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?
 A: 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?
A: 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 
$$x=\boxed{4}$$
