How to find integral $\underbrace{\int\sqrt{2+\sqrt{2+\sqrt{2+\cdots+\sqrt{2+x}}}}}_{n}dx,x>-2$

Question

Find the integral $$\int\underbrace{\sqrt{2+\sqrt{2+\sqrt{2+\cdots+\sqrt{2+x}}}}}_{n}dx,x>-2$$

where $n$ define the number of the square

I know this if $0 \le x\le 2$, then let $$x=2\cos{t},0\le t\le\dfrac{\pi}{2}$$ so $$\sqrt{2+x}=\sqrt{2+2\cos{t}}=2\cos{\dfrac{t}{2}}$$ so $$\sqrt{2+\sqrt{2+x}}=2\cos{\dfrac{t}{2^2}}$$ so $$\int\sqrt{2+\sqrt{2+\sqrt{2+\cdots+\sqrt{2+x}}}}dx=\int2\cos{\dfrac{t}{2^n}}(-2\sin{t})dt$$

and for $x\ge 2$ case, I let $x=\cosh{t}$, but for $-2\le x\le 0$ case, I can't do it.

Answer 1

I think you have the wrong approach, take a look at this

$$f(x) = \sqrt{2+\sqrt{2+...+\sqrt{2+x}}} $$

Square both sides

$$(f(x))^2 = 2+\sqrt{2+...+\sqrt{2+x}} $$

$$(f(x))^2 = 2+f(x) $$

$$(f(x))^2 -f(x)-2= 0 $$

This has two solutions $f(x)=2$ and $f(x)=-1$

Note that $f(x)$ cannot be -1, since square root is always positive for real numbers, and it is given that $x > -2$

Hence you have $f(x)=2$, which is independent of $x$

The answer is therefore finally 2x

Answer 2

For the finitely many nested radical case, $$-4\int \cos\left(\frac{t}{2^n}\right)\sin t\,dt=-2\int \left(\sin(2^{-n}t+t)-\sin(2^{-n}t-t)\right)\,dt$$ $$=-2\left(-\frac{\cos(t(2^{-n}+1))}{2^{-n}+1}+\frac{\cos(t(2^{-n}-1))}{2^{-n}-1}\right)+C$$

Substitute $t=\arccos(x/2)$ to obtain the answer.

The substitution is valid for $-2\le x\le 2$.

I hope this addresses the issue.

Answer 3

You were too timid: For $-2\leq x\leq2$ use the substitution $$x=2\cos t\qquad(-\pi\leq t\leq 0)\ .$$ Then everything goes through as before: $$\sqrt{2+x}=\sqrt{2+2\cos t}=2\cos{t\over2},\quad \sqrt{2+\sqrt{2+x}}=\sqrt{2+\cos{t\over2}}=2\cos{t\over4}\ ,$$ etcetera.