MX2 Integration Marathon 2021 (1 Viewer)

stupid_girl

Active Member
I made a typo in Wolfram earlier. It does give a nice answer.
$\bg_white \int_{-1}^{1} \frac{(x-1)e^x}{x^2+e^{2x}}$

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vernburn

Active Member
Not sure why Wolfram can't give a nice answer.
$\bg_white \int_{-1}^{1} \frac{(x-1)e^x}{x^2+e^{2x}}$
Let $\bg_white u = -x \implies du = -dx$:
$\bg_white I =\int_{-1}^{1} \frac{(x-1)e^x}{x^2+e^{2x}}dx$
$\bg_white =-\int_{1}^{-1} \frac{(-u-1)e^{-u}}{u^2+e^{-2u}}du$
$\bg_white =\int_{1}^{-1}\frac{(u+1)e^u}{e^{2u}u^2+1}du$
Notice that $\bg_white (u+1)e^u$ is the derivative of $\bg_white ue^u$ and hence the integral is in the form: $\bg_white \int \frac{f'(x)}{\left[f(x)\right]^2+1}dx = \tan^{-1}f(x)+C$. Thus,
$\bg_white I = \left[\tan^{-1}\left(ue^u\right)\right]_{1}^{-1}$
$\bg_white =-\tan^{-1}\left(\frac{1}{e}\right)-\tan^{-1}(e)$
$\bg_white =\boxed{-\frac{\pi}{2}}$ (using the well-known identity $\bg_white \tan^{-1}x+\tan^{-1}\frac{1}{x} = \frac{\pi}{2}$)

s97127

Member
Let $\bg_white u = -x \implies du = -dx$:
$\bg_white I =\int_{-1}^{1} \frac{(x-1)e^x}{x^2+e^{2x}}dx$
$\bg_white =-\int_{1}^{-1} \frac{(-u-1)e^{-u}}{u^2+e^{-2u}}du$
$\bg_white =\int_{1}^{-1}\frac{(u+1)e^u}{e^{2u}u^2+1}du$
Notice that $\bg_white (u+1)e^u$ is the derivative of $\bg_white ue^u$ and hence the integral is in the form: $\bg_white \int \frac{f'(x)}{\left[f(x)\right]^2+1}dx = \tan^{-1}f(x)+C$. Thus,
$\bg_white I = \left[\tan^{-1}\left(ue^u\right)\right]_{1}^{-1}$
$\bg_white =-\tan^{-1}\left(\frac{1}{e}\right)-\tan^{-1}(e)$
$\bg_white =\boxed{-\frac{\pi}{2}}$ (using the well-known identity $\bg_white \tan^{-1}x+\tan^{-1}\frac{1}{x} = \frac{\pi}{2}$)
how come a medical student like yourself still has so much interest in math after finishing HSC?

tito981

Active Member
how come a medical student like yourself still has so much interest in math after finishing HSC?
once u learn math properly u never forget.

CM_Tutor

Moderator
Moderator
I made a typo in Wolfram earlier. It does give a nice answer.
$\bg_white \int_{-1}^{1} \frac{(x-1)e^x}{x^2+e^{2x}}$
Interesting... integration calculator doesn't

idkkdi

Well-Known Member
how come a medical student like yourself still has so much interest in math after finishing HSC?
I have a feeling @vernburn likes maths more than the stuff he is going to do in med.

YonOra

Well-Known Member
Med is science, so how could u not get bored

Qeru

Well-Known Member
The ultimate tedious integral:

$\bg_white \int \sqrt[5]{\tan{x}} dx$

CoolKids101

New Member
The ultimate tedious integral:

$\bg_white \int \sqrt[5]{\tan{x}} dx$
You should attend the integration bee that's being held at UNSW lol

vernburn

Active Member
I have a feeling @vernburn likes maths more than the stuff he is going to do in med.
It does look like it doesn’t it! Jkjk
In reality, I really only like integration now because there is a certain elegance to it and skill required (and it’s still fresh in my head). I find the rest of maths quite boring and stale imho. My attempts at the above integrals are really just due to boredom (not long till uni starts now though).

The ultimate tedious integral:

$\bg_white \int \sqrt[5]{\tan{x}} dx$
I may be bored but not enough to attempt this monster!

Qeru

Well-Known Member
It does look like it doesn’t it! Jkjk
In reality, I really only like integration now because there is a certain elegance to it and skill required (and it’s still fresh in my head). I find the rest of maths quite boring and stale imho. My attempts at the above integrals are really just due to boredom (not long till uni starts now though).
Doesn't look like you were bored of maths considering you got a 99 for 4U. There had to be some interest there lol

CM_Tutor

Moderator
Moderator
The ultimate tedious integral:

$\bg_white \int \sqrt[5]{\tan{x}} dx$
This can easily be converted to a much more recognisable integral with the substitution:

\bg_white \begin{align*} \text{Let:} \qquad u^5 &= \tan^2{x} \\ \text{Implicitly differentiating with respect to x:} \qquad 5u^4\frac{du}{dx} &= 2\tan{x} \times \sec^2{x} \\ &= 2\sqrt{\tan^2{x}}(\tan^2{x} + 1) \\ &= 2\sqrt{u^5}(u^5+ 1) \\ \frac{5u^4}{2u^2\sqrt{u}(u^5 + 1)}du &= dx \\ dx &= \frac{5u^2}{2\sqrt{u}(u^5 + 1)}du \end{align*}

From which it follows that:

\bg_white \begin{align*} I &= \int{\sqrt[5]{\tan{x}}}\ dx \\ &= \int{\left(u^{\frac{5}{2}}\right)^{\frac{1}{5}} \times \frac{5u^2}{2\sqrt{u}(u^5 + 1)}}\ du \qquad \text{where u^5=\tan^2{x}} \\ &= \int{\sqrt{u} \times \frac{5u^2}{2\sqrt{u}(u^5 + 1)}}\ du \\ &= \frac{5}{2}\int{\frac{u^2}{u^5 + 1}}\ du \end{align*}

and the problem is reduced to a very (very) messy partial fractions problem. According to the integral calculator, and after cleaning up its awful formatting / presentation, the indefinite integral is:

$\bg_white I = \int{\sqrt[5]{\tan{x}}}\ dx$

$\bg_white = \frac{\sqrt{5}-1}{8}\ln\left[2\sqrt[5]{\tan^4{x}}-\left(\sqrt{5}-1\right)\sqrt[5]{\tan^2{x}}+2\right] + \frac{\sqrt{10+2\sqrt{5}}}{4}\tan^{-1}\left[\frac{4\sqrt[5]{\tan^2{x}}-\sqrt{5}-1}{\sqrt{10-2\sqrt{5}}}\right]$

$\bg_white - \left(\frac{\sqrt{5}+1}{8}\ln\left[2\sqrt[5]{\tan^4{x}}+\left(\sqrt{5}-1\right)\sqrt[5]{\tan^2{x}}+2\right] + \frac{\sqrt{10-2\sqrt{5}}}{4}\tan^{-1}\left[\frac{4\sqrt[5]{\tan^2{x}}+\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}}\right]\right)$

$\bg_white + \frac{1}{2}\ln\left(\sqrt[5]{\tan^2{x}}+1\right)+C \qquad \text{for some constant C}$

stupid_girl

Active Member
I wrote this in the past but I have now forgotten how to solve it.
$\bg_white \int_{0}^{1}\log_{2}\left(\frac{\left(16^{\sqrt{x}}+k\cdot4^{\sqrt{x}}+4\right)\left(16^{1-\sqrt{x}}+k\cdot4^{1-\sqrt{x}}+4\right)}{\left(4^{\sqrt{x}}+k\cdot2^{\sqrt{x}}+4\right)\left(4^{1-\sqrt{x}}+k\cdot2^{1-\sqrt{x}}+4\right)}\right)dx=1$

Edit: I remember how to solve it now.

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CM_Tutor

Moderator
Moderator
I wrote this in the past but I have now forgotten how to solve it.
$\bg_white \int_{0}^{1}\log_{2}\left(\frac{\left(16^{\sqrt{x}}+k\cdot4^{\sqrt{x}}+4\right)\left(16^{1-\sqrt{x}}+k\cdot4^{1-\sqrt{x}}+4\right)}{\left(4^{\sqrt{x}}+k\cdot2^{\sqrt{x}}+4\right)\left(4^{1-\sqrt{x}}+k\cdot2^{1-\sqrt{x}}+4\right)}\right)dx=1$
Is the question here to prove this result - i.e. that the integral is 1 irrespective of the value of k - or is this an equation that needs to be solved to find the value(s) of k?

stupid_girl

Active Member
Is the question here to prove this result - i.e. that the integral is 1 irrespective of the value of k - or is this an equation that needs to be solved to find the value(s) of k?
The integral is 1 irrespective of the value of k.

s97127

Member
I wrote this in the past but I have now forgotten how to solve it.
$\bg_white \int_{0}^{1}\log_{2}\left(\frac{\left(16^{\sqrt{x}}+k\cdot4^{\sqrt{x}}+4\right)\left(16^{1-\sqrt{x}}+k\cdot4^{1-\sqrt{x}}+4\right)}{\left(4^{\sqrt{x}}+k\cdot2^{\sqrt{x}}+4\right)\left(4^{1-\sqrt{x}}+k\cdot2^{1-\sqrt{x}}+4\right)}\right)dx=1$

Edit: I remember how to solve it now.
can you post the solution here? This gives me a headache for the last few days

Qeru

Well-Known Member
can you post the solution here? This gives me a headache for the last few days
Yeah lol, some factoring trick I imagine, or King Rule but I dont see how that is used.

stupid_girl

Active Member
can you post the solution here? This gives me a headache for the last few days
What tricks have you tried?

The following techniques are NOT required.
integration by parts
differentiation under the integral sign
hyperbolic function

s97127

Member
What tricks have you tried?

The following techniques are NOT required.
integration by parts
differentiation under the integral sign
hyperbolic function
I've tried all the techniques above. I think the problem will be solved by substituting x with another variable t and we got I + I = 2 so I = 1.
Am i on the right path?

stupid_girl

Active Member
By adjusting the coefficient of k to 1, you should be able to take out log_2 (2), which is just 1. Integrating dx from 0 to 1 gives you 1.

The remaining factors have the same structure and you can offset them by suitable substitution.