Rightarrow 1-u=e^{kx} \Rightarrow -ke^{kx}=k(u-1)\\ du=-ke^{kx}dx \Rightarrow \frac{du}{k(u-1)}=dx \\ I=\frac{1}{k} \int \frac{du}{u(u-1)} \\ = \frac{1}{k} \int \frac{1}{u-1}-\frac{1}{u} du \\ = \frac{1}{k} \ln \left|\frac{u-1}{u}\right| +C \\=\frac{1}{k} \ln \left|1-\frac{1}{1-e^{kx}}\right|+C$

Carrotsticks · Sep 18, 2012

Re: MX2 Integration Marathon

Try dividing numerator and denominator by e^{-kx} ....

asianese · Sep 18, 2012

Re: MX2 Integration Marathon

lol

barbernator · Sep 18, 2012

Re: MX2 Integration Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=\int (sin^{-1}(x))^{2}~dx" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\int (sin^{-1}(x))^{2}~dx" title="\int (sin^{-1}(x))^{2}~dx" /></a>

nightweaver066 · Sep 18, 2012

Re: MX2 Integration Marathon

barbernator said:
<a href="http://www.codecogs.com/eqnedit.php?latex=\int (sin^{-1}(x))^{2}~dx" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\int (sin^{-1}(x))^{2}~dx" title="\int (sin^{-1}(x))^{2}~dx" /></a>

$= x(sin^{-1}x)^2 - \int \frac{2x}{\sqrt{1 - x^2}} \times sin^{-1}xdx$

$= x(sin^{-1}x)^2 + 2 \int sin^{-1}x d(\sqrt{1 - x^2})$

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

$= x(sin^{-1}x)^2 + 2\sqrt{1 - x^2}sin^{-1}(x) - 2x + c$

Question:
$$Show that $ \int ^1 _0 \frac{ln(1 + x)}{1 + x^2} = \frac{\pi}{8}ln2 $ using the substitution $ x = \frac{1 - u}{1 + u}$

Aesytic · Sep 18, 2012

Re: MX2 Integration Marathon

let u=arcsin(x)

I = integral(u^2) dx
= integral(u^2) du * dx/du

u=arcsin(x)
x=sinu
dx/du = cosu

.'. I = integral (u^2 * cosu) du
integrating by parts,
m=u^2
m'=2u
n'=cosu
n=sinu
I = u^2*sinu - integral(2u*sinu) du
= u^2*sinu - 2*integral(u*sinu) du
Using parts again,
p=u
p'=1
q'=sinu
q=-cosu
I = u^2*sinu - 2(-u*cosu - integral(-cosu) du)
= u^2*sinu + 2u*cosu - 2*integral(cosu) du
= u^2*sinu + 2u*cosu - 2sinu
= x(arcsin(x))^2 + 2arcsin(x)*cos(arcsin(x)) - 2x + C

not sure if this is the fastest way to do it

HeroicPandas · Sep 18, 2012

Re: MX2 Integration Marathon

Here is a great question:

Integrate e^[e^(x) + x].dx

EDIT: AWWWWW SOMEONE ALREADY POSTED UP THE QUESTION

Drongoski · Sep 19, 2012

Re: MX2 Integration Marathon

HeroicPandas said:
Here is a great question:

Integrate e^[e^(x) + x].dx

EDIT: AWWWWW SOMEONE ALREADY POSTED UP THE QUESTION

$\int e^{e^x +x} dx = \int e^{e^x} e^xdx = \int e^{e^x}d e^x = e^{e^x} + C$

Edit

Can you. Heroic, or someone post the next question please. I forgot I need to do this in a Marathon.

HeroicPandas · Sep 20, 2012

Re: MX2 Integration Marathon

Drongoski said:
$\int e^{e^x +x} dx = \int e^{e^x} e^xdx = \int e^{e^x}d e^x = e^{e^x} + C$

Edit

Can you. Heroic, or someone post the next question please. I forgot I need to do this in a Marathon.

Ok, i just learnt this today and its funny LOL

Integrate: ROOT[ 1 + sin2x ].dx

Sy123 · Sep 20, 2012

Re: MX2 Integration Marathon

HeroicPandas said:
Ok, i just learnt this today and its funny LOL

Integrate: ROOT[ 1 + sin2x ].dx

$\int \sqrt{1+\sin 2x} \ dx \\ \\ = \int \sqrt{\sin^2 x + 2\sin x \cos x + \cos^2 x} \ dx \\ \\ = \int \sqrt{(\sin x + \cos x)^2} \ dx \\ \\ =\int \sin x + \cos x \ dx \\ = \sin x - \cos x +C$

Also if its possible, could someone when they post their integrals put a tag on whether a 3U student can do this?
Fanks (I dont know of any good 4U integrals, but here is a 3u one)

$\int \frac{1}{\tan^{-1}x (1+x^2)} \ dx$

Just one I thought of on top of my head

seanieg89 · Sep 20, 2012

Re: MX2 Integration Marathon

Sy123 said:
$\int \sqrt{1+\sin 2x} \ dx \\ \\ = \int \sqrt{\sin^2 x + 2\sin x \cos x + \cos^2 x} \ dx \\ \\ = \int \sqrt{(\sin x + \cos x)^2} \ dx \\ \\ =\int \sin x + \cos x \dx \\ = \sin x - \cos x +C$

Within a restricted domain

. Think about square roots carefully.

HeroicPandas · Sep 20, 2012

Re: MX2 Integration Marathon

seanieg89 said:
Within a restricted domain . Think about square roots carefully.

Yes u got it, i was gonna put in borders to see if anyone can cancel the absolute but NAHHHH looks ugly without latex

and good job on seeing that 1 is not a 1 ahah

HeroicPandas · Sep 20, 2012

Re: MX2 Integration Marathon

Sy123 said:
$\int \sqrt{1+\sin 2x} \ dx \\ \\ = \int \sqrt{\sin^2 x + 2\sin x \cos x + \cos^2 x} \ dx \\ \\ = \int \sqrt{(\sin x + \cos x)^2} \ dx \\ \\ =\int \sin x + \cos x \ dx \\ = \sin x - \cos x +C$

Also if its possible, could someone when they post their integrals put a tag on whether a 3U student can do this?
Fanks (I dont know of any good 4U integrals, but here is a 3u one)

$\int \frac{1}{\tan^{-1}x (1+x^2)} \ dx$

Just one I thought of on top of my head

Is the answer ln(arctanx) +c

Let u = arctanx

du = dx/ (1 + x^2)

.'. I = integral du/u

= ln(u) + C

= ln ( arctanx) +C

OK i dont know if this is 3u or 4u, if this is 3unit i will stop posting up 3unit integration:

EDIT: Sorry i confused myself when making up aquestion lol

Sy123 · Sep 21, 2012

Re: MX2 Integration Marathon

seanieg89 said:
Within a restricted domain . Think about square roots carefully.

Oops, lol. Hmm..

So the domain is for all x
$\sin x + \cos x \geq 0$
Using auxilliary angles

$\sqrt{2} \sin (x + \frac{\pi}{4}) \geq 0$

So our domain is

$k\pi \leq x+ \pi/4 \leq (k+1) \pi \\ \\ k\pi - \frac{\pi}{4} \leq x \leq (k+1)\pi -\frac{\pi}{4} \ \ \ \ k \ \epsilon \ \ \mathbb{Z}$

?

HSC 2012-14 MX2 Integration Marathon (archive) (4 Viewers)

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