# HSC 2018-2019 MX2 Integration Marathon (1 Viewer)

##### -insert title here-
Re: HSC 2018 MX2 Integration Marathon

That being said is there a way of doing it using only 4U methods
not without assuming the standard series expansion of e

#### leehuan

##### Well-Known Member
Re: HSC 2018 MX2 Integration Marathon

not without assuming the standard series expansion of e
Which isn't even 4U.

Savage,

#### sharky564

##### Member
Re: HSC 2018 MX2 Integration Marathon

Which isn't even 4U.

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

##### -insert title here-
Re: HSC 2018 MX2 Integration Marathon

We can prove series expansion of $e$ using 4U maths.
using a highly unmotivated proof, sure.... xd

##### -insert title here-
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

##### Member
Re: HSC 2018 MX2 Integration Marathon

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

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#### fan96

##### 617 pages
Re: HSC 2018 MX2 Integration Marathon

\bg_white \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}

\bg_white \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:

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

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

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

So,

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

Substituting, we get

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

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

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

So we have

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

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#### jathu123

##### Active Member
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

##### Member
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

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##### -insert title here-
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

##### -insert title here-
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)
may as well just jump ahead and multiply by $\bg_white \frac{\sqrt{x} + \sqrt{x+1}}{\sqrt{x} + \sqrt{x+
1}}$

#### supR

##### Active Member
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
Yeah I'm seeing that and it's really confusing

Sent from my iPhone using Tapatalk

#### HeroWise

##### Active Member
Re: HSC 2018 MX2 Integration Marathon

is it 13/15 pi u^2 ??

#### fan96

##### 617 pages
Re: HSC 2018 MX2 Integration Marathon

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

\bg_white \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}

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#### fan96

##### 617 pages
Re: HSC 2018 MX2 Integration Marathon

Can someone give me a hint for this question?

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

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#### fan96

##### 617 pages
Re: HSC 2018 MX2 Integration Marathon

These are nice practice:

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

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

#### leehuan

##### Well-Known Member
Re: HSC 2018 MX2 Integration Marathon

These are nice practice:

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

\bg_white \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*}

\bg_white \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

##### Active Member
Re: HSC 2018 MX2 Integration Marathon

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

#### leehuan

##### Well-Known Member
Re: HSC 2018 MX2 Integration Marathon

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

\bg_white \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*}

\bg_white \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)