Substitution Integral: $\int_0^3\frac {\sqrt{x}}{\sqrt{x}+\sqrt{3-x}} dx$ $$\int_0^3\frac {\sqrt{x}}{\sqrt{x}+\sqrt{3-x}} dx   $$ I tried with substitution $t=3-x$ but I don't know what to do next.
Also how do I solve this limit: $$\lim_{x\to\infty}\frac{1}{x}\int_0^x\frac{dt}{2+\cos t} $$ I saw that most problems like this are solved by putting the period of the function in front,would it work here?

Hint:
(1)  $\displaystyle \frac{\sqrt{x}}{\sqrt{x}+\sqrt{3-x}}=\frac{\sqrt{x}(\sqrt{x}+\sqrt{3-x})}{(\sqrt{x}+\sqrt{3-x})(\sqrt{x}+\sqrt{3-x})}$
(2) As the integrand is even and has a period of $2\pi$, $\displaystyle\int_{k\pi}^{(k+1)\pi}\frac{dt}{2+\cos t} =\int_{0}^{\pi}\frac{dt}{2+\cos t} $ for all integers $k$. 
Let $\displaystyle\int_{0}^{\pi}\frac{dt}{2+\cos t} =l$. As the integrand is strictly positive, $\displaystyle\int_{0}^{x}\frac{dt}{2+\cos t} $ is strictly increasing. 
If $n\pi\le x<(n+1)\pi$,
$$nl\le \int_{0}^{x}\frac{dt}{2+\cos t}<(n+1)l $$
By letting $u=\tan\frac{t} {2}$,
$$\int_{0}^{\pi}\frac{dt}{2+\cos t} =2\int_0^\infty\frac{du}{3+u^2}=\frac{\pi}{\sqrt{3}}$$
So,
$$\frac{n\pi}{\sqrt{3}}\le \int_{0}^{x}\frac{dt}{2+\cos t}<\frac{(n+1)\pi}{\sqrt{3}} $$
$$\frac{x-\pi}{\sqrt{3}}< \int_{0}^{x}\frac{dt}{2+\cos t}<\frac{x+\pi}{\sqrt{3}} $$
$$\frac{1-\frac{\pi}{x}}{\sqrt{3}}< \frac{1}{x}\int_{0}^{x}\frac{dt}{2+\cos t}<\frac{1+\frac{\pi}{x}}{\sqrt{3}} $$
So, $\displaystyle \lim_{x\to\infty}\left[\frac{1}{x}\int_{0}^{x}\frac{dt}{2+\cos t}\right]=\frac{1}{\sqrt{3}}$.
