MX2 Integration Marathon (1 Viewer)

Paradoxica · Feb 11, 2018

Re: HSC 2018 MX2 Integration Marathon

leehuan said:
That being said is there a way of doing it using only 4U methods

not without assuming the standard series expansion of e

leehuan · Feb 11, 2018

Re: HSC 2018 MX2 Integration Marathon

Paradoxica said:
not without assuming the standard series expansion of e

Which isn't even 4U.

Savage,

sharky564 · Mar 18, 2018

Re: HSC 2018 MX2 Integration Marathon

leehuan said:
Which isn't even 4U.

Savage,

We can prove series expansion of $e$ using 4U maths.

Paradoxica · Mar 18, 2018

Re: HSC 2018 MX2 Integration Marathon

sharky564 said:
We can prove series expansion of $e$ using 4U maths.

using a highly unmotivated proof, sure.... xd

Paradoxica · Mar 18, 2018

Re: HSC 2018 MX2 Integration Marathon

but in the hsc such a question would not be likely to appear, all that work just for a simple integral....

mrbunton · Apr 2, 2018

Re: HSC 2018 MX2 Integration Marathon

only requires u substitution.
$\int \frac{1}{\sqrt{x^2+x}}$

fan96 · Apr 3, 2018

Re: HSC 2018 MX2 Integration Marathon

$\begin{aligned} &\int \frac{1}{\sqrt{x^2+x}} \,dx \\ &= \int \frac{1}{\sqrt{\left(x+\frac{1}{2}\right)^2 - \frac{1}{4}}}\,dx\end{aligned}$

$\begin{aligned}\mathrm{Let}\,\, x + \frac{1}{2} &= \frac{1}{2} \sec \theta \\ x &= \frac{1}{2}(\sec \theta - 1) \end{aligned}$

Justifying this substitution:

$x = (1/2)(\sec \theta - 1)$ has a solution for $\theta$ only when $x \geq 0$ and $x \leq -1$ .

The domain of the integrand is $x > 0$ and $x < -1$ .

$\theta$ does not exist only when $-1 < x < 0$ . But since this range is undefined in the integrand to begin with, the substitution is valid.

So,

$dx = \frac 1 2 \sec \theta \tan \theta\, d\theta$

Substituting, we get

$\begin{aligned}&\int \frac{\frac 1 2 \sec \theta \tan \theta}{\sqrt{\frac 1 4 \sec^2 \theta - \frac 1 4}} \,d\theta \\ &=\int \frac{\frac 1 2 \sec \theta \tan \theta}{\frac 1 2 \sqrt{\tan^2 \theta}} \,d\theta \\ &= \int \sec \theta \frac{ \tan \theta}{|\tan \theta|} \,d\theta \\ &= \frac{ \tan \theta}{|\tan \theta|} \int \sec \theta \,d\theta \\ &= \left(\frac{ \tan \theta}{|\tan \theta|} \right)\log |\sec \theta + \tan \theta| + C\end{aligned}$

Not sure if that second last step was valid though.

(edit: oops forgot to sub back in)

And $\sec \theta = 2x +1$ so

$\cos \theta = \frac{1}{2x + 1}$

and by considering the appropriate right angled triangle, $\tan \theta = 2 \sqrt{x^2+x}$ . Since that's always positive, the sign function on the left can be eliminated.

So we have

$\log|2x + 1 + 2\sqrt{x^2+x}| + C$

jathu123 · Apr 3, 2018

Re: HSC 2018 MX2 Integration Marathon

an easier way is to just use the substitution u = sqrt(x)
the integrand will be 2/sqrt(u^2+1), which is trivial to integrate (those usual log ones)

mrbunton · Apr 3, 2018

Re: HSC 2018 MX2 Integration Marathon

correct but the two tan signs are generally said to cancel each other out; although i understand it can vary from +-1. that being said if u look at example at the integral tanx/|tanx| it is a horizontal line being split up into positive and negative in pi/2 segments at a time and hence u should not be allowed to take it out theoretically (although area stays the same; definite integral get affected from my understanding)but discontinuous graphs in integration are probably beyond the scope of even ext2(when dealing with indefinite integrals btw; or i havent been taught it yet). Its easier just to cancel the tan out imo at this level.

im not completely sure but in the solution i got had the tan cancelled out

Paradoxica · Apr 4, 2018

Re: HSC 2018 MX2 Integration Marathon

is anyone else seeing a bunch of extra addition symbols lodged between every character or is that just me

Paradoxica · Apr 4, 2018

HSC 2018 MX2 Integration Marathon

jathu123 said:
an easier way is to just use the substitution u = sqrt(x)
the integrand will be 2/sqrt(u^2+1), which is trivial to integrate (those usual log ones)

may as well just jump ahead and multiply by $\frac{\sqrt{x} + \sqrt{x+1}}{\sqrt{x} + \sqrt{x+ 1}}$

supR · Apr 12, 2018

Re: HSC 2018 MX2 Integration Marathon

Paradoxica said:
is anyone else seeing a bunch of extra addition symbols lodged between every character or is that just me

Yeah I'm seeing that and it's really confusing

Sent from my iPhone using Tapatalk

CapitalSwine · Apr 12, 2018

Re: HSC 2018 MX2 Integration Marathon

How to do this question https://imgur.com/a/jrRra

HeroWise · Apr 12, 2018

Re: HSC 2018 MX2 Integration Marathon

is it 13/15 pi u^2 ??

fan96 · Apr 12, 2018

Re: HSC 2018 MX2 Integration Marathon

CapitalSwine said:
How to do this question https://imgur.com/a/jrRra

We can get the volume of the solid by taking the volume of $x = y^2-2$ from 0 to 1 (rotated about the y-axis) and then subtracting the volume of the cylinder inside, which has radius and height 1.

$\begin{aligned} V &= \pi \int_0^1 (y^2-2)^2\, dy - \pi(1)^2(1) \\ &= \pi\left[\frac 1 5 y^5 - \frac 4 3 y^3 + 4y\right]^1_0 - \pi \\ &= \pi \left(\frac 1 5 - \frac 4 3 + 4-1\right) \\ &= \left(\frac{28}{15} \pi \right)\mathrm{u^3}\end{aligned}$

fan96 · May 7, 2018

Re: HSC 2018 MX2 Integration Marathon

Can someone give me a hint for this question?

$$Show that$\,\, \frac{\pi}{12} < \int_0^{\frac{\pi}{4}} \sec^{\frac 5 2} x\,dx - \frac{1}{3}\left(2^{\frac 5 4}\right) < \frac{\pi}{12}\left(2^{\frac 1 4}\right).$

fan96 · May 25, 2018

Re: HSC 2018 MX2 Integration Marathon

These are nice practice:

$\int_0^{\frac\pi 6}\frac{x \sec^2x}{\tan x + \sqrt 3}\,dx = \frac{\pi}{6}\log\left(\frac{2\sqrt3}{3}\right)$

$\int_0^{1}\frac{\log(1+x)}{1+x^2}\,dx = \frac\pi8\log2$

leehuan · Jul 11, 2018

Re: HSC 2018 MX2 Integration Marathon

fan96 said:
These are nice practice:

$\int_0^{\frac\pi 6}\frac{x \sec^2x}{\tan x + \sqrt 3}\,dx = \frac{\pi}{6}\log\left(\frac{2\sqrt3}{3}\right)$

$\text{Let }I = \int_0^{\pi/6} \ln (\tan x + \sqrt{3} ) \,dx$

$\begin{align*} I &= \int_0^{\pi/6} \ln \left( \frac{ \frac{1} {\sqrt3} - \tan x }{1 + \frac{\tan x}{\sqrt3} } + \sqrt{3} \right)dx\\ &= \int_0^{\pi/6 } \ln \left( \frac{4}{\sqrt3 + \tan x} \right)dx\\ \therefore 2 I &= \int_0^{\pi / 6}\ln 4\, dx\\ I &= \frac\pi6 \ln 2\end{align*}$

$\begin{align*} \int_0^{\pi/6} \frac{x \sec^2 x}{\tan x + \sqrt3}\,dx &= x\ln (\tan x + \sqrt{3}) \bigg|_0^{\pi/6} - I\\ &= \frac\pi6 \ln \left( \frac4{\sqrt3} \right) - \frac\pi6 \ln 2\\ &= \frac\pi 6 \ln \left( \frac{2\sqrt3}{3} \right) \end{align*}$

The second one is done similarly but with sub instead of IBP

jathu123 · Jul 14, 2018

Re: HSC 2018 MX2 Integration Marathon

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\cos x}{e^{\sinh x}+1} $ d$x \qquad $where $ \sinh x = \frac{1}{2}\left ( e^x-e^{-x} \right )$

leehuan · Jul 14, 2018

Re: HSC 2018 MX2 Integration Marathon

jathu123 said:
$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\cos x}{e^{\sinh x}+1} $ d$x \qquad $where $ \sinh x = \frac{1}{2}\left ( e^x-e^{-x} \right )$

$$\noindent Take $f(x)$ to be any arbitrary odd function well defined for all $ -\frac\pi2 \leq x \leq \frac\pi2\\ $and let $ I = \int_{-\pi/2}^{\pi/2} \frac{\cos x}{e^{f(x)}+1}\,dx.$

$$\noindent Considering the substitution $u=-x,\\ \begin{align*}I&= \int_{-\pi/2}^{\pi/2} \frac{\cos (-u)}{e^{f(-u)} + 1}\, du\\ &= \int_{-\pi/2}^{\pi/2} \frac{\cos u}{e^{-f(u)}+1}du\\ &= \int_{-\pi/2}^{\pi/2} \frac{e^{f(u)}\cos u}{1+ e^{f(u)}}du\end{align*}$

$$\noindent Performing a dummy variable switch, we have$\\ \begin{align*}2I &= \int_{-\pi/2}^{\pi/2} \frac{(e^{f(x)}+1)\cos x}{e^{f(x)}+1}dx\\ I &= \int_0^{\pi/2} \cos x \, dx\\ &= 1\end{align*}$

Note that sinh(-x) = 2^-1.(exp(-x) - exp(x)) = -2^-1.(exp(x)-exp(-x)) = -sinh(x)

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