# Find $x>0$ for which the integral value $\int_0^{\sqrt {x}} \sin (\frac{2\pi t}{t+2}) dt$ is the largest

Find $$x>0$$ for which the integral value of $$\int_0^{\sqrt {x}} \sin \left(\frac{2\pi t}{t+2}\right) dt$$ is the largest.

My try:
Let: $$f(x)=\int_0^{\sqrt {x}} \sin \left(\frac{2\pi t}{t+2}\right) dt$$ and $$F'(t)=\sin \left(\frac{2\pi t}{t+2}\right) dt$$ Then: $$f(x)=F(\sqrt {x})-F(0)$$ $$f'(x)=\frac{F'(\sqrt {x})}{2\sqrt {x}}-F'(0)\cdot 0=\frac{F'(x)}{2\sqrt {x}}=\frac{\sin \left(\frac{2\pi \sqrt {x}}{\sqrt {x}+2}\right)}{2\sqrt {x}}$$In this moment I am wanting to find $$x$$ for which $$f'(x)=0$$ and $$f'$$ changes the sign because it is a maximum for $$f$$. However I have a sine and I don't know how to do it in the simplest way, because the study of all ranges of sine variation is a terrible calculation.

Can you help me?

• In the last expression, replace $\sqrt{x}$ by $y$, say, and analyse the corresponding function of $y$ in $(0,\infty)$. It looks doable. Jun 20 '19 at 10:21
• @ViktorGlombik yes, but when I calculate only $f'(x)=0$ I don't know if in this $x$ is maximum or minimum Jun 20 '19 at 10:37
• I am getting infinite values. Jun 20 '19 at 10:41
• @AjayMishra WolframAlpha says otherwise. Jun 20 '19 at 10:43
• I didn't get that at all. Jun 20 '19 at 10:46

Given function, $$f(x) = \int_0^{\sqrt{x}} \sin\bigg( \cfrac{2 \pi t}{t + 2} \bigg) dt$$
On differentiating, $$f'(x) = \cfrac{1}{2 \sqrt{x} } \sin\bigg( \cfrac{2 \pi \sqrt{x}}{\sqrt{x} + 2} \bigg)$$For stationary points, $$f'(x) = 0$$. Here, in order to satisfy the condition, $$x = \bigg( \cfrac{2n}{2-n} \bigg)^2$$
Where $$n \ne 0$$ and $$\in \mathbb I$$ . But, remember at these values of $$x$$, there is stationary points, in order to find the maxima you can plug these values in $$f'(x)$$ and use first derivative test to find that. Can you do this from now?
You did almost half of the solution After using $$f'(x)=0$$ we get $$\sin \bigg(\frac{2\pi \sqrt x}{\sqrt x +2}\bigg)=0$$ Which yields $$\frac{2\pi \sqrt x}{\sqrt x +2}=n\pi$$ or $$2\sqrt x=n(\sqrt x+2)$$ By rearranging one can see that $$x$$ must be an integer in order to satisfy above equality because $$n$$ is integer And this equation satisfies only when $$x=0,4$$ for some $$n\in Z$$ But as $$x>0$$ Thus, The only solution is $$x=\boxed{4}$$
• Where it was stated that $x$ must be integer? Jun 20 '19 at 10:57