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.
 A: We can make the $u$-substitution $$u=5-\sqrt{x}$$ and solve for $x$ here. We get $x=(u-5)^2$ so $$\mathrm{d}x=(2u-10) \ \mathrm{d}u$$ Now to find the new bounds we use $u=5-\sqrt{x}$. So the bounds are $u=5-\sqrt{0} =5$ and $u=5-\sqrt{5}$. The integral is now $$\int_5^{5-\sqrt{5}}\! \sqrt{u} \cdot (2u-10)\ \mathrm{d}u$$ Which is easy to solve.
A: Hint: substitute $u=5-\sqrt{x}$
A: Hint: The substitution $u=5-\sqrt x$, i.e. $x=(5-u)^2$ yields 
$$
\int \sqrt u (-2(5-u))\,du
$$
for some new limits that I'm not going to compute right now.
A: Here, is another easier method 
Let $\sqrt{5-\sqrt{x}}=u\implies \frac{-1}{4\sqrt x\sqrt{5-\sqrt x}}\ dx=du$ or $dx=4u(u^2-5)\ du$, 
$$\int_0^5 \sqrt{5-\sqrt x}\ dx=\int_{\sqrt 5}^{\sqrt{5-\sqrt 5}} u(4u(u^2-5))\ du$$
$$=4\int_{\sqrt 5}^{\sqrt{5-\sqrt 5}} (u^4-5u^2)\ du$$
$$=4\left(\frac{u^5}{5}-\frac{5u^3}{3}\right)_{\sqrt 5}^{\sqrt{5-\sqrt 5}}$$
$$=4\left(\frac{(5-\sqrt 5)^{5/2}}{5}-\frac{5(5-\sqrt 5)^{3/2}}{3}-\frac{25\sqrt5}{5}+\frac{25\sqrt 5}{3}\right)$$
$$=\color{red}{\left(\frac{(5-\sqrt 5)^{5/2}}{5}-\frac{5(5-\sqrt 5)^{3/2}}{3}+\frac{10\sqrt5}{3}\right)}$$
