Evaluating $\int {\frac{1}{2+\sqrt{1-x}+\sqrt{1+x}}}\mathrm dx$

I would like to evaluate: $$\int {\frac{1}{2+\sqrt{1-x}+\sqrt{1+x}}}\mathrm dx$$

$$\frac{1}{2+\sqrt{1-x}+\sqrt{1+x}}=\frac{\sqrt{1-x}+\sqrt{1+x}-2}{2(\sqrt{1-x^2}-1)}$$

The substitution $x \rightarrow \sin(x)$ or $\cos(x)$ can only simplify the denominator, and $x \rightarrow \sqrt{1+x}$ or $\sqrt{1-x}$ is also useless... Can you help me find a useful substitution?

$$x=\cos(2t)$$ $$\int {\frac{1}{2+\sqrt{1-x}+\sqrt{1+x}}}\mathrm dx=-\int {\frac{\sqrt{2}\sin(t)\cos(t)}{\sqrt{2}+\sin(t)+\cos(t)}}\mathrm dt$$

$$u=\tan(t/2)$$

$$-4\sqrt{2}\int \frac{u(1-u^2)}{(1+u^2)^2((\sqrt{2}-1)u^2+2u+1+\sqrt{2})}\mathrm du$$

But now it looks even more complicated... ?

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Use $1-\cos(x) = 2\sin^2(\frac{x}{2})$ and $1+\cos(x) = 2 \cos^2(\frac{x}{2})$. – Sasha Sep 24 '11 at 13:38
I tried $u = \sqrt{1-x}$, $u^2 = 1-x$, $2u\;du = -dx$. That reduced it to an expression in which only one radical appeared: $\sqrt{2-u^2}$. Then I tried $v=\sqrt{2-u^2}$, and that transformed it to exactly the same expression with $v$ in place of $u$. I'm not sure I've seen exactly that happen before, although I wouldn't be surprised if I have. – Michael Hardy Sep 24 '11 at 13:44
It seems that the integral is not really simplified after using $x=\cos(2t)$ and $u=\tan(t/2)$, as I wrote it above (if there is no mistake in my calculus)... What can I do? – Chon Sep 24 '11 at 16:00
I see you added my answer into your question. You say it "looks more complicated", but it fits right into the standard algorithm involving partial fractions: $\frac{\text{numerator}}{(1+u^2)^2((\sqrt{2}-1)u^2+2u+1+\sqrt{2})} = \frac{Au+B}{1+u^2} + \frac{Cu+D}{(1+u^2)^2}+\frac{Eu+F}{(\sqrt{2}-1)u^2+2u+1+\sqrt{2}}$. – Michael Hardy Sep 24 '11 at 17:03
...AND: $(\sqrt{2}-1)u^2 + 2u + (\sqrt{2}+1)$ is a perfect square, since it's $\Big( \sqrt{\sqrt{2}-1}\; u + \sqrt{\sqrt{2}+1}\Big)^2$. – Michael Hardy Sep 24 '11 at 17:06

$$\frac{2-(x+2) \sqrt{1-x}+(x-2) \sqrt{1+x}+2 \sqrt{1-x^2}}{2 x^2}?$$

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P.S. Repeated rationalization (i.e., the clever use of the difference-of-squares identity) is the key. – J. M. Sep 24 '11 at 13:46

The method posted by Sasha and J.M. (are they both the same thing?) should do it, but just for fun, let's try another. \begin{align} u & = \sqrt{1-x} \\ u^2 & = 1-x \\ 2u\;du & -dx \\ 2-u^2 & = 1+x \end{align} $$\int \frac{dx}{2 + \sqrt{1-x} + \sqrt{1+x}} = \int \frac{-2u\;du}{2+u + \sqrt{2-u^2}}.$$ Now write $$u = \sqrt{2}\sin\theta,\quad du = \sqrt{2}\cos\theta\;d\theta,$$ and we get $$\int\frac{-2\sqrt{2}\sin\theta\cos\theta\;d\theta}{2 + \sqrt{2}\sin\theta+\sqrt{2}\cos\theta} = \int\frac{-2\sin\theta\cos\theta\;d\theta}{\sqrt{2}+\sin\theta+\cos\theta}.$$ Finally, a tangent half-angle substitution reduces this to an integral of a rational function, and then one can use partial fractions if necessary.

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Integral= $$\int\frac{dx}{2+\sqrt{1-x}+\sqrt{1+x}}$$
$$=\int\frac{\sqrt{1-x}-\sqrt{1+x}}{(2+\sqrt{1-x}+\sqrt{1+x})(\sqrt{1-x}-\sqrt{1+x})}dx$$

Substitution:

$z=2+\sqrt{1-x}+\sqrt{1+x}$

$$dz=\frac{\sqrt{1-x}-\sqrt{1+x}}{2\sqrt{1-x^2}}dx$$

$\sqrt{1-x^2}=(1/2)(z^2-4z+2)$

$\sqrt{1-x}-\sqrt{1+x}=\sqrt{4z-z^2}$

Integral=

$$=\int\frac{z^2-4z+2}{z\sqrt{4z-z^2}}dz$$ $$=\int\frac{z-4}{\sqrt{4z-z^2}}dz+\int\frac{2dz}{z\sqrt{4z-z^2}}$$ $$=\int\frac{-(1/2)(-2z+4)}{\sqrt{4z-z^2}}dz-\int\frac{2dz}{\sqrt{4z-z^2}}+\int\frac{2dz}{z\sqrt{4z-z^2}}$$

For the third integral you may use the substitution $z=1/t$

We have,Integral

$$=-\sqrt{4z-z^2}-4\sin^{-1}\frac{\sqrt{z}}{2}-\sqrt{\frac{4-z}{z}}+C$$

Where $z=2+\sqrt{1-x}+\sqrt{1+x}$

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To get the expression for $\sqrt{1-x^2}$ square both sides of $z-2=\sqrt{1-x}+\sqrt{1+x}$. To get the expression for $\sqrt{x-1}+\sqrt{x+1}$ calculate $(\sqrt{1-x}+\sqrt{1+x})^2-(\sqrt{1-x}-\sqrt{1+x})^2$and simplify – Anamitra Palit Dec 7 '11 at 8:45
As an alternative method of integration you may multiply the Nr and the Dr by $\cos\theta-\sin\theta$ and proceed with the substitution:$x=\cos2\theta$ – Anamitra Palit Dec 7 '11 at 9:05

Let $w_+ = \sqrt{1+x}$ and $w_- = \sqrt{1-x}$. Then

$$\begin{eqnarray} \frac{1}{2+w_+ + w_-} &=& \frac{(2 - w_+ + w_- )(2 + w_+ - w_- )(2 - w_+ - w_- )}{(2 + w_+ + w_- )(2 - w_+ + w_- )(2 + w_+ - w_- )(2 - w_+ - w_- )} \\ &=& \frac{4 w_- w_+ - 2 x w_- - 4 w_- +2 x w_+ - 4 w_+ + 4}{4 x^2} \end{eqnarray}$$

This can now be integrate term-wise.

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\newcommand{\ct}[0]{\color{grey}{\text{constant}}} \newcommand{\b}[1]{\left(#1\right)} \begin{align} \int {\frac{1}{2+\sqrt{1-x}+\sqrt{1+x}}}dx&=\int\frac{-2\sin(2t)}{2+\sqrt2\sin t+\sqrt2\cos t}dt\tag{1}\\ &=-2\int\frac{\sin(2t)}{2+2\sin\b{t+\frac{\pi}4}}dt\\ &=-2\int\frac{\sin(2t)}{2+2\cos\b{\frac{\pi}4-t}}dt\\ &=-\frac12\int\sin(2t)\sec^2\b{\frac{\pi}8-\frac t2}dt\\ &=\int\sin(2\b{\pi/4-2u})\sec^2(u)du\tag{2}\\ &=\int\cos(4u)\sec^2(u)du\\ &=\cos(4u)\tan(u)+4\int\sin(4u)\tan(u)du\tag{3}\\ \int\sin(4u)\tan(u)du&=\int(2\cos(2u)-\cos(4u)-1)du\tag{4}\\ &=\sin(2u)-\frac14\sin(4u)-u+\ct\\ &=\sin\b{\frac{\pi}4-t}-\frac14\cos(2t)+\frac t2+\ct\\ &=\frac1{\sqrt2}\b{\sin(t)-\cos(t)}-\frac14\cos(2t)+\frac t2+\ct\\ &=\frac12\b{\sqrt{1-x}-\sqrt{1+x}}-\frac14x+\frac 14\arccos(x)+\ct\\ \cos(4u)\tan(u)&=\sin(2t)\tan\b{\pi/8-t/2}\\&=\sin(2t)\sqrt{\frac{1-\cos(\pi/4-t)}{1+\cos(\pi/4-t)}}\\&=\sin(2t)\sqrt{\frac{1-\sin(t+\pi/4)}{1+\sin(t+\pi/4)}}\\&=\sqrt{1-x^2}\sqrt{\frac{\sqrt2-\sqrt{1-x}-\sqrt{1+x}}{\sqrt2+\sqrt{1-x}+\sqrt{1+x}}} \end{align} So: $$\large\int \frac{1}{2+\sqrt{1-x}+\sqrt{1+x}}dx=\sqrt{1-x^2}\sqrt{\frac{\sqrt2-\sqrt{1-x}-\sqrt{1+x}}{\sqrt2+\sqrt{1-x}+\sqrt{1+x}}}+2\sqrt{1-x}-2\sqrt{1+x}-x+\arccos(x)+\color{grey}{\rm constant}$$

$(1)$ let $x=\cos(2t)\mid dx=-2\sin(2t)$
$(2)$ let $u=\pi/8-t/2\iff t=\pi/4-2u\mid dt=-2du$
$(3)$ use integration by parts
$(4)$ use \begin{align}\sin(4u)\tan(u)&=2\sin(2u)\cos(2u)\tan(u)\\&=4\sin^2(u)\cos(2u)\\&=2\cos(2u)(1-\cos(2u))\\&=2\cos(2u)-2\cos^2(2u)\\&=2\cos(2u)-\cos(4u)-1\end{align}

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