HSC 2015 MX2 Integration Marathon (archive) (3 Viewers)

braintic · Feb 7, 2015

Re: MX2 2015 Integration Marathon

InteGrand said:
Also, isn't the principle square root of a complex number a+bi (a, b real, a not negative if b = 0) defined as the one with positive real part (this definition fails if the complex number is a negative real number).

Defined by whom?

InteGrand · Feb 7, 2015

Re: MX2 2015 Integration Marathon

braintic said:
Defined by whom?

Defined in mathematics (not sure which person defined it like this first), like how square root of a positive number is defined as the positive value, e.g. principal value of $\sqrt{4}$ is defined to be 2, not -2.

Wikipedia explains it here, http://en.wikipedia.org/wiki/Square_root#Principal_square_root_of_a_complex_number, and I've seen it on other sites, but haven't checked any complex analysis textbooks yet, I'll try and find some maybe.

There is a geometric interpretation on the second answer here: http://math.stackexchange.com/questions/44406/how-do-i-get-the-square-root-of-a-complex-number

braintic · Feb 8, 2015

Re: MX2 2015 Integration Marathon

InteGrand said:
Defined in mathematics (not sure which person defined it like this first), like how square root of a positive number is defined as the positive value, e.g. principal value of $\sqrt{4}$ is defined to be 2, not -2.

Wikipedia explains it here, http://en.wikipedia.org/wiki/Square_root#Principal_square_root_of_a_complex_number, and I've seen it on other sites, but haven't checked any complex analysis textbooks yet, I'll try and find some maybe.

There is a geometric interpretation on the second answer here: http://math.stackexchange.com/questions/44406/how-do-i-get-the-square-root-of-a-complex-number

Just wondering what is the point of defining a principal square root in the first place? Why should one take precedence over the other, especially when the choice seems completely arbitrary?

InteGrand · Feb 8, 2015

Re: MX2 2015 Integration Marathon

braintic said:
Just wondering what is the point of defining a principal square root in the first place?

It's probably done so that there is a meaning to what is referred to by the expression $\sqrt{a+bi}$ that mathematicians can agree on, which is probably useful in complex analysis, like how it's useful with real numbers. If we didn't have principal square roots for positive real numbers, we'd have to keep saying which one we're referring to if we were using one in an equation or something, whereas with a principal root defined, we don't need to do this. This may also be useful for complex numbers for those doing complex analysis.

braintic said:
Why should one take precedence over the other, especially when the choice seems completely arbitrary?

Well, for square roots of positive numbers, it's more natural to define the positive square root as the principal value. Otherwise, we'd have $\sqrt{4}$ referring to -2 and $-\sqrt{4}$ refer to +2. While there wouldn't be anything mathematically wrong with this (it's just notation), it's more natural and less confusing to have the notation that has a negative sign to be the negative number, so we define the principal value of the square root of a positive number as the positive square root.

Defining the principal value of a complex number as the one with positive real part is maybe done because then the definition of the principal value of the square root of a positive real number is consistent with this, so the complex number definition works for all complex numbers except for the negative real numbers (since these have square roots with 0 real part). For a negative number $-b$ , maybe the principal square root is taken to be $\sqrt{b}i$ (where $\sqrt{b}$ refers to the principal value, which is the positive square root), since $i$ is defined as the principal value of $\sqrt{-1}$ .

seanieg89 · Feb 8, 2015

Re: MX2 2015 Integration Marathon

Yeah for negative reals, the principal square root is just an imaginary number with positive imaginary part.

More generally, the principal argument of a complex number z is the argument in the interval (-pi,pi]. (Note that any complex number will have at least one value of its arg in this range.)

This gives us the principal logarithm

Generally, we have

$\log(z)=\log(|z|)+i\arg(z)$

$z^w=\exp(w\log(z))$

as the means of defining complex powers as a multi-function. But if we replace arg by principal arg, log becomes single-valued and we get a notion of arbitrary principal powers. Certain familiar properties of positive real exponentiation break though, but it is still convenient to have an single-valued operator sometimes. (Say we only wanted to prove the existence of something, and our work required the extraction of a root for example.)

sirable1 · Feb 8, 2015

Re: MX2 2015 Integration Marathon

$\int \cos^6 x \sin^5 x \ dx$

sirable1 · Feb 8, 2015

Re: MX2 2015 Integration Marathon

$\int \cos^6 x \sin^6 x \ dx$

nightweaver066 · Feb 8, 2015

Re: MX2 2015 Integration Marathon

sirable1 said:
$\int \cos^6 x \sin^5 x \ dx$

$\begin{align*} \int \cos^6 x \sin^5x\, dx &= -\int \cos^6 x (1 - \cos^2 x)^2 \, d(\cos x)\\ &= -\int \cos^6 x - 2\cos^8 x + \cos^{10} x \, d(\cos x)\\ &= -\frac{1}{7}\cos^7 x + \frac{2}{9} \cos^9 x - \frac{1}{11}\cos^{11} x + c \end{align*}$

nightweaver066 · Feb 8, 2015

Re: MX2 2015 Integration Marathon

sirable1 said:
$\int \cos^6 x \sin^6 x \ dx$

$\begin{align*} \int \cos^6 x \sin^6 x \, dx &= \int 2^{-6} \sin^6 2x \, dx\\ &= 2^{-6} \int (\sin^3 2x)^2 \, dx\\ &= 2^{-6} \int (\frac{3\sin 2x - \sin 6x}{4})^2 \, dx\\ &= 2^{-10} \int 9\sin^2 2x - 6\sin 2x \sin 6x + \sin^2 6x \, dx\\ &= 2^{-10} \int \frac{9}{2}(1 - \cos 4x) + 3(\cos 8x - \cos 4x) + \frac{1}{2}(1 - \cos 12 x) \, dx\\ &= 2^{-10} \left(5x - \frac{9}{8} \sin 4x + \frac{3}{8} \sin 8x - \frac{3}{4} \sin 4x - \frac{1}{24} \sin 12x\right)\\ &= 2^{-10} \left(5x - \frac{15}{8} \sin 4x+ \frac{3}{8} \sin 8x - \frac{1}{24}\sin 12x \right) + c \end{align*}$

braintic · Feb 9, 2015

Re: MX2 2015 Integration Marathon

seanieg89 said:
extraction of a root

That reminds me ... time to see a dentist.

FrankXie · Feb 11, 2015

Re: MX2 2015 Integration Marathon

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

VBN2470 · Feb 11, 2015

Re: MX2 2015 Integration Marathon

FrankXie said:
$\int\frac{x^2-1}{x^4+1}dx$

$Express the denominator of the integrand as a product of two quadratic factors (by completing the square), then use partial fractions to get to the standard $\frac{f'(x)}{f(x)}$ form to get $\frac{1}{2\sqrt2}\ln\frac{x^2-\sqrt2x+1}{x^2+\sqrt2x+1}+C.$$

nightweaver066 · Feb 11, 2015

Re: MX2 2015 Integration Marathon

FrankXie said:
$\int\frac{x^2-1}{x^4+1}dx$

$= \int \frac{1-\frac{1}{x^2}}{\left(x+\frac{1}{x}\right)^2 - 2} \, dx$

$= \int \frac{1}{u^2 - 2} \, du \qquad \text{ by letting } u = x + \frac{1}{x}$

etc

FrankXie · Feb 11, 2015

Re: MX2 2015 Integration Marathon

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

nightweaver066 · Feb 11, 2015

Re: MX2 2015 Integration Marathon

FrankXie said:
$\int\frac{x^3}{\sqrt{1+2x-x^2}}dx$

$= \int \frac{x^3}{\sqrt{2 - (x - 1)^2}} \, dx$

$= \int \frac{(\sqrt{2}\sin \theta + 1)^3}{\sqrt{2}\cos\theta} \times \sqrt{2} \cos \theta \, d\theta \qquad \text{ using } x - 1 = \sqrt{2}\sin \theta$

$= \int (\sqrt{2}\sin \theta + 1)^3 \, d\theta$

$= \int 2\sqrt{2}\sin^3\theta + 6\sin^2\theta + 3\sqrt{2}\sin\theta + 1 \, d\theta$

$\text{Since } \sin3\theta = 3\sin\theta - 4\sin^3\theta.....$

etc.

soz should stop answering. far too bored lul

FrankXie · Feb 12, 2015

Re: MX2 2015 Integration Marathon

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

InteGrand · Mar 25, 2015

Re: MX2 2015 Integration Marathon

FrankXie said:
$\int\frac{dx}{x^3\sqrt{1+2x-x^2}}$

Firstly, we note that $\int \sqrt{t^2 - a^2}\text{ d}t=\frac{1}{2}\left[t\sqrt{t^2 - a^2}-a^2\ln(t+\sqrt{t^2 - a^2})\right]+c$ (we will use this later; this integral can be found, for example, by using integration by parts: integrate the 1 and differentiate the $\sqrt{t^2 - a^2}$ ).

Let $I = \int\frac{\text{d}x}{x^3\sqrt{1+2x-x^2}}, x^3\sqrt{1+2x-x^2} \neq 0$ .

Then $I=\int\frac{\text{d}x}{x^3\sqrt{x^2}\sqrt{ \frac{1}{x^2}+\frac{2}{x}-1}}$ .

For $x>0$ , $\sqrt{x^2} =x$ , so for $x>0$ , we have
$I=\int\frac{\text{d}x}{x^4 \sqrt{\frac{1}{x^2}+\frac{2}{x}-1}}$

$=\int\frac{\frac{1}{x^2}}{x^2 \sqrt{\frac{1}{x^2}+\frac{2}{x}-1}}\text{ d}x$ .

Let $u=\frac{1}{x}$ . Then $-\text{d}u=\frac{1}{x^2}\text{ d}x$ and $x^2 = \frac{1}{u^2}$ , so we get

$-I=\int\frac{\text{d}u}{\frac{1}{u^2} \sqrt{u^2 + 2u -1}}=\int\frac{u^2 }{\sqrt{u^2 + 2u -1}}\text{ d}u$ .

Then

$-I=\int\frac{u^2 +2u -1}{\sqrt{u^2 + 2u -1}}\text{ d}u - \int\frac{2u -1}{\sqrt{u^2 + 2u -1}}\text{ d}u$

$=\int \sqrt{u^2 + 2u -1} \text{ d}u - \int\frac{2u +2}{\sqrt{u^2 + 2u -1}} \text{ d}u + \int\frac{3}{\sqrt{u^2 + 2u -1}}\text{ d}u$

$=\int \sqrt{(u+1)^2 -2} \text{ d}u - 2\sqrt{u^2 + 2u -1} + \int\frac{3}{\sqrt{(u+1)^2 -2}}\text{ d}u$

$= \frac{1}{2}\left[(u+1)\sqrt{u^2 + 2u -1}-2\ln(u+1+\sqrt{u^2 + 2u -1})\right]- 2\sqrt{u^2 + 2u -1}+3 \ln (u+1+\sqrt{u^2+2u-1})+c$

$=\frac{1}{2}((u+1)\sqrt{u^2 + 2u -1})- 2\sqrt{u^2 + 2u -1}+2 \ln (u+1+\sqrt{u^2+2u-1})+c$

$=\frac{1}{2}\left [ \left(\frac{1}{x}+1 \right)\sqrt{\frac{1}{x^2} + \frac{2}{x} -1} \right ]- 2\sqrt{\frac{1}{x^2} + \frac{2}{x} -1}+2 \ln \left(\frac{1}{x}+1+\sqrt{\frac{1}{x^2} + \frac{2}{x} -1}\right)+c$

$\Rightarrow I = -\frac{1}{2}\left [ \left(\frac{1}{x}+1 \right)\sqrt{\frac{1}{x^2} + \frac{2}{x} -1} \right ]+ 2\sqrt{\frac{1}{x^2} + \frac{2}{x} -1}-2 \ln \left(\frac{1}{x}+1+\sqrt{\frac{1}{x^2} + \frac{2}{x} -1}\right)+c$ , for some constant c.

For $x<0$ , we would get the negative of this, since at the start, we would have had $\sqrt{x^2} = -x$ .

So overall, we have, for $x^3\sqrt{1+2x-x^2} \neq 0$ :

$I = \frac{|x|}{x}\left \{-\frac{1}{2}\left [ \left(\frac{1}{x}+1 \right)\sqrt{\frac{1}{x^2} + \frac{2}{x} -1} \right ]+ 2\sqrt{\frac{1}{x^2} + \frac{2}{x} -1}-2 \ln \left(\frac{1}{x}+1+\sqrt{\frac{1}{x^2} + \frac{2}{x} -1}\right)\right\}+c$

swagswagyoloswag · Mar 25, 2015

Re: MX2 2015 Integration Marathon

turntaker said:
Nice!! Well done

I tried doing this by letting t = tanx and t^2 = tan^(2)x which you can then find sin^2x in terms of t. I got a logarithm function

Ekman · Mar 29, 2015

Re: MX2 2015 Integration Marathon

Next Question:

$\int_{0}^{100\pi} \sqrt{1-\cos(2x)} dx$

swagswagyoloswag · Mar 30, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
Next Question:

$\int_{0}^{100\pi} \sqrt{1-\cos(2x)} dx$

$\int_{0}^{100\pi} \sqrt{cos^{2}(x) + sin^{2}(x) - cos^{2}(x) + sin^{2}(x)} dx \\ \int_{0}^{100\pi} \sqrt{{2}} \sqrt{sin^{2}(x)} dx \\ int_{0}^{100\pi} \sqrt{{2}} \|sin(x)| \\ [SIZE=7]$Now, the graph of \|sin(x)| will repeat every pi/2 interval as it is a reflection of the negative y values across the x-axis. For every interval of the area[/SIZE] under \|sinx|, there are 100pi/{pi/2} = 200 pieces.$ \\ $therefore, the integral can be represented as 200*\\sqrt{{2}} \int_{0}^{\pi/2} \sin(x) dx \\ = 200\sqrt{{2}} * (1) = 200\sqrt{{2}}$$

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