MX2 Integration Marathon 2021 (1 Viewer)

idkkdi

Well-Known 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.
if ur stupid, what am i.

F

-insert title here-
Define

$\bg_white \psi(a,k,u,v) := \log_a\left(a^{u} + k + a^{v}\right)$

The original question takes the argument a=2.

For brevity, we will suppress the first two arguments⁰, as it turns out they are irrelevant to the final answer.

(This is a thing in higher mathematics lmao get used to it)

The original question is equivalent¹ to:

$\bg_white \int_0^1 \left(\psi\left(2\sqrt{x},2-2\sqrt{x}\right)+\psi\left(2-2\sqrt{x},2\sqrt{x}\right)- \psi\left(\sqrt{x},2-\sqrt{x}\right) - \psi\left(1-\sqrt{x},1+\sqrt{x}\right) + \nolinebreak 1 \right) \mathrm{d}x$

It is clear that ψ(u,v) = ψ(v,u), from the definition, so we can collapse the first two terms together.

Break off the +1 term and ignore it for now, as that is what gives the final answer.

Using the substitution x = t², the integral becomes:

$\bg_white 2 \int_0^1 \left(2\psi\left(2t,2-2t\right) - \psi\left(t,2-t\right) - \psi\left(1+t,1-t\right) \right) t \mathrm{d}t$

Using the boundary preserving transformation t→1-t, this becomes:

$\bg_white 2 \int_0^1 \left(2\psi\left(2-2t,2t\right) - \psi\left(1-t,1+t\right) - \psi\left(2-t,t\right) \right) (1-t) \mathrm{d}t$

Use symmetry in u,v, once again, to reorder the arguments to be the same as before.

Take the average of this and the previous step, to obtain:

$\bg_white \int_0^1 \left(2\psi\left(2t,2-2t\right) - \psi\left(t,2-t\right) - \psi\left(1+t,1-t\right) \right) \mathrm{d}t$

We consider only the last two terms for now.

Define:

$\bg_white A := \int_0^1 \psi\left(t,2-t\right) \mathrm{d}t; \quad B := \int_0^1 \psi\left(1+t,1-t\right) \mathrm{d}t$

By the same boundary preserving transformation as before, it is obvious that A = B.

Use the scaling transformation t→2t:

$\bg_white A = B = 2\int_0^{\frac{1}{2}} \psi\left(2t,2-2t\right) \mathrm{d}t$

Use the same "boundary preserving transformation" to change the interval [0,½] → [½,1]

$\bg_white B = 2\int_{\frac{1}{2}}^1 \psi\left(2-2t,2t\right) \mathrm{d}t = 2\int_{\frac{1}{2}}^1 \psi\left(2t,2-2t\right) \mathrm{d}t$

Finally, we have:

$\bg_white A + B = 2\int_0^1 \psi\left(2t,2-2t\right) \mathrm{d}t = \int_0^1 2\psi\left(2t,2-2t\right) \mathrm{d}t$

Returning to the step before we defined A and B, we obtain:

$\bg_white \int_0^1 2\psi\left(2t,2-2t\right) \mathrm{d}t -(A+B) = \int_0^1 2\psi\left(2t,2-2t\right)-2\psi\left(2t,2-2t\right) \mathrm{d}t = 0$

The +1 term is trivially integrated to obtain 1, which is the final answer.

⁰ ψ(u,v) = ψ(a,k,u,v)

¹ left as an exercise to the reader

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-insert title here-
the real answer to this conundrum is that by using a substitution which maps a finite open interval to ℝ, you cannot invert the substitution to obtain an antiderivative valid over all of ℝ. the substitution is only bijective locally, not globally, so a true inverse does not exist, hence the antiderivative is only valid over specific subsets of ℝ. in order to obtain an antiderivative that is valid on all of ℝ, you need a different method.

more to read here:

stupid_girl

Active Member
$\bg_white \int_{1}^{100}\frac{x^{3}-x\cos^{2}\left(2\ln x\right)}{x^{4}+x^{2}\sin\left(4\ln x\right)+\cos^{2}\left(2\ln x\right)}dx=\frac{11\pi}{8}+\frac{1}{2}\tan^{-1}\left(\frac{1+10000\tan\left(4\ln10\right)}{10000}\right)$

Qeru

Well-Known Member
$\bg_white \int_{1}^{100}\frac{x^{3}-x\cos^{2}\left(2\ln x\right)}{x^{4}+x^{2}\sin\left(4\ln x\right)+\cos^{2}\left(2\ln x\right)}dx=\frac{11\pi}{8}+\frac{1}{2}\tan^{-1}\left(\frac{1+10000\tan\left(4\ln10\right)}{10000}\right)$
Let $\bg_white u=\tan^{-1}\left(\frac{1+x^2\tan{(2\ln{x})}}{x^2}\right)$

From derivative calc:
$\bg_white \frac{du}{dx}=\frac{4x^3\sec^2(2\ln x)-2x}{x^4\tan^2(2\ln x)+2x^2\tan(2\ln x)+x^4+1}$

$\bg_white \frac{du}{dx}=\frac{4x^3-2x\cos^2(2\ln x)}{x^4\sin^2(2\ln x)+2x^2\sin(2\ln x)\cos(2\ln x)+(x^4+1)\cos^2(2\ln x)}$

$\bg_white \frac{du}{dx}=\frac{4x^3-2x\cos^2(2\ln x)}{x^4+x^2\sin(4\ln x)+\cos^2(2\ln x)}$

$\bg_white I=\int_{1}^{100}\frac{x^{3}-x\cos^{2}\left(2\ln x\right)}{x^{4}+x^{2}\sin\left(4\ln x\right)+\cos^{2}\left(2\ln x\right)}dx$

$\bg_white I=\int_{1}^{100}\frac{2x^{3}-x\cos^{2}\left(2\ln x\right)}{x^{4}+x^{2}\sin\left(4\ln x\right)+\cos^{2}\left(2\ln x\right)}dx-\int_{1}^{100}\frac{x^{3}}{x^{4}+x^{2}\sin\left(4\ln x\right)+\cos^{2}\left(2\ln x\right)}dx$

$\bg_white I=\frac{1}{2}\int_{1}^{100}1du-\int_{1}^{100}\frac{x^{3}}{x^{4}+x^{2}\sin\left(4\ln x\right)+\cos^{2}\left(2\ln x\right)}dx$

$\bg_white I=\frac{1}{2}\tan^{-1}\left(\frac{1+10000\tan\left(4\ln10\right)}{10000}\right)-\frac{\pi}{8}-\int_{1}^{100}\frac{x^{3}}{x^{4}+x^{2}\sin\left(4\ln x\right)+\cos^{2}\left(2\ln x\right)}dx$

No clue how to do the second integral.

stupid_girl

Active Member
Let $\bg_white u=\tan^{-1}\left(\frac{1+x^2\tan{(2\ln{x})}}{x^2}\right)$

From derivative calc:
$\bg_white \frac{du}{dx}=\frac{4x^3\sec^2(2\ln x)-2x}{x^4\tan^2(2\ln x)+2x^2\tan(2\ln x)+x^4+1}$

$\bg_white \frac{du}{dx}=\frac{4x^3-2x\cos^2(2\ln x)}{x^4\sin^2(2\ln x)+2x^2\sin(2\ln x)\cos(2\ln x)+(x^4+1)\cos^2(2\ln x)}$

$\bg_white \frac{du}{dx}=\frac{4x^3-2x\cos^2(2\ln x)}{x^4+x^2\sin(4\ln x)+\cos^2(2\ln x)}$

$\bg_white I=\int_{1}^{100}\frac{x^{3}-x\cos^{2}\left(2\ln x\right)}{x^{4}+x^{2}\sin\left(4\ln x\right)+\cos^{2}\left(2\ln x\right)}dx$

$\bg_white I=\int_{1}^{100}\frac{2x^{3}-x\cos^{2}\left(2\ln x\right)}{x^{4}+x^{2}\sin\left(4\ln x\right)+\cos^{2}\left(2\ln x\right)}dx-\int_{1}^{100}\frac{x^{3}}{x^{4}+x^{2}\sin\left(4\ln x\right)+\cos^{2}\left(2\ln x\right)}dx$

$\bg_white I=\frac{1}{2}\int_{1}^{100}1du-\int_{1}^{100}\frac{x^{3}}{x^{4}+x^{2}\sin\left(4\ln x\right)+\cos^{2}\left(2\ln x\right)}dx$

$\bg_white I=\frac{1}{2}\tan^{-1}\left(\frac{1+10000\tan\left(4\ln10\right)}{10000}\right)-\frac{\pi}{8}-\int_{1}^{100}\frac{x^{3}}{x^{4}+x^{2}\sin\left(4\ln x\right)+\cos^{2}\left(2\ln x\right)}dx$

No clue how to do the second integral.
You choose the correct substitution but calculate du/dx incorrectly.
$\bg_white \frac{du}{dx}=\frac{2x^3-2x\cos^2(2\ln x)}{x^4+x^2\sin(4\ln x)+\cos^2(2\ln x)}$

stupid_girl

Active Member
By the way, choosing the correct substitution is not the end of story.

Remember to split the interval.
$\bg_white \int_{1}^{100}\frac{x^{3}-x\cos^{2}\left(2\ln x\right)}{x^{4}+x^{2}\sin\left(4\ln x\right)+\cos^{2}\left(2\ln x\right)}dx=\int_{1}^{e^{\frac{\pi}{4}}}\frac{x^{3}-x\cos^{2}\left(2\ln x\right)}{x^{4}+x^{2}\sin\left(4\ln x\right)+\cos^{2}\left(2\ln x\right)}dx+\int_{e^{\frac{\pi}{4}}}^{e^{\frac{3\pi}{4}}}\frac{x^{3}-x\cos^{2}\left(2\ln x\right)}{x^{4}+x^{2}\sin\left(4\ln x\right)+\cos^{2}\left(2\ln x\right)}dx+\int_{e^{\frac{3\pi}{4}}}^{e^{\frac{5\pi}{4}}}\frac{x^{3}-x\cos^{2}\left(2\ln x\right)}{x^{4}+x^{2}\sin\left(4\ln x\right)+\cos^{2}\left(2\ln x\right)}dx+\int_{e^{\frac{5\pi}{4}}}^{100}\frac{x^{3}-x\cos^{2}\left(2\ln x\right)}{x^{4}+x^{2}\sin\left(4\ln x\right)+\cos^{2}\left(2\ln x\right)}dx$

If you are wondering how these break points are chosen, you may try to graph $\bg_white u=\tan^{-1}\left(\frac{1+x^2\tan{(2\ln{x})}}{x^2}\right)$.

stupid_girl

Active Member
$\bg_white \int_{0}^{2}\left(\sqrt{4^{x+1}-8^{x}}\sqrt{4^{x}-5\left(2^{x}\right)+7}-\sqrt[3]{16^{x}}\sqrt[3]{2-2^{x}}\right)dx=\frac{6}{\ln2}$