<s>[INST] 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?
 [/INST] 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, $$I+I=\int_0^3\dfrac{\sqrt x}{\sqrt{x}+\sqrt{3-x}}\ dx+\int_0^3\dfrac{\sqrt{3-x}}{\sqrt{3-x}+\sqrt x}\ dx\\
\implies2I=\int_0^3\dfrac{\sqrt x+\sqrt{3-x}}{\sqrt{3-x}+\sqrt x}\ dx\\
\implies2I=\int_0^3dx=x\big|_0^3=3\\
\implies I=\dfrac{3}{2}\\
\implies\boxed{\int_0^3\dfrac{\sqrt x}{\sqrt{x}+\sqrt{3-x}}\ dx=\dfrac{3}{2}}.
$$
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