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

aDimitri · May 27, 2014

Re: MX2 Integration Marathon

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

not sure if i got this right, but i got

$\\ \frac{1}{6}\left({-\sqrt{2}\tan^{-1}{\left (\frac{1}{2}\right )}} - \ln{\left | \frac{\sqrt{2} + 2}{2}\right |} + \ln{\left | \frac{\sqrt{2} - 2}{2}\right |} \right )$

i substituted u = sinx then did partial fractions with $\frac{1}{(1-u)(1+u)(2+u^2)}$

Sy123 · May 27, 2014

Re: MX2 Integration Marathon

aDimitri said:
not sure if i got this right, but i got

$\\ \frac{1}{6}\left({-\sqrt{2}\tan^{-1}{\left (\frac{1}{2}\right )}} - \ln{\left | \frac{\sqrt{2} + 2}{2}\right |} + \ln{\left | \frac{\sqrt{2} - 2}{2}\right |} \right )$

i substituted u = sinx then did partial fractions with $\frac{1}{(1-u)(1+u)(2+u^2)}$

That looks like a different form to what I had but correct, well done, all I did was:

$\frac{\sec x}{3 - \cos^2 x} = \frac{1}{\cos x (3-\cos^2x)} = \frac{1}{3} \left(\frac{1}{\cos x} + \frac{\cos x}{2 + \sin^2 x} \right)$

Which is easily integratable

aDimitri · May 27, 2014

Re: MX2 Integration Marathon

Ah right, that makes a lot of sense. I got it to that stage but I always forget that partial fractions is an option with trig

aDimitri · Jun 2, 2014

Re: MX2 Integration Marathon

$\text {Prove, without the use of complex numbers that} \\ \\ \int_0^{2\pi} {\ln{\left (\frac {\left (1+sinx \right )^{1+cosx}}{1+cosx}\right )}}dx = 0$

aDimitri · Jun 2, 2014

Re: MX2 Integration Marathon

$\int{\left (\frac {1}{lnx} + ln (ln (x))\right ) dx$

seanieg89 · Jun 3, 2014

Re: MX2 Integration Marathon

A function defined on the reals is said to be T-periodic if f(x+T)=f(x) for all x. (T is a positive real number).

Show that the integral of a continuous T-periodic function over ANY interval of length T is the same.

aDimitri · Jun 3, 2014

Re: MX2 Integration Marathon

ok so what we are trying to prove is that
$\int_0^T{f(x)dx} = \int_{(k-1)T}^{kT}{f(x)dx}$
i'm not sure how valid my method is for this but

$RHS = I = \int_{(k-1)T}^{kT}{f(x)dx} \\ \\ u = x - (k-1)T \\ du = dx \\ u_1 = T \\ u_2 = 0 \\ I = \int_0^T{f(u+(k-1)T)du} \\ \text{Now since f(x) = f(x+T) = f(x+2T) = ... = f(x+kT)} \\ I = \int_0^T{f(u)du} = LHS \\ \text{Hence, the integral of the periodic function across any length T is constant}$

seanieg89 · Jun 3, 2014

Re: MX2 Integration Marathon

aDimitri said:
ok so what we are trying to prove is that
$\int_0^T{f(x)dx} = \int_{(k-1)T}^{kT}{f(x)dx}$
i'm not sure how valid my method is for this but

$RHS = I = \int_{(k-1)T}^{kT}{f(x)dx} \\ \\ u = x - (k-1)T \\ du = dx \\ u_1 = T \\ u_2 = 0 \\ I = \int_0^T{f(u+(k-1)T)du} \\ \text{Now since f(x) = f(x+T) = f(x+2T) = ... = f(x+kT)} \\ I = \int_0^T{f(u)du} = LHS \\ \text{Hence, the integral of the periodic function across any length T is constant}$

Good work, but notice that your working depends on k being an integer. So you have only proven the claim for intervals like [7T,8T] or [-3T,-2T], but not for intervals like [1.5T,2.5T].

(The notation [a,b] refers to the closed interval a =< x =< b)

aDimitri · Jun 4, 2014

Re: MX2 Integration Marathon

Yeah I realised that after but am yet to solve this problem, will have a crack at it tomorrow morning at school

EDIT:
ok i think i've got it now!

$\text{RTP:} \; I = \int_a^{a+T}{f(x)dx} = \int_0^T{f(x)dx} \: \: , \: \: a \in \mathbb{R} \\ \equiv \int_a^{a+T}{f(x)dx} - \int_0^T{f(x)dx} = 0 \\ \\ LHS = F(a+T) - F(a) - F(T) + F(0) \; (\text{Where F(x) is the antiderivative function of f(x)}) \\ \\ = F(a+T) - F(T) + F(0) - F(a) \\ \\ = \int_T^{a+T}{f(x)dx} - \int_0^a{f(x)dx} \\ \text{By substituting x = u + t, this reduces to} \: \int_0^a{f(u+T)du} - \int_0^a{f(x)dx} \\ = 0 \: \text{Since f(x) = f(x+T)} \\ = RHS$

Davo_01 · Jun 7, 2014

Re: MX2 Integration Marathon

aDimitri said:
$\int{\left (\frac {1}{lnx} + ln (ln (x))\right ) dx$

$\dfrac{d}{dx}xln(lnx)=\dfrac{1}{lnx}+ln(lnx)$

$\therefore \int{\left (\frac {1}{lnx} + ln (ln (x))\right ) dx =xln(ln(x))+c$

$Other than this method we can apply IBP to $ \int ln(lnx)dx $ which will cancel out the other integral.$

aDimitri · Jun 7, 2014

Re: MX2 Integration Marathon

Davo_01 said:
$\dfrac{d}{dx}xln(lnx)=\dfrac{1}{lnx}+ln(lnx)$

$\therefore \int{\left (\frac {1}{lnx} + ln (ln (x))\right ) dx =xln(ln(x))+c$

$Other than this method we can apply IBP to $ \int ln(lnx)dx $ which will cancel out the other integral.$

Yeah if you intuitively can see it's xln(lnx) then you're lucky, but what IBP did you use on ln(lnx)? That seems a lot faster than my method.

Davo_01 · Jun 7, 2014

Re: MX2 Integration Marathon

aDimitri said:
Yeah if you intuitively can see it's xln(lnx) then you're lucky, but what IBP did you use on ln(lnx)? That seems a lot faster than my method.

$$Using$\; u=ln(lnx) \rightarrow du=\dfrac{dx}{xln(x)}\; $and$\; dv=dx \rightarrow v=x$

$\therefore \int ln(lnx)dx=xln(lnx)-\int \dfrac{dx}{ln(x)}$

$\int\left(\dfrac{1}{ln(x)}+ ln(lnx)\right)dx=xln(lnx)+c$

Sy123 · Jun 8, 2014

Re: MX2 Integration Marathon

$\\ \ \int_0^{n\pi} |x\sin x| \ dx \ $where$ \ n \ $is an integer$$

dunjaaa · Jun 8, 2014

Re: MX2 Integration Marathon

Screen shot 2014-06-08 at 3.50.07 AM.png

Nice question!

Fade1233 · Jun 20, 2014

Re: MX2 Integration Marathon

Good question.

Can't use latex sorry.

integral95 · Jun 30, 2014

Re: MX2 Integration Marathon

For parts i and ii

$I1 = \int_{0}^{\frac{\pi}{4}}\frac{1-cos2x}{sin2x} \\ = \int_{0}^{\frac{\pi}{4}}\frac{2sin^22x}{sin2x} = \int_{0}^{\frac{\pi}{4}} tan x = ln(\sqrt{2}) \\ \\ \\ I^{2r+1} - I^{2r-1} = \int_{0}^{\frac{\pi}{4}} \frac{cos2(2r-1)x-cos2(2r+1)x)}{sin2x} \\ \int_{0}^{\frac{\pi}{4}} \frac{-2sin(x(2r-1)+x(2r+1))))sin(x(2r-1)-x(2r+1)))}{sin2x} \ \ \ $using part a)$ \\ = \int_{0}^{\frac{\pi}{4}}\frac{-2sin(4rx)sin(-2x)}{sin2x} = \int_{0}^{\frac{\pi}{4}}2sin(4rx) dx = \frac{1-cos(\pi r)}{2r} = \frac{1-(-1)^r)}{2r} \\$

Sy123 · Jul 1, 2014

Re: MX2 Integration Marathon

$\int (x-1) \sqrt[3]{x-2} \ dx$

aDimitri · Jul 1, 2014

Re: MX2 Integration Marathon

Sy123 said:
$\int (x-1) \sqrt[3]{x-2} \ dx$

$= \frac{3}{28}(x-2)^{\frac{4}{3}}(4x-1) + C$

dunjaaa · Jul 13, 2014

Re: MX2 Integration Marathon

Screen shot 2014-07-13 at 8.37.58 PM.png

integral95 · Jul 28, 2014

Re: MX2 Integration Marathon

$$Show that$ \ \ \int_{0}^{1}x^m(1-x)^n dx = \frac{m!n!}{(m+n+1)!}$

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

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