HSC 2012-14 MX2 Integration Marathon (archive) (13 Viewers)

Sy123 · Aug 25, 2013

Re: MX2 Integration Marathon

seanieg89 said:
$\lim_{t\rightarrow 1}\int_{\sqrt{t}}^t \frac{x}{x^n-1}\, dx=\lim_{t\rightarrow 1}\int_{1/2}^1 \frac{t^{2s}\log(t)}{t^{ns}-1}\, dx =\int_{1/2}^1 \lim_{t\rightarrow 1} \left(\frac{t^{2s}\log(t)}{t^{ns}-1}\right)\, ds\\ = \int_{1/2}^1 \frac{ds}{ns} =\frac{1}{n}\log(2).$

(After interchanging the limit and the integral, we use L'Hospitals to evaluate the inner limit.)

I'm afraid I don't understand your second step, could you please explain it?

EDIT: NVM, I get it, nice!

seanieg89 · Aug 25, 2013

Re: MX2 Integration Marathon

Sy123 said:
I'm afraid I don't understand your second step, could you please explain it?

EDIT: NVM, I get it, nice!

Cool just saw your edit. Yep, the 1/n factor screamed L'Hospital's to me. It is not completely trivial that we can bring the limit inside the integral, but either the monotone convergence theorem or the dominated convergence theorem lets us do this just fine.

Sy123 · Sep 1, 2013

Re: MX2 Integration Marathon

A cool question my friend made

$Prove$

$\int_0^1 (e^{x^2} + e^{-x^2} ) \ dx > 2$

HeroicPandas · Sep 1, 2013

Re: MX2 Integration Marathon

$$By graphing: e^{x^2} > 2 - e^{-x^2} \\ \\ ~~~~~~~~~~~\therefore e^{x^2}+e^{-x^2} >2 \\ \\ \therefore $\int_0^1\left ( e^{x^2}+e^{-x^2} \right )dx >\int_0^12dx \\ \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~=2$

Question:

$$Show that $\int \frac{dx}{1+cosx}dx = tan\left ( \frac{x}{2} \right ) +C $ without using t-formula$

Sy123 · Sep 1, 2013

Re: MX2 Integration Marathon

HeroicPandas said:
$$By graphing: e^{x^2} > 2 - e^{-x^2} \\ \\ ~~~~~~~~~~~\therefore e^{x^2}+e^{-x^2} >2 \\ \\ \therefore $\int_0^1\left ( e^{x^2}+e^{-x^2} \right )dx >\int_0^12dx \\ \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~=2$

Question:

Find $\int \frac{dx}{1+cosx}$ without using t-formula

Yep I guess that works, you could also use the identity:

$a+1/a \geq 2$

===========

HeroicPandas · Sep 1, 2013

Re: MX2 Integration Marathon

Sy123 said:
Yep I guess that works, you could also use the identity:

$a+1/a \geq 2$

===========

ahhh

chris_power_96 · Sep 1, 2013

Re: MX2 Integration Marathon

HeroicPandas said:
$$Show that $\int \frac{dx}{1+cosx}dx = tan\left ( \frac{x}{2} \right ) +C $ without using t-formula$

use double angle formula for cosx

Sy123 · Sep 1, 2013

Re: MX2 Integration Marathon

Alternatively multiply top and bottom by (1-cos x) to yield 2 'standard' integrals

$\int x^3 e^{x^2} \ dx$

integral95 · Sep 1, 2013

Re: MX2 Integration Marathon

Sy123 said:
$\int x^3 e^{x^2} \ dx$

Let u = x^2
then integration by parts

There fore I = 1/2 (x^2.e^x^2 - e^x^2)+C

Drongoski · Sep 1, 2013

Re: MX2 Integration Marathon

$\int \frac {1}{1 + cos x} dx = \int \frac {1}{2cos^2 (\frac {x}{2})} 2d(\frac {x}{2}) = \int sec^2 (\frac {x}{2}) d(\frac {x}{2}) = tan \frac {x}{2} + C$

chris_power_96 · Sep 1, 2013

Re: MX2 Integration Marathon

integrate sin(x^2)

HeroicPandas · Sep 1, 2013

Re: MX2 Integration Marathon

chris_power_96 said:
use double angle formula for cosx

yep

Sy123 said:
Alternatively multiply top and bottom by (1-cos x) to yield 2 'standard' integrals

Nice, more methods are great

Drongoski said:
$\int \frac {1}{1 + cos x} dx = \int \frac {1}{2cos^2 (\frac {x}{2})} 2d(\frac {x}{2}) = \int sec^2 (\frac {x}{2}) d(\frac {x}{2}) = tan \frac {x}{2} + C$

Yes!

integral95 · Sep 1, 2013

Re: MX2 Integration Marathon

chris_power_96 said:
integrate sin(x^2)

Use integration by parts (lol)
I = xsin(x^2) - integral (2xcos(x^2))
let u = x^2 for the integral
final answer should be
I = xsin(x^2) - sin(x^2) + C

P.S: How do you use the latex lol

chris_power_96 · Sep 1, 2013

Re: MX2 Integration Marathon

MethewYan said:
Use integration by parts (lol)
I = xsin(x^2) - integral (2xcos(x^2))
let u = x^2 for the integral
final answer should be
I = xsin(x^2) - sin(x^2) + C

P.S: How do you use the latex lol

2nd part of second line is wrong should be 2x^2cos(x^2)

Sy123 · Sep 1, 2013

Re: MX2 Integration Marathon

MethewYan said:
Use integration by parts (lol)
I = xsin(x^2) - integral (2xcos(x^2))
let u = x^2 for the integral
final answer should be
I = xsin(x^2) - sin(x^2) + C

P.S: How do you use the latex lol

By parts yields:

$x\sin(x^2) - \int 2x^2 \cos(x^2) \ dx$

your integral had only a (2x) instead of (2x^2)

===

The integral in fact cannot be found in terms of elementary function.

====

$\int \frac{\cos^3 x - \sin x \cos^2 x + 1}{\sin x \cos^2 x + \cos^3 x + \sin x \cos x} \ dx$

Sy123 · Sep 3, 2013

Re: MX2 Integration Marathon

$\int_0^{\frac{\pi}{2}} \sqrt{1 - \sin(2x)} \ dx$

RealiseNothing · Sep 3, 2013

Re: MX2 Integration Marathon

Sy123 said:
$\int_0^{\frac{\pi}{2}} \sqrt{1 - \sin(2x)} \ dx$

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

$\sqrt{(sin(x)-cos(x))^2} = sin(x) - cos(x)$

$\int_0^{\frac{\pi}{2}} sin(x) - cos(x) dx = -cos(x) - sin(x)$

Now since $sin(x) - cos(x) < 0$ for $0 \leq x < \frac{\pi}{4}$ we will just use the limits $\frac{\pi}{4}$ to $\frac{\pi}{2}$ and multiply by 2 due to the symmetry of the curve about the line $x = \frac{\pi}{4}{$

$2[-cos(\frac{\pi}{2}) - sin(\frac{\pi}{2}) + cos(\frac{\pi}{4}) + sin(\frac{\pi}{4})]$

$2[-0 - 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}]$

$2[\sqrt{2}-1]$

Sy123 · Sep 3, 2013

Re: MX2 Integration Marathon

RealiseNothing said:
$\sqrt{1-sin(2x)} = \sqrt{sin^2(x) + cos^2(x) - 2sin(x)cos(x)} = \sqrt{(sin(x)-cos(x))^2}$

$\sqrt{(sin(x)-cos(x))^2} = sin(x) - cos(x)$

$\int_0^{\frac{\pi}{2}} sin(x) - cos(x) dx = -cos(x) - sin(x)$

Now since $sin(x) - cos(x) < 0$ for $0 \leq x < \frac{\pi}{4}$ we will just use the limits $\frac{\pi}{4}$ to $\frac{\pi}{2}$ and multiply by 2 due to the symmetry of the curve about the line $x = \frac{\pi}{4}{$

$2[-cos(\frac{\pi}{2}) - sin(\frac{\pi}{2}) + cos(\frac{\pi}{4}) + sin(\frac{\pi}{4})]$

$2[-0 - 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}]$

$2[\sqrt{2}-1]$

Yep well done

$\int_{-\infty}^{\infty} \frac{dx}{(x^2+a^2)(x^2+b^2)}$

chris_power_96 · Sep 4, 2013

Re: MX2 Integration Marathon

Sy123 said:
Yep well done

$\int_{-\infty}^{\infty} \frac{dx}{(x^2+a^2)(x^2+b^2)}$

-pi/ab?

Sy123 · Sep 4, 2013

Re: MX2 Integration Marathon

chris_power_96 said:
-pi/ab?

That is incorrect
Your answer should be positive anyway

HSC 2012-14 MX2 Integration Marathon (archive) (13 Viewers)

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