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

juantheron · Oct 11, 2014

Re: MX2 Integration Marathon

$Calculation \; of \; Integral \; $\displaystyle \int\frac{\sin^2(nx)}{\sin^2 (x)}dx$\;, where \; $n\in \mathbb{N}$

Carrotsticks · Oct 11, 2014

Re: MX2 Integration Marathon

juantheron said:
$$Calculate\; the \; Integrals$

$(a)\;\; \displaystyle \int_{0}^{\frac{\pi}{2}}\sin^{n}(x)\cdot \sin (nx)dx$\;\;\;\;\;\;(b)\;\; \int_{0}^{\frac{\pi}{2}}\cos^{n}(x)\cdot \cos (nx)dx$

$Where \; $n\in \mathbb{N}$

I'll just do the right one because the left one works pretty much the same way.

Notice that the integral satisfies the recurrence $I_n = \frac{1}{2} I_{n-1}$ and the problem comes along from there.

Sy123 · Nov 1, 2014

Re: MX2 Integration Marathon

juantheron said:
$Calculation \; of \; Integral \; $\displaystyle \int\frac{\sin^2(nx)}{\sin^2 (x)}dx$\;, where \; $n\in \mathbb{N}$

Sy123 · Nov 1, 2014

Re: MX2 Integration Marathon

$\\ \int_0^{\pi /4} \frac{dx}{1+ \sin^2 x}$

integral95 · Nov 1, 2014

Re: MX2 Integration Marathon

Sy123 said:
$\\ \int_0^{\pi /4} \frac{dx}{1+ \sin^2 x}$

$\int_0^{\pi /4} \frac{dx}{1+ \sin^2 x} = \int_0^{\pi /4} \frac{sec^2xdx}{sec^2x +tan^2x} $ \ \ u = tan x $\\ \\ \\ \Rightarrow \ \frac{1}{2}\int_0^{1}\frac{du}{u^2+\frac{1}{2}} = \frac{\sqrt{2}}{2} tan^{-1}\sqrt{2}$

Sy123 · Nov 2, 2014

Re: MX2 Integration Marathon

$\\ \int_0^{\pi /2} \frac{dx}{\sin^4x + \cos^4x}$

FrankXie · Nov 10, 2014

Re: MX2 Integration Marathon

Sy123 said:
$\\ \int_0^{\pi /2} \frac{dx}{\sin^4x + \cos^4x}$

My method--which I reckon kind of ugly--is:
After twice applications of double angle formula, convert the integrand into a fraction of $\cos4x$. Then use the $t$-formula. Care must be taken for tangent function has infinity discontinuity at $\frac{\pi}{2}. So I split the interval into two parts. Finally I got my answer $\frac{\pi}{\sqrt2}$.

Sy123 · Nov 10, 2014

Re: MX2 Integration Marathon

FrankXie said:
My method--which I reckon kind of ugly--is:
After twice applications of double angle formula, convert the integrand into a fraction of $\cos4x$. Then use the $t$-formula. Care must be taken for tangent function has infinity discontinuity at $\frac{\pi}{2}. So I split the interval into two parts. Finally I got my answer $\frac{\pi}{\sqrt2}$.

You can do LaTeX by surrounding the tex in $and$ tags, no need to put $ signs like mathstackexchange

I would just use this:

$\\ 1 = (\sin^2 x + \cos^2 x)^2 = \sin^4 x + \cos^4 x + 2\sin^2 x \cos^2 x \\ \\ \Rightarrow \ \frac{1}{\sin^4 x + \cos^4 x} = \frac{1}{1 - \frac{1}{2} \sin^2 2x} \\ \\ = \frac{2}{2 - \sin^2 2x} = \frac{2\sec^2x}{2\sec^2 2x - \tan^2 2x} = \frac{2 \sec^2 2x}{\tan^2 2x + 2} \\ \\ u = \tan 2x \\ du = 2\sec^2 2x$

$\\ $And you need to split the integral by$ \ \frac{\pi}{4} \ $so the substitution is valid (monotone)$$

FrankXie · Nov 10, 2014

Re: MX2 Integration Marathon

Sy123 said:
You can do LaTeX by surrounding the tex in $and$ tags, no need to put $ signs like mathstackexchange

I would just use this:

$\\ 1 = (\sin^2 x + \cos^2 x)^2 = \sin^4 x + \cos^4 x + 2\sin^2 x \cos^2 x \\ \\ \Rightarrow \ \frac{1}{\sin^4 x + \cos^4 x} = \frac{1}{1 - \frac{1}{2} \sin^2 2x} \\ \\ = \frac{2}{2 - \sin^2 2x} = \frac{2\sec^2x}{2\sec^2 2x - \tan^2 2x} = \frac{2 \sec^2 2x}{\tan^2 2x + 2} \\ \\ u = \tan 2x \\ du = 2\sec^2 2x$

$\\ $And you need to split the integral by$ \ \frac{\pi}{4} \ $so the substitution is valid (monotone)$$

Yup, nearly the same as my working. as a test of typseting LaTex, I got $\frac{\pi}{\sqrt2}$ what was your answer

Sy123 · Nov 10, 2014

Re: MX2 Integration Marathon

FrankXie said:
Yup, nearly the same as my working. as a test of typseting LaTex, I got $\frac{\pi}{\sqrt2}$ what was your answer

Yea its something like that
With integration I just try to figure out how to do the integration but I don't really do the computing lol

e.g. "oh yea this integral you just do this substitution then alter that then partial fractions and compute"

Sinyeta · Nov 10, 2014

Re: MX2 Integration Marathon

hello there

http://community.boredofstudies.org/3/non-school/330782/bos-boring-after-hsc.html#post6797283

FrankXie · Nov 11, 2014

Re: MX2 Integration Marathon

Try this one
$\int(2x^2-1)e^{-x^2}dx$

dan964 · Nov 11, 2014

Re: MX2 Integration Marathon

FrankXie said:
Try this one
$\int(2x^2-1)e^{-x^2}dx$

FrankXie · Nov 11, 2014

Re: MX2 Integration Marathon

$\int\frac{x^4}{x^6-1}dx$

Sy123 · Nov 11, 2014

Re: MX2 Integration Marathon

FrankXie said:
Try this one
$\int(2x^2-1)e^{-x^2}dx$

Can that be found in terms of elementary functions?

seanieg89 · Nov 11, 2014

Re: MX2 Integration Marathon

Sy123 said:
Can that be found in terms of elementary functions?

Yep

$-xe^{-x^2}$

by inspection/product rule.

For those, its pretty obvious the same exponential has to be in there because we cant obtain that as the derivative of something else. Unless its super hard, you can often guess it from there.

dan964 · Nov 11, 2014

Re: MX2 Integration Marathon

seanieg89 said:
Yep

$-xe^{-x^2}$

by inspection/product rule.

For those, its pretty obvious the same exponential has to be in there because we cant obtain that as the derivative of something else. Unless its super hard, you can often guess it from there.

yes it can, I did it earlier.

seanieg89 · Nov 11, 2014

Re: MX2 Integration Marathon

dan964 said:
yes it can, I did it earlier.

I do realise that this particular question can be done, seeing as I literally did it in the post you are quoting.

If you think you have found an elementary primitive of $e^{-x^2}$ on the other hand (which is what I am claiming is impossible when I say we can't obtain it as the derivative of something else), I would be willing to bet a large sum of money that you are wrong.

FrankXie · Nov 11, 2014

Re: MX2 Integration Marathon

Dont forget to try this one

FrankXie said:
$\int\frac{x^4}{x^6-1}dx$

seanieg89 · Nov 11, 2014

Re: MX2 Integration Marathon

FrankXie said:
Dont forget to try this one

I didn't forget, I just didn't bother. Any rational function can just be done by partial fractions, which isn't terribly exciting.

$I=\int \frac{1}{2(x^2-1)}+\frac{1}{4(x^2+x+1)}+\frac{1}{4(x^2-x+1)}\, dx$

$=\frac{1}{4}\log(\frac{1-x}{1+x})+\frac{1}{2 \sqrt{3}}(\arctan(\frac{2x+1}{ \sqrt{3}})+\arctan(\frac{2x-1}{ \sqrt{3}}))+C.$

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