How do I integrate $\sqrt{4-\sqrt{x}}$?

Question

I'm trying to integrate $\int\sqrt{4 - \sqrt{x}} \, dx$. I thought I should do a $u$-sub with $u=\sqrt{x}$, or maybe $u=4-\sqrt{x}$, but then I get $du=\frac{1}{2\sqrt{x}}\,dx$ or $du=-\frac{1}{2\sqrt{x}} \, dx$ which isn't part of what I'm trying to integrate.

This is in the section of the book on u-subs so I'm sure we're supposed to do a u-sub, but I don't know what else to try. Please help!

Answer 1

$$I = \int \sqrt{4-\sqrt{x}} dx$$ Set $4 - \sqrt{x} = t$ i.e. $x = (4-t)^2$. We then have $$I = \int \sqrt{t} 2 (t-4)dt$$ Now you should be able to integrate this. Move your mouse over the gray area for the complete answer.

\begin{align}I & = \int \sqrt{t} 2 (t-4)dt = 2 \int t^{3/2} dt - 8 \int t^{1/2} dt = 2 \dfrac{t^{5/2}}{5/2} - 8 \dfrac{t^{3/2}}{3/2} + \text{constant}\\& = \dfrac45 (4-\sqrt{x})^{5/2} - \dfrac{16}3 (4-\sqrt{x})^{3/2} + \text{constant}\end{align}