The task i'm trying to solve is:
How many solutions (roots) does equation have:
$$\int\limits_x^{x+\frac{1}{2}} \cos \left( \frac{t^2}{3} \right) dt = 0$$
on the segment [0, 3] ?
By the moment i've found that cosine can be presented as series (the queston Maclaurin Series of $\int_0^x \cos t^2\,dt$ is about quite similar integral):
$$\cos(\frac{t^2}{3}) = \sum_{n=0}^{\infty}\frac{(-1)^n (\frac{t^2}{3})^{2n}}{2n!} = \sum_{n=0}^{\infty}\frac{(-1)^n {t}^{4n}}{9^n2n!} $$
So:
$$ \int\limits_x^{x+\frac{1}{2}} \cos \left( \frac{t^2}{3} \right) dt = \sum_{n=0}^{\infty} \frac{(-1)^n}{2n!} \int\limits_x^{x+\frac{1}{2}} \frac{t^{4n}}{9^n} dt = \sum_{n=0}^{\infty} \frac{(-1)^n}{9^n2n!} \int\limits_x^{x+\frac{1}{2}}{t^{4n}}dt = \sum_{n=0}^{\infty} \frac{(-1)^n}{9^n2n!} \frac{t^{4n+1}}{4n+1} \Biggr|_{x}^{x+\frac{1}{2}} $$
Further fundamental theorem of calculus can be applied. But i'm stuck on the part related to the number of solutions on the segment.
Is the approach above correct to solve this kind of problem at all?
Any help will be appreciated.
Thank you in advance
Take a look at the graph of $cos(\frac{t^2}{3})$.
This function is decreasing from $t=0$ up to some point $3+\varepsilon$ that you can compute. This means that the value
$$
\int_x^{x+\frac{1}{2}}cos(\frac{t^2}{3})dt
$$
is decreasing from $0$ to $\frac{5}{2}+\varepsilon$. It is obvious (and can be proven by the positivity/negativity of $cos(\frac{t^2}{3})$) that at $x=0$ the integral is positive, at $x=\frac{5}{2}+\varepsilon$ it is negative, and it is continuous in x, so there will be one zero. Due to monotonicity (it is decreasing) that zero will be the only zero in the interval. After that, in the part with $x= \frac{5}{2}+\varepsilon$ up to $x=3$ the integral stays strictly negative so there will be no additional zeroes.
All of these properties have to be proven, I have just given an informal proof using the graph of the function, but it should help to move you in the right direction.