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

Question

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?

Answer 1

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 $$ 7 x^2 - \sqrt{49 x^4 -50 x^2 +1} = x \times 7x - \sqrt{\left(1-x^2\right)\left(1-(7x)^2\right)} $$

Answer 2

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 $

Answer 3

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.