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

asianese · Jun 9, 2013

Re: MX2 Integration Marathon

chris_power_96 said:
integrate from 0 to pi/2 (1+sin2x)^.5

$\begin{align*} \int_0^{\pi/2} \sqrt{\sin^2x+2\sin x \cos x +\cos^2x} \mathop{\mathrm{d}x}&= \int_0^{\pi/2} \sqrt{(\sin x +\cos x)^2}\mathop{\mathrm{d}x} \\ &= \int_0^{\pi/2} |\sin x + \cos x| \mathop{\mathrm{d}x} \\ &=\sqrt2 \int_0^{\pi/2} |\sin(x+\dfrac{\pi}{4})| \mathop{\mathrm{d}x} \end{align*}$

Draw a picture for $\left|\sin\left(x+\dfrac{\pi}{4}\right)\right|$ and sum areas.

$\int \tan^{-1}\sqrt x \mathop{\mathrm{d}x}$

$\int \dfrac{\sin x +\cos x}{\sqrt[3]{\sin x - \cos x}} \mathop{\mathrm{d}x}$

braintic · Jun 9, 2013

Re: MX2 Integration Marathon

asianese said:
$\begin{align*} \int_0^{\pi/2} \sqrt{\sin^2x+2\sin x \cos x +\cos^2x} \mathop{\mathrm{d}x}&= \int_0^{\pi/2} \sqrt{(\sin x +\cos x)^2}\mathop{\mathrm{d}x} \\ &= \int_0^{\pi/2} |\sin x + \cos x| \mathop{\mathrm{d}x} \\ &=\sqrt2 \int_0^{\pi/2} |\sin(x+\dfrac{\pi}{4})| \mathop{\mathrm{d}x} \end{align*}$

Draw a picture for $\left|\sin\left(x+\dfrac{\pi}{4}\right)\right|$ and sum areas.

I think you've somewhat overcomplicated this integral.

The integral is done in the 1st quadrant, so the absolute values are unnecessary.
And sin x + cos x integrates as it stands - there is no need to transform it.

asianese · Jun 9, 2013

Re: MX2 Integration Marathon

braintic said:
I think you've somewhat overcomplicated this integral.

The integral is done in the 1st quadrant, so the absolute values are unnecessary.
And sin x + cos x integrates as it stands - there is no need to transform it.

Yeah you're right. works out the same.

HeroicPandas · Jun 9, 2013

Re: MX2 Integration Marathon

asianese said:
$[tex]\int \tan^{-1}\sqrt x \mathop{\mathrm{d}x}$
$\int \dfrac{\sin x +\cos x}{\sqrt[3]{\sin x - \cos x}} \mathop{\mathrm{d}x}$

1. Let u^2 = x
2. IBP

$I = 2\left ( \frac{u^2}{2}tan^{-1}u - \frac{1}{2}\int \frac{u^2du}{1+u^2} \right ) +C \\ \\ I = 2 \left ( \frac{u^2}{2}tan^{-1}u - \frac{1}{2} \left ( u - tan^{-1}u \right )\right ) +C \\ \\ I = x tan^{-1} {\sqrt{x}} + tan^{-1}{\sqrt{x}} - \sqrt{x} +C?$

Next one

$\int \dfrac{\sin x +\cos x}{\sqrt[3]{\sin x - \cos x}} \mathop{\mathrm{d}x}$

Let u^3 = sinx - cosx

3u^2 du = (cosx + sinx)dx

therefore, $I = 3\int udu = 3\frac{(sinx-cosx)^\frac{2}{3}}{2} +C$

problème

$$Find $\int_{-2}^{2} \frac{x^2}{e^x +1}dx$

seanieg89 · Jun 9, 2013

Re: MX2 Integration Marathon

HeroicPandas said:
1. Let u^2 = x
2. IBP

$I = 2\left ( \frac{u^2}{2}tan^{-1}u - \frac{1}{2}\int \frac{u^2du}{1+u^2} \right ) +C \\ \\ I = 2 \left ( \frac{u^2}{2}tan^{-1}u - \frac{1}{2} \left ( u - tan^{-1}u \right )\right ) +C \\ \\ I = x tan^{-1} {\sqrt{x}} + tan^{-1}{\sqrt{x}} - \sqrt{x} +C?$

Next one

$\int \dfrac{\sin x +\cos x}{\sqrt[3]{\sin x - \cos x}} \mathop{\mathrm{d}x}$

Let u^3 = sinx - cosx

3u^2 du = (cosx + sinx)dx

therefore, $I = 3\int udu = 3\frac{(sinx-cosx)^\frac{2}{3}}{2} +C$

problème

$$Find $\int \frac{x^2}{e^x +1}dx$

Pretty sure no elementary primitive exists. Can people please be more careful when posting questions?

HeroicPandas · Jun 9, 2013

Re: MX2 Integration Marathon

seanieg89 said:
Pretty sure no elementary primitive exists. Can people please be more careful when posting questions?

woops, from -2 to 2

Makematics · Jun 9, 2013

Re: MX2 Integration Marathon

HeroicPandas said:
woops, from -2 to 2

omg i love this question! it's in cambridge 3U and 4U though lol

already done it several times... will leave for someone else

chris_power_96 · Jun 9, 2013

Re: MX2 Integration Marathon

HeroicPandas said:
problème

$$Find $\int_{-2}^{2} \frac{x^2}{e^x +1}dx$

Use the property that makes it the integral of 0 to 2 f(x)+f(-x), then times the f(-x) by e^x on both sides and add the f(x) and f(-x) and factorise out the e^x +1, making it the integral of x^2?

Sy123 · Jun 9, 2013

Re: MX2 Integration Marathon

$\int \frac{(x+7)^5}{(x+2)^5} \ dx$

HeroicPandas · Jun 9, 2013

Re: MX2 Integration Marathon

chris_power_96 said:
Use the property that makes it the integral of 0 to 2 f(x)+f(-x), then times the f(-x) by e^x on both sides and add the f(x) and f(-x) and factorise out the e^x +1, making it the integral of x^2?

yes

braintic · Jun 9, 2013

Re: MX2 Integration Marathon

Sy123 said:
$\int \frac{(x+7)^5}{(x+2)^5} \ dx$

Write as [1+5/(x+2)]^5 and expand.
Is there an easier method?

Makematics · Jun 9, 2013

Re: MX2 Integration Marathon

braintic said:
Write as [1+5/(x+2)]^5 and expand.
Is there an easier method?

you could let u=x+2 which will still result in an expansion, although it will be slightly easier

Sy123 · Jun 9, 2013

Re: MX2 Integration Marathon

braintic said:
Write as [1+5/(x+2)]^5 and expand.
Is there an easier method?

Oops, that is a typo, it should read (x+7)^5/(x+2)^7

Which can be resolved through substitution.

======

I'm running out of cool integrals =(

HeroicPandas · Jun 9, 2013

Re: MX2 Integration Marathon

$$Find $\int \frac{x^4}{1-x^2}dx$

asianese · Jun 9, 2013

Re: MX2 Integration Marathon

Sy123 said:
I'm running out of cool integrals =(

Uw0tm8

Makematics · Jun 9, 2013

Re: MX2 Integration Marathon

Sy123 said:
I'm running out of cool integrals =(

Is this possible?

braintic · Jun 9, 2013

Re: MX2 Integration Marathon

Makematics said:
you could let u=x+2 which will still result in an expansion, although it will be slightly easier

Actually, its exactly the same - it just looks easier.

Makematics · Jun 9, 2013

Re: MX2 Integration Marathon

braintic said:
Actually, its exactly the same - it just looks easier.

How? using my way i will have to expand (u+5)^5, while you will have to expand something with fractions, which are harder to deal with.

Makematics · Jun 9, 2013

Re: MX2 Integration Marathon

HeroicPandas said:
$$Find $\int \frac{x^4}{1-x^2}dx$

Using long division followed by a simple application of the heaviside method, the answer is

$-\frac{1}{3}x^3-x+\frac{1}{2}\ln\left |\frac{x+1}{x-1} \right |+c$

braintic · Jun 9, 2013

Re: MX2 Integration Marathon

Makematics said:
How? using my way i will have to expand (u+5)^5, while you will have to expand something with fractions, which are harder to deal with.

The fraction is still a unit that remains intact - it doesn't need to be broken up. As long as you think of the fraction as a variable itself, its exactly the same.
Seriously, if anyone sees expanding [1+5/(x+2)]^5 as being more difficult that (u+5)^5, they are going to struggle with a lot in Extension 2.

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