HSC 2015 MX2 Integration Marathon (archive) (2 Viewers)

Ekman · Oct 1, 2015

Re: MX2 2015 Integration Marathon

Paradoxica said:
$\int_{0}^{1}{\frac{dx}{(1+x^n)^2 \sqrt[n]{1+x^n}}}$

$$ Substitute: $ x= (\tan{\theta})^{\frac{2}{n}}$

$\therefore \int_{0}^{1}{\frac{dx}{(1+x^n)^2 \sqrt[n]{1+x^n}}} dx = \frac{2}{n} \int_{0}^{\frac{\pi}{4}} \frac{(\tan{\theta})^{\frac{2}{n}-1}}{(\sec{\theta})^{\frac{2}{n} + 2}} d\theta$

$= \frac{2}{n} \int_{0}^{\frac{\pi}{4}} \frac{(\sin{\theta})^{\frac{2}{n}}}{\tan{\theta} \sec^{2}{\theta}} d\theta = \frac{2}{n} \int_{0}^{\frac{\pi}{4}} \cos^{3}{\theta}(\sin{\theta})^{\frac{2}{n}-1} d\theta$

$= \frac{2}{n} \left[\frac{(\sin{\theta})^{\frac{2}{n}}}{\frac{2}{n}} - \frac{(\sin{\theta})^{\frac{2}{n} + 2}}{\frac{2}{n}(n+1)} \right]_{0}^{\frac{\pi}{4}} = \left(\frac{1}{2} \right)^{\frac{1}{n} + 1} \cdot \frac{2n+1}{n+1}$

Paradoxica · Oct 2, 2015

Re: MX2 2015 Integration Marathon

Woooo! Trials Complete! Time to celebrate with another integral.

$\int{\frac{1}{x}\sqrt{\frac{x+a}{x-a}}}dx$
$a$ is a constant.

InteGrand · Oct 2, 2015

Re: MX2 2015 Integration Marathon

Paradoxica said:
Woooo! Trials Complete! Time to celebrate with another integral.

$\int{\frac{1}{x}\sqrt{\frac{x+a}{x-a}}}dx$
$a$ is a constant.

$Let $I = \int {\frac{1}{x}\sqrt{\frac{x+a}{x-a}}}\text{ d}x, x \neq 0$$

$If $a=0$, $I=\int \frac{1}{x}\text{ d}x = \ln |x|+c$.$

$If $a\neq 0$, write $\frac{1}{x}\sqrt{\frac{x+a}{x-a}} =\frac{1}{x}\frac{\sqrt{x+a}}{\sqrt{x-a}}= \frac{x+a}{x\sqrt{x^2 - a^2}}$ (multiplying numerator and denominator by $\sqrt{x+a}$). So$

$I=\int \frac{x+a}{x\sqrt{x^2 - a^2}}\text{ d}x$

$I=\int \frac{a}{x\sqrt{x^2 - a^2}} \text{ d}x + \int \frac{x}{x\sqrt{x^2 - a^2}}\text{ d}x$

$I=J + \int \frac{1}{\sqrt{x^2 - a^2}}\text{ d}x $, where $J=\int \frac{a}{x\sqrt{x^2 - a^2}} \text{ d}x$

$$I=J + \ln \left | x+\sqrt{x^2 - a^2}\right |$.$

$To find $J$, sub. $x=a\sec t$, $t\in \left [ 0,\pi \right ] \backslash \left \{ \frac{\pi}{2} \right\}$. Then $\mathrm{d}x = a\sec t \tan t \text{ d}t$ and $x^2 - a^2 = a^2 \left(\sec^2 -1\right)=a^2 \tan^2 t$.$

$Hence $J=\int \frac{a}{a \sec t |a||\tan t|}\times a\sec t \tan t \text{ d}t$$

$$\Rightarrow J =\int \frac{\mathrm{sgn} (a)}{|\tan t|}\times \tan t \text{ d}t $ ($\mathrm{sgn}$ is the sign function)$

$\Rightarrow J =\int \mathrm{sgn}(a)\times \mathrm{sgn} (\tan t) \text{ d}t$

$Note that $\tan t > 0 \Longleftrightarrow 0 \leq t < \frac{\pi}{2} \Longleftrightarrow \mathrm{sgn}(\tan t) = 1$ iff $\mathrm{sgn} (x) = \mathrm{sgn} (a)$. Also $\tan t < 0 \Longleftrightarrow \frac{\pi}{2} < t \leq \pi \Longleftrightarrow \mathrm{sgn}(\tan t)=-1$ iff $\mathrm{sgn} (x) = -\mathrm{sgn} (a)$.$

$Since $\cos t = \frac{a}{x}$ and $0\leq t \leq \pi$, we have $t = \cos ^{-1} \left(\frac{a}{x} \right)$. Therefore, we have$

$$I=\begin{cases} \ln |x|+c_1 & \text{ if } a=0 \\ \ln \left | x+\sqrt{x^2 - a^2}\right | + \mathrm{sgn}(a)\cos^{-1} \left ( \frac{a}{x} \right )+c_2& \text{ if } \mathrm{sgn}(x)= \mathrm{sgn}(a) \text{ and }a\neq 0 \\ \ln \left | x+\sqrt{x^2 - a^2}\right | - \mathrm{sgn}(a)\cos^{-1} \left ( \frac{a}{x} \right )+c_3 & \text{ if } \mathrm{sgn}(x)= -\mathrm{sgn}(a) \text{ and }a\neq 0 \end{cases}$.$

$This simplifies to$

$$I=\begin{cases} \ln |x|+c_1 & \text{ if } a=0 \\ \ln \left | x+\sqrt{x^2 - a^2}\right | + \mathrm{sgn}(x)\cos^{-1} \left ( \frac{a}{x} \right )+c_2& \text{ if } a\neq 0 \end{cases}$.$

Paradoxica · Oct 2, 2015

Re: MX2 2015 Integration Marathon

InteGrand said:
$Let $I = \int {\frac{1}{x}\sqrt{\frac{x+a}{x-a}}}\text{ d}x, x \neq 0$$

$If $a=0$, $I=\int \frac{1}{x}\text{ d}x = \ln |x|+c$.$

$If $a\neq 0$, write $\frac{1}{x}\sqrt{\frac{x+a}{x-a}} =\frac{1}{x}\frac{\sqrt{x+a}}{\sqrt{x-a}}= \frac{x+a}{x\sqrt{x^2 - a^2}}$ (multiplying numerator and denominator by $\sqrt{x+a}$). So$

$I=\int \frac{x+a}{x\sqrt{x^2 - a^2}}\text{ d}x$

$I=\int \frac{a}{x\sqrt{x^2 - a^2}} \text{ d}x + \int \frac{x}{x\sqrt{x^2 - a^2}}\text{ d}x$

$I=J + \int \frac{1}{\sqrt{x^2 - a^2}}\text{ d}x $, where $J=\int \frac{a}{x\sqrt{x^2 - a^2}} \text{ d}x$

$$I=J + \ln \left | x+\sqrt{x^2 - a^2}\right |$.$

$To find $J$, sub. $x=a\sec t$, $t\in \left [ 0,\pi \right ] \backslash \left \{ \frac{\pi}{2} \right\}$. Then $\mathrm{d}x = a\sec t \tan t \text{ d}t$ and $x^2 - a^2 = a^2 \left(\sec^2 -1\right)=a^2 \tan^2 t$.$

$Hence $J=\int \frac{a}{a \sec t |a||\tan t|}\times a\sec t \tan t \text{ d}t$$

$$\Rightarrow J =\int \frac{\mathrm{sgn} (a)}{|\tan t|}\times \tan t \text{ d}t $ ($\mathrm{sgn}$ is the sign function)$

$\Rightarrow J =\int \mathrm{sgn}(a)\times \mathrm{sgn} (\tan t) \text{ d}t$

$Note that $\tan t > 0 \Longleftrightarrow 0 \leq t < \frac{\pi}{2} \Longleftrightarrow \mathrm{sgn}(\tan t) = 1$ iff $\mathrm{sgn} (x) = \mathrm{sgn} (a)$. Also $\tan t < 0 \Longleftrightarrow \frac{\pi}{2} < t \leq \pi \Longleftrightarrow \mathrm{sgn}(\tan t)=-1$ iff $\mathrm{sgn} (x) = -\mathrm{sgn} (a)$.$

$Since $\cos t = \frac{a}{x}$ and $0\leq t \leq \pi$, we have $t = \cos ^{-1} \left(\frac{a}{x} \right)$. Therefore, we have$

$$I=\begin{cases} \ln |x|+c_1 & \text{ if } a=0 \\ \ln \left | x+\sqrt{x^2 - a^2}\right | + \mathrm{sgn}(a)\cos^{-1} \left ( \frac{a}{x} \right )+c_2& \text{ if } \mathrm{sgn}(x)= \mathrm{sgn}(a) \text{ and }a\neq 0 \\ \ln \left | x+\sqrt{x^2 - a^2}\right | - \mathrm{sgn}(a)\cos^{-1} \left ( \frac{a}{x} \right )+c_3 & \text{ if } \mathrm{sgn}(x)= -\mathrm{sgn}(a) \text{ and }a\neq 0 \end{cases}$.$

$This simplifies to$

$$I=\begin{cases} \ln |x|+c_1 & \text{ if } a=0 \\ \ln \left | x+\sqrt{x^2 - a^2}\right | + \mathrm{sgn}(x)\cos^{-1} \left ( \frac{a}{x} \right )+c_2& \text{ if } a\neq 0 \end{cases}$.$

um... my solution says it should be $\ln(x+\sqrt{x^2 - a^2})-\tan^{-1}(\frac{a}{\sqrt{x^2 - a^2}})$

Differentiating this in Wolfram Alpha also verifies this fact.

InteGrand · Oct 2, 2015

Re: MX2 2015 Integration Marathon

Paradoxica said:
um... my solution says it should be $\ln(x+\sqrt{x^2 - a^2})-\tan^{-1}(\frac{a}{\sqrt{x^2 - a^2}})$

Differentiating this in Wolfram Alpha also verifies this fact.

$The answers are equivalent. This is because $\cos ^{-1} \frac{a}{x}=\tan ^{-1} \frac{\sqrt{x^2 - a^2}}{a} $ (for one branch, if $a>0$, namely the branch where $x>a>0$ $(\mathrm{sgn}(x)=1)$), and $\tan ^{-1} \frac{ \sqrt{x^2 - a^2} }{a}$ differs from $-\tan^{-1} \frac{a}{\sqrt{x^2 - a^2}}$ by a constant (as primitives do) (both branches should end up differing by a constant if $\mathrm{sgn}(x)$ is used). I tried differentiating my answer (admittedly by hand rather than Wolfram Alpha) and it worked.$

Paradoxica · Oct 2, 2015

Re: MX2 2015 Integration Marathon

$\int{\sqrt{\frac{x+k}{x}}}dx$

$\int{\sqrt{\frac{x}{x+k}}}dx$

In both questions, k is a real, non-zero constant.

kawaiipotato · Oct 2, 2015

Re: MX2 2015 Integration Marathon

$= \int \frac{x+k}{\sqrt{x^2 + kx}}dx$
$= \sqrt{x^2 + kx} + k \int \frac{1}{\sqrt{(x+ \frac{k}{2})^2 - \frac{k^2 }{4} } } dx$
$= \sqrt{x^2 + kx} + k \ln \left ( x + \frac{k}{2} + \sqrt{x^2 + kx} \right ) + $ C $$
That took so long to type on mobile

porcupinetree · Oct 2, 2015

Re: MX2 2015 Integration Marathon

Paradoxica said:
$\int{\sqrt{\frac{x+k}{x}}}dx$

$\int{\sqrt{\frac{x}{x+k}}}dx$

In both questions, k is a real, non-zero constant.

I have a question for you Paradoxica, it's easy but oh well:

$\int \sqrt{\frac{x-1}{17-x}}\ dx$

porcupinetree · Oct 2, 2015

Re: MX2 2015 Integration Marathon

porcupinetree said:
I have a question for you Paradoxica, it's easy but oh well:

$\int \sqrt{\frac{x-1}{17-x}}\ dx$

Here's what I did:

Edit: I don't know if there's any mistakes, there might be

Paradoxica · Oct 2, 2015

Re: MX2 2015 Integration Marathon

porcupinetree said:
Here's what I did:

Edit: I don't know if there's any mistakes, there might be

That's what I would have done.

$-\sqrt{-x^2+18x-17} + \sin^{-1}{\frac{x-9}{8}}+C$

porcupinetree · Oct 2, 2015

Re: MX2 2015 Integration Marathon

Here's another one inspired from the one in the 4U paper today - it's a little bit different though

$\int \frac{\sqrt{cosx}\ sinx}{1 -sec^{2}x}\ dx$

InteGrand · Oct 2, 2015

Re: MX2 2015 Integration Marathon

porcupinetree said:
Here's another one inspired from the one in the 4U paper today - it's a little bit different though

$\int \frac{\sqrt{cosx}\ sinx}{1 -sec^{2}x}\ dx$

$Let $u=\sqrt{\cos x}$, $0<x<\frac{\pi}{2}$, so $u^2 = \cos x \Rightarrow \sec ^2 x = \frac{1}{u^4}$. Also, $x=\cos ^{-1} u^2 \Rightarrow \mathrm{d}x = \frac{-2u}{\sqrt{1-u^4}}\text{ d}u$. Furthermore, $\sin x = \sqrt{1-\cos^2 x}=\sqrt{1-u^4}$ (as $\sin x>0$ in this domain). Therefore, the integral, $I$, is given by$

$I = \int \frac{u\cdot \sqrt{1-u^4}}{1 - \frac{1}{u^4}}\times \frac{-2}{\sqrt{1-u^4}}\text{ d}u$

$= -\int \frac{2u}{1 - \frac{1}{u^4}}\text{ d}u$

$= -\int \frac{2u^5}{u^4 - 1}\text{ d}u$

$= -2\int \left(u + \frac{u}{u^4 -1} \right) \text{ d}u$

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

$= -u^2+ \frac{1}{2}\ln \left (\frac{1+u^2}{1-u^2} \right)+C$

$$ = -\cos x+ \frac{1}{2}\ln \left (\frac{1+\cos x}{1-\cos x} \right)+C$.$

Paradoxica · Oct 2, 2015

Re: MX2 2015 Integration Marathon

InteGrand said:
$Let $u=\sqrt{\cos x}$, $0<x<\frac{\pi}{2}$, so $u^2 = \cos x \Rightarrow \sec ^2 x = \frac{1}{u^4}$. Also, $x=\cos ^{-1} u^2 \Rightarrow \mathrm{d}x = \frac{-2u}{\sqrt{1-u^4}}\text{ d}u$. Furthermore, $\sin x = \sqrt{1-\cos^2 x}=\sqrt{1-u^4}$ (as $\sin x>0$ in this domain). Therefore, the integral, $I$, is given by$

$I = \int \frac{u\cdot \sqrt{1-u^4}}{1 - \frac{1}{u^4}}\times \frac{-2}{\sqrt{1-u^4}}\text{ d}u$

$= -\int \frac{2u}{1 - \frac{1}{u^4}}\text{ d}u$

$= -\int \frac{2u^5}{u^4 - 1}\text{ d}u$

$= -2\int \left(u + \frac{u}{u^4 -1} \right) \text{ d}u$

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

$= -u^2+ \frac{1}{2}\ln \left (\frac{1+u^2}{1-u^2} \right)+C$

$$ = -\cos x+ \frac{1}{2}\ln \left (\frac{1+\cos x}{1-\cos x} \right)+C$.$

Your answer doesn't even make intuitive sense. There should be a good number of surdic trigonometry terms in it, since that was present in the beginning.

Differentiating our answers in W|A shows yours to be incorrect and mine to be correct. But don't worry, I also went down the line of missing out that extra u factor in the numerator. I stopped before I went too far though.

InteGrand · Oct 2, 2015

Re: MX2 2015 Integration Marathon

Paradoxica said:
Your answer doesn't even make intuitive sense. There should be a good number of surdic trigonometry terms in it, since that was present in the beginning.

Differentiating our answers in W|A shows yours to be incorrect and mine to be correct. But don't worry, I also went down the line of missing out that extra u factor in the numerator. I stopped before I went too far though.

Lol my bad, somehow missed including the u when replacing the dx, even though I had it when I found dx... X_X

Paradoxica · Oct 2, 2015

Re: MX2 2015 Integration Marathon

$\int{\sqrt{\frac{x+k}{x}}}dx$

...k is a real, non-zero constant.

Bump.

InteGrand · Oct 2, 2015

Re: MX2 2015 Integration Marathon

Paradoxica said:
$\int{\sqrt{\frac{x+k}{x}}}dx$

...k is a real, non-zero constant.

Bump.

Done above by kawaiipotato. The other integral is similar.

Drsoccerball · Oct 2, 2015

Re: MX2 2015 Integration Marathon

$\int{\frac{e^{x^3}}{x} \cdot (6x^3 lnx +2)}dx$

Paradoxica · Oct 2, 2015

Re: MX2 2015 Integration Marathon

InteGrand said:
Done above by kawaiipotato. The other integral is similar.

My answer says otherwise, but ok....

Drsoccerball said:
$\int{\frac{e^{x^3}}{x} \cdot (6x^3 lnx +2)}dx$

InteGrand · Oct 3, 2015

Re: MX2 2015 Integration Marathon

Paradoxica said:
My answer says otherwise, but ok....

$As in similar to find, I wasn't commenting on whether the answers were similar. If $I_1 = \int \sqrt{\frac{x+k}{x}}\text{ d}x = \int \frac{x+k}{\sqrt{x^2 + kx}}\text{ d}x$ and $I_2 = \int \sqrt{\frac{x}{x+k}}\text{ d}x = \int \frac{x}{\sqrt{x^2 +kx}}$, then $I_1 = I_2 + k\int \frac{1}{\sqrt{x^2 + kx}}\text{ d}x$, so the two are obviously related. And both involved multiplying top and bottom by the suitable radical at the start, so similar process.$

Paradoxica · Oct 3, 2015

Re: MX2 2015 Integration Marathon

$\int{\frac{(a\sin{x})^2 + (b\cos{x})^2}{(a^2 \sin{x})^2+(b^2\cos{x})^2}}dx$

$a,b$ are real non-zero constants

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