MHB Integral Compute: $\int \sqrt{2+\sqrt{2+\cdots+\sqrt{2+x}}}\ dx$

[anemone]

Compute the integral $I=\int \sqrt{2+\sqrt{2+\cdots+\sqrt{2+x}}}\,dx$ where the expression contains $n\ge 1$ square roots.

 

[Prove It]

Compute the integral $I=\int \sqrt{2+\sqrt{2+\cdots+\sqrt{2+x}}}\,dx$ where the expression contains $n\ge 1$ square roots.

Just a comment - the integrand looks like a variation on an approximation for $\displaystyle \begin{align*} \pi \end{align*}$...

 

[anemone]

Hint:

Spoiler

Use trigonometric substitution.

 

[chisigma]

Compute the integral $I=\int \sqrt{2+\sqrt{2+\cdots+\sqrt{2+x}}}\,dx$ where the expression contains $n\ge 1$ square roots.

[sp]The solution is strictly connected to the identity...

$\displaystyle 2\ \cos \frac{\theta}{2}= \sqrt {2 + 2\ \cos \theta} = \sqrt{2 + \sqrt {2 + 2\ \cos 2 \theta}} = \sqrt{2 + \sqrt{2 + ... + \sqrt{2 + 2\ \cos 2^{n-1}\ \theta}}}\ (1)$

... and from (1) we derive...

$\displaystyle f_{n} (x) = 2\ \cos \frac{\cos^{-1} \frac{x}{2}}{2^{n}}\ (2)$

... so that with a little of patience we find...

$\displaystyle \int f_{n}(x)\ d x = \frac{2^{n+1}}{4^{n}-1}\ \{ \sqrt{4 - x^{2}}\ \sin (2^{-n}\ \cos^{-1} \frac{x}{2}) + 2^{n}\ x\ \cos (2^{-n}\ \cos^{-1} \frac{x}{2})\} + c\ (3) $ [/sp]

Kind regards

$\chi$ $\sigma$

 

[anemone]

Thank you chisigma for your solution and thanks for participating too!:)

A solution of other for sharing:

Spoiler

With the substitution $x=2\cos t$, we have

$\begin{align*}\sqrt{2+\sqrt{2+\cdots+\sqrt{2+x}}}&=\sqrt{2+\sqrt{2+\cdots+\sqrt{2+2\cos t}}}\\&=\sqrt{2+\sqrt{2+\cdots+2\cos \dfrac{t}{2}}}\\&=2\cos \dfrac{t}{2^n}\end{align*}$

The integral becomes

$\begin{align*}I&=-4\int \sin t \cos \dfrac{t}{2^n} \,dt\\&=-2\int \left(\sin \left(\dfrac{2^n+1}{2^n} \right)t -\sin \left(\dfrac{2^n-1}{2^n} \right)t \right) \,dt\\&=\dfrac{2^{n+1}}{2^n+1}\cos \left( \dfrac{2^n+1}{2^n}\cos^{-1}\dfrac{x}{2}\right)-\dfrac{2^{n+1}}{2^n-1}\cos \left( \dfrac{2^n-1}{2^n}\cos^{-1}\dfrac{x}{2}\right)+C\end{align*}$

 

[kaliprasad]

Thank you chisigma for your solution and thanks for participating too!:)

A solution of other for sharing:

Spoiler

With the substitution $x=2\cos t$, we have

$\begin{align*}\sqrt{2+\sqrt{2+\cdots+\sqrt{2+x}}}&=\sqrt{2+\sqrt{2+\cdots+\sqrt{2+2\cos t}}}\\&=\sqrt{2+\sqrt{2+\cdots+2\cos \dfrac{t}{2}}}\\&=2\cos \dfrac{t}{2^n}\end{align*}$

The integral becomes

$\begin{align*}I&=-4\int \sin t \cos \dfrac{t}{2^n} \,dt\\&=-2\int \left(\sin \left(\dfrac{2^n+1}{2^n} \right)t -\sin \left(\dfrac{2^n-1}{2^n} \right)t \right) \,dt\\&=\dfrac{2^{n+1}}{2^n+1}\cos \left( \dfrac{2^n+1}{2^n}\cos^{-1}\dfrac{x}{2}\right)-\dfrac{2^{n+1}}{2^n-1}\cos \left( \dfrac{2^n-1}{2^n}\cos^{-1}\dfrac{x}{2}\right)+C\end{align*}$

Spoiler

above is true only if $-2 \le x \le 2$ . This condition was not mentioned

 

[anemone]

Hi kaliprasad,

I think you're right, we can only tell $x\ge -2$ but we cannot prove that $x\le 2$...:(

 

[Saitama]

Hi kaliprasad,

I think you're right, we can only tell $x\ge -2$ but we cannot prove that $x\le 2$...:(

kaliprasad meant that the substitution is valid only if $-2 \le x \le 2$ and this was not mentioned in the problem, for $x \ge 2$, the substitution $x=2\cosh t$ would work.

 

[anemone]

kaliprasad meant that the substitution is valid only if $-2 \le x \le 2$ and this was not mentioned in the problem, for $x \ge 2$, the substitution $x=2\cosh t$ would work.

My point is, if we could show that $x \le 2$, we then could safely opt for that trig substitution but since we could not, then there is a flaw in the proposed solution.:)