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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... ?

  • 0
    I suppose I should add that since that last one is a perfect square, the partial fraction should be $\frac{E}{\sqrt{\sqrt{2}-1}\;u + \sqrt{\sqrt{2} + 1}} + \frac{F}{(\sqrt{\sqrt{2}-1}\;u + \sqrt{\sqrt{2} + 1})^2}$.2011-09-24

5 Answers 5

7

Would it help you greatly if you transform the integrand to

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

  • 0
    P.S. Repeated rationalization (i.e., the clever use of the difference-of-squares identity) is the key.2011-09-24
6

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.

4

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

  • 0
    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$2011-12-07
3

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.

2

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