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This is a problem that my calculus professor gave to his students many years ago.

$$\int\ln x \arccos\left( 7x^2-\sqrt{49x^4-50x^2+1}\right) dx$$

Wolfram doesn't find any solution in terms of standard mathematical functions. I'm sure that this integral has a solution otherwise my professor wouldn't have assigned it.

Could someone help me?

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    $\begingroup$ If WolframAlpha doesn't know, then I see no reason for this to have a closed form. More likely I would imagine that this was not the intended problem. Either there was a typo, or you are missing information e.g. bounds or that what you were asked for was the derivative of this. $\endgroup$ – Simply Beautiful Art Sep 2 '19 at 16:00
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    $\begingroup$ @SimplyBeautifulArt The answer below proves that human brains are still superior at pattern recognition. $\endgroup$ – Deepak Sep 2 '19 at 16:18
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    $\begingroup$ Well relying on wolfram might not be the best idea, it seems that it also fails on some rather easier integrals, such as: $$\int \frac{\sin x}{1+\sqrt{\sin(2x)}}dx, \quad \int \frac{x}{\sqrt{e^x+\left(x+2\right)^2}} dx$$ $\endgroup$ – Zacky Sep 2 '19 at 17:20
  • $\begingroup$ @Zacky very interesting integrals. Where can I see the solution of the two? $\endgroup$ – Leprep98 Sep 2 '19 at 20:26
  • $\begingroup$ For the first one I would recommend to consider: $$I=\int \frac{\sin x}{1+\sqrt{\sin(2x)}}dx, \quad J=\int \frac{\cos x}{1+\sqrt{\sin(2x)}}dx $$ Then evaluate: $I+J$ and $I-J$ and extract $I$ afterwards. As a bigger hint, note that $$\sin(2x)=(\sin x+\cos x)^2-1=1-(\sin x-\cos x)^2$$ You must realise which substitution to use now for $I+J$ and which one for $I-J$. $$\int \frac{x}{\sqrt{e^x+\left(x+2\right)^2}} dx=\int \frac{xe^{-x/2}}{\sqrt{1+\left(\color{red}{(x+2)e^{-x/2}}\right)^2}} dx$$ Try to substitute the red thing. $\endgroup$ – Zacky Sep 2 '19 at 20:44
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I'll try to complete the other answer. Note that we only have to deal with a single integral since they're similar.

For start notice: $$\int \arccos(ax)dx=x\arccos(ax)-\sqrt{\frac{1}{a^2}-x^2}+C$$ So integrating by parts we get: $$\small\int \ln x\arccos(ax)dx=\ln x\left(x\arccos(ax)-\sqrt{\frac{1}{a^2}-x^2}\right)-\int\arccos(ax)dx+\int \sqrt{\frac{1}{a^2}-x^2}\ \frac{dx}{x}$$ What is left is: $$\int \sqrt{\frac{1}{a^2}-x^2}\ \frac{dx}{x}=\int \sqrt{\frac{1}{(ax)^2}-1}\ dx\overset{ax=1/t}=-\frac1a \int \frac{\sqrt{t^2-1}}{t^2}dt$$ $$\overset{IBP}=\frac1{at}\sqrt{t^2-1}+\frac1a \int \frac{1}{\sqrt{t^2-1}}dt=a x^2\sqrt{\frac{1}{(ax)^2}-1}+\frac1a \operatorname{arccosh} \left(\frac{1}{ax}\right)+C $$ This is only the approach, what's left to do is to get the final result, by taking $a=1$ and $a=7$.

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I'm not ready to give the full answer, but I guess, I can help a bit. Try using formula $$ \arccos(x) + \arccos(y) = \arccos\left(xy - \sqrt{\left(1-x^2\right)\left(1-y^2\right)}\right) $$ (for ref. see Wikipedia). To do this, write $$ 7 x^2 - \sqrt{49 x^4 -50 x^2 +1} = x \times 7x - \sqrt{\left(1-x^2\right)\left(1-(7x)^2\right)} $$

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  • $\begingroup$ Formula $\cos ^{-1}\left(7 x^2-\sqrt{49 x^4-50 x^2+1}\right)=\cos ^{-1}(x)+\cos ^{-1}(7 x)$ is true for: $x\geq 0$ and $x\in \mathbb{R}$ $\endgroup$ – Mariusz Iwaniuk Sep 2 '19 at 16:29
  • $\begingroup$ @Mariusz Iwaniuk, very good point, many thanks! I guess, we need absolute value on the right-hand side ($\arccos(|x|)+\arccos(|7x|)$), since the left-hand side contains only x squared thus is insensitive to the x's sign. $\endgroup$ – guest Sep 2 '19 at 16:44
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Sufficient care has to be taken before the assignment of equality sign.

Using https://en.m.wikipedia.org/wiki/Inverse_trigonometric_functions#Principal_values,

$0\le\arccos(x)\le\pi$

$\implies0\le\arccos(x)+\arccos(y)\le2\pi$

$\arccos(x)+\arccos(y)$ will be $$=\arccos(xy-\sqrt{(1-x^2)(1-y^2)})$$ iff $\arccos(x)+\arccos(y)\le\pi$

$\iff\dfrac\pi2-\arcsin(x)+\cdots\le\pi$

$\iff\arcsin(x)\ge-\arcsin(y)=\arcsin(-y)$

As $\arcsin$ is an increasing function, we need $$x\ge-y\iff x+y\ge0$$

Otherwise $$\arccos(x)+\arccos(y)=2\pi-\arccos(xy-\sqrt{(1-x^2)(1-y^2)})$$

Can you identify $y$ here?

See also: Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $

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Cross-posted on AoPS here. Answered by ysharifi. I will just put it here in length.

That is much neater than your other integral. Assuming you want to stay in real numbers, first we need to have $x > 0$ because of the presence of $\ln x.$ Secondly we also need to have $$49x^4-50x^2+1=(1-x^2)(1-49x^2) \ge 0,$$ and so we have either $0 < x \le \frac{1}{7}$ or $x > 1.$ Thirdly, we also need to have $-1 \le 7x^2-\sqrt{49x^4-50x^2+1} \le 1$ and so $7x^2-1 \le \sqrt{49x^4-50x^2+1},$ which does not hold for $x > 1.$ So the domain of the function you want to integrate is $0 < x \le \frac{1}{7}.$ Now see that $$\cos^{-1}(7x^2-\sqrt { 49x^4-50x^2+1})=\cos^{-1}(7x^2-\sqrt{1-x^2}\sqrt{1-49x^2})$$ $$=\cos^{-1}(\cos(\cos^{-1}x)\cos(\cos^{-1}(7x))-\sin(\cos^{-1}x)\sin(\cos^{-1}(7x)))$$ $$=\cos^{-1}(\cos(\cos^{-1}x+\cos^{-1}(7x)))=\cos^{-1}x+\cos^{-1}(7x).$$ So $$I:=\int \ln x \cos^{-1}(7x^2-\sqrt{49x^4-50x^2+1})\ \mathrm dx=\int \ln x \ (\cos^{-1}x+\cos^{-1}(7x))\ \mathrm dx.$$ The rest of the solution is straightforward. Use integration by parts with $\ln x=u$ and $(\cos^{-1}x+\cos^{-1}(7x)) \ \mathrm dx=\mathrm dv.$ Then $$\mathrm du=\frac{\mathrm dx}{x}, \ \ \ \ \ \ v=x\cos^{-1}x-\sqrt{1-x^2}+x\cos^{-1}(7x)-\frac{1}{7}\sqrt{1-49x^2}$$ and hence $$I=\ln x \left(x\cos^{-1}x-\sqrt{1-x^2}+x\cos^{-1}(7x)-\frac{1}{7}\sqrt{1-49x^2}\right)-\int \left(\cos^{-1}x+\cos^{-1}(7x)-\frac{\sqrt{1-x^2}}{x}-\frac{\sqrt{1-49x^2}}{7x}\right)\ \mathrm dx$$ $$=(\ln x-1) \left(x\cos^{-1}x-\sqrt{1-x^2}+x\cos^{-1}(7x)-\frac{1}{7}\sqrt{1-49x^2}\right)+\int \left(\frac{\sqrt{1-x^2}}{x}+\frac{\sqrt{1-49x^2}}{7x}\right)\mathrm dx. \ \ \ \ \ \ \ \ \ \ (1)$$ Also, given $a > 0,$ the substitution $1-a^2x^2=t^2$ gives $$\int \frac{\sqrt{1-a^2x^2}}{x} \ \mathrm dx=\sqrt{1-a^2x^2}+\ln x-\ln(1+\sqrt{1-a^2x^2})+ \text{constant}. \ \ \ \ \ \ \ \ \ \ \ (2)$$ Now $(1),(2)$ together complete the solution.

Brilliantly done; as always by him.

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