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?
 A: 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, $\displaystyle \lim_{x\to\infty}\left[\frac{1}{x}\int_{0}^{x}\frac{dt}{2+\cos t}\right]=\frac{1}{\sqrt{3}}$.
A: Let $$F (x)=\int_0^x\frac{dt}{1+\cos(t)} $$
the limit is then
$$\lim_{x \to 0}\frac {F (x)-F (0)}{x-0}=F'(0) $$
$$\frac {1}{1+\cos (0)}=\frac {1}{2} $$
for the integral, put $$x=3\cos^2 (t). $$
It becomes
$$I_1=\int_0^\frac \pi 2 \frac {6\cos^2 (t)\sin (t)dt}{\cos (t)+\sin(t)} $$
now with $$x=3\sin^2 (t) $$, it becomes
$$I_2=\int_0^\frac \pi 2 \frac {6\sin^2 (t)\cos (t)dt}{\sin (t)+\cos(t)} $$
the substitution $t=\frac \pi 2-u$ yields to $$I_1=I_2$$
but
$$I_1+I_2=3\int_0^\frac \pi 2\sin (2u)du =3$$
thus
$$I_1=I_2=\frac 32$$
A: Only hints for $\displaystyle \frac{1}{x}\int_0^x \frac{dt}{2+\cos(t)}=g(x)/x$. Let $\displaystyle f(x)=\frac{1}{2+\cos(t)}$. Note that $f$ is periodic, with period $2\pi$. Let $x$ large, and define $n$ by $2\pi n\leq x<(2n+2)\pi$. Then $g(x)$ is the sum of $\int_{2k\pi}^{(2k+2)\pi} f(t)dt$, for $k=0,\cdots n-1$ and of $\int_{2n\pi}^{x} f(t)dt$. The change of variable $t=2k\pi+u$ gives that $\int_{2k\pi}^{(2k+2)\pi} f(t)dt=\int_{0}^{2\pi} f(u)du=L$. You get that $g(x)=nL+\int_{2n\pi}^{x} f(t)dt$ Now it is easy to bound $\int_{2n\pi}^{x} f(t)dt$ by $2\pi$, and to show that if we divide by $x$, it has limit $0$, and to show that $n/x$ has a limit, and we are done.  
A: Most answers have handled the first part of the problem correctly, so I deal with the limit evaluation only. The function $g(t) =1/(2+\cos t) $ is periodic with period $2\pi$ and hence its average value over $(0,\infty)$ ie $$\lim_{x\to\infty} \frac{1}{x}\int_{0}^{x}g(t)\,dt$$ is same as its average value on $[0,2\pi]$ ie $$\frac{1}{2\pi}\int_{0}^{2\pi}\frac{dt}{2+\cos t} = \frac{1}{\pi}\int_{0}^{\pi}\frac{dt}{2+\cos t} =\frac{1}{\sqrt{3}}$$

The integral formula $$\int_{0}^{\pi}\frac{dt}{a+b\cos t} =\frac{\pi} {\sqrt{a^{2}-b^{2}}}$$ for $a>|b|$ is easily proved by the substitution $$(a+b\cos t) (a-b \cos x)=a^{2}-b^{2}$$ The result regarding average value of periodic functions is not that difficult to prove. 
A: Form the properties of definite integral we know that:
$$\int_b^af(x)\ dx=\int_b^af(a+b-x)\ dx\\
\implies \int_0^af(x)\ dx=\int_0^af(a-x)\ dx$$
Applying this property to your problem:
Let, 
\begin{align*}I&=\int_0^3\dfrac{\sqrt x}{\sqrt{x}+\sqrt{3-x}}\ dx\\
&=\int_0^3\dfrac{\sqrt{3-x}}{\sqrt{3-x}+\sqrt{3-(3-x)}}\ dx\\
I&=\int_0^3\dfrac{\sqrt{3-x}}{\sqrt{3-x}+\sqrt x}\ dx
\end{align*}
Now, $$I+I=\int_0^3\dfrac{\sqrt x}{\sqrt{x}+\sqrt{3-x}}\ dx+\int_0^3\dfrac{\sqrt{3-x}}{\sqrt{3-x}+\sqrt x}\ dx\\
\implies2I=\int_0^3\dfrac{\sqrt x+\sqrt{3-x}}{\sqrt{3-x}+\sqrt x}\ dx\\
\implies2I=\int_0^3dx=x\big|_0^3=3\\
\implies I=\dfrac{3}{2}\\
\implies\boxed{\int_0^3\dfrac{\sqrt x}{\sqrt{x}+\sqrt{3-x}}\ dx=\dfrac{3}{2}}.
$$
