Solve $\int_0^y\frac x{\cos(x)^2}dx = \frac12$

Question

My son as an exercise had to solve for $y$ in

$$\int_0^y\frac x{\cos(x)^2}dx = \frac12$$

and he didn't manage to do it, and neither do I. You can find the primitive of the integrand, turning the equation into

$$y\tan y + \log\left(\cos y\right) = \frac12,$$

but that doesn't seem easy to solve either.

Is there some kind of a closed-form solution? Note that this is secondary school, so I expect it to be possible without complex integration, though if you find a solution using complex integration I would be interested in seeing that as well.

Answer 1

Certainly, one way to look for solutions is by using graphs. The red curve is $f(x) = \int_0^x t\sec^2 t\, dt$ and the blue line is just $g(x) = \frac12$. It is worth noting that $f$ is an even function, and so the solutions occur in pairs, as the image below suggests.

Answer 2

Making the problem more general, we want to find the zero of function $$f(y)=y \tan (y)+\log (\cos (y)) -k$$ This is a transcendental equation which requires numerical methods or approximations.

We can start using a series expansion $$f(y)+k=\sum_{n=1}^\infty (-1)^n\frac{ 4^{n-1} (2 n-1) (E_{2 n-1}(0)-E_{2 n-1}(1))}{(2n)!}\,y^{2n}$$ where appear Euler polynomials. The first coefficients are $$\left\{\frac{1}{2},\frac{1}{4},\frac{1}{9},\frac{17}{360}, \frac{31}{1575},\frac{691}{85050},\frac{10922}{3274425} ,\frac{929569}{681080400},\frac{3202291}{5746615875},\cdots\right\}$$

If you plot the function and this truncated expansion, you will notice that they match pretty well up to $y=1.25$ (at this point, the difference is $0.054$ (which is not much since the functio value is $2.608$).

Now, we can try series reversion to get, as an estimate, $$y=t-\frac{t^3}{4}+\frac{31 t^5}{288}-\frac{317 t^7}{5760}+\frac{890453 t^9}{29030400}-\frac{896029 t^{11}}{49766400}+O\left(t^{13}\right)$$ where $t=\sqrt{2k}$.

Trying for $k=\frac 12$ (which nicely corresponds to $t=1$) gives $$y=\frac{284012273}{348364800}=0.815273\cdots$$ while the exact solution given by Newton method for example is $y=0.822034\cdots$.

If I had used all the terms corresponding to the above table, the result would have been $$y=\frac{66180692011972831}{80343513169920000}=0.823722\cdots$$

Edit

Anothe possibility is to use Padé approximants instead of Taylor series. A very simple one would lead to $$k=-\frac{9 y^2 \left(y^2-10\right)}{2 \left(7 y^4-54 y^2+90\right)}$$ and we just need to solve a quadratic equation in $y^2$.

Tis would give $$y=\frac{1}{2} \sqrt{3 \left(6-\sqrt{26}\right)}=0.822031\cdots$$

And we can do much better.

Update

In his answer, @Bob Dobbs proposed to solve the cubic in $y^2$; it would write $$4 y^6+9 y^4+18 y^2-18=0$$ The solution is simple $$y=\frac{1}{2} \sqrt{2 \sqrt{15} \sinh \left(\frac{1}{3} \sinh ^{-1}\left(\sqrt{15}\right)\right)-3}=0.830347$$

Answer 3

Let $F(y)=y\tan y+\log(\cos y).$ Then $F$ is continuous on $[0,{\pi\over 2})$ and $F(0)=0.$ On the other hand $$F\left ({\pi\over 3}\right )={\pi \over \sqrt{3}}-\log 2> \sqrt{3}-1>{1\over 2}$$ Therefore, there exists a solution $y_0$ such that $0<y_0<{\pi\over 3}.$

Answer 4

Maclaurin series $\ln\cos y=\sum_{n=1}^{\infty}\frac{(-1)^{n}2^{2n}(2^{2n}-1)B_{2n}}{(2n)!(2n)}y^{2n}$ and $\tan y=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}2^{2n}(2^{2n}-1)B_{2n}}{(2n)!}y^{2n-1}$ give $$y\tan y+\ln\cos y=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}2^{2n}(2^{2n}-1)(2n-1)B_{2n}}{(2n)!(2n)}y^{2n}=\frac12y^2+\frac14y^4+\frac19y^6+\frac{17}{360}y^8+...$$ I can't revert series like Bob (see the comment of Bob in Claude's answer), so I solved approximately by solving $\frac12y^2+\frac14y^4=\frac12$ which gave $y\approx\sqrt{\sqrt3 -1}\approx 0.855$. To get a better approximation I must solve $\frac12y^2+\frac14y^4+\frac19y^6=\frac12.$ But, I think I need a lot of time to do it by hand.