Integration of a square root inside another square root function. Integrate the following $$\int_0^5 \sqrt{5-\sqrt{x}} \ \mathrm{d}x$$ I have done the following: 
I got the expression under the square root to equal $$5-x^{1/2}$$ so the integral then becomes $$\int_0^5 \left(5-x^{1/2}\right)^{1/2}\ \mathrm{d}x$$ Here is where I am stuck. Do I need to use some variation of the binomial theorem here?
I hope my use of MathJax was "up-to-code" here. I apologize if not.

Here, is another easier method 
Let $\sqrt{5-\sqrt{x}}=u\implies \frac{-1}{4\sqrt x\sqrt{5-\sqrt x}}\ dx=du$ or $dx=4u(u^2-5)\ du$, 
$$\int_0^5 \sqrt{5-\sqrt x}\ dx=\int_{\sqrt 5}^{\sqrt{5-\sqrt 5}} u(4u(u^2-5))\ du$$
$$=4\int_{\sqrt 5}^{\sqrt{5-\sqrt 5}} (u^4-5u^2)\ du$$
$$=4\left(\frac{u^5}{5}-\frac{5u^3}{3}\right)_{\sqrt 5}^{\sqrt{5-\sqrt 5}}$$
$$=4\left(\frac{(5-\sqrt 5)^{5/2}}{5}-\frac{5(5-\sqrt 5)^{3/2}}{3}-\frac{25\sqrt5}{5}+\frac{25\sqrt 5}{3}\right)$$
$$=\color{red}{\left(\frac{(5-\sqrt 5)^{5/2}}{5}-\frac{5(5-\sqrt 5)^{3/2}}{3}+\frac{10\sqrt5}{3}\right)}$$
