8
$\begingroup$

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... ?

Improve this question
$\endgroup$
7
  • 1
    $\begingroup$ Use $1-\cos(x) = 2\sin^2(\frac{x}{2})$ and $1+\cos(x) = 2 \cos^2(\frac{x}{2})$. $\endgroup$ – Sasha Sep 24 '11 at 13:38
  • $\begingroup$ 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. $\endgroup$ – Michael Hardy Sep 24 '11 at 13:44
  • $\begingroup$ 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? $\endgroup$ – Chon Sep 24 '11 at 16:00
  • $\begingroup$ 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: $\frac{\text{numerator}}{(1+u^2)^2((\sqrt{2}-1)u^2+2u+1+\sqrt{2})} = \frac{Au+B}{1+u^2} + \frac{Cu+D}{(1+u^2)^2}+\frac{Eu+F}{(\sqrt{2}-1)u^2+2u+1+\sqrt{2}}$. $\endgroup$ – Michael Hardy Sep 24 '11 at 17:03
  • $\begingroup$ ...AND: $(\sqrt{2}-1)u^2 + 2u + (\sqrt{2}+1)$ is a perfect square, since it's $\Big( \sqrt{\sqrt{2}-1}\; u + \sqrt{\sqrt{2}+1}\Big)^2$. $\endgroup$ – Michael Hardy Sep 24 '11 at 17:06
8
$\begingroup$

Would it help you greatly if you transform the integrand to

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

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ P.S. Repeated rationalization (i.e., the clever use of the difference-of-squares identity) is the key. $\endgroup$ – J. M. isn't a mathematician Sep 24 '11 at 13:46
6
$\begingroup$

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.

Improve this answer
$\endgroup$
4
$\begingroup$

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

Improve this answer
$\endgroup$
2
  • $\begingroup$ 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 $\endgroup$ – Anamitra Palit Dec 7 '11 at 8:45
  • $\begingroup$ As an alternative method of integration you may multiply the Nr and the Dr by $\cos\theta-\sin\theta$ and proceed with the substitution:$x=\cos2\theta$ $\endgroup$ – Anamitra Palit Dec 7 '11 at 9:05
3
$\begingroup$

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.

Improve this answer
$\endgroup$
2
$\begingroup$

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

$\endgroup$
0
$\begingroup$

Just to give a complete answer using the method of the square. Using the easily proved equation

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

one can get

\begin{align*} \int_{-1}^{1} \frac{\mathrm{d}x}{\sqrt{1+x} + \sqrt{1-x} + 2} &= \int_{-1}^{1} \frac{\mathrm{d}x}{\sqrt{2\left (1+\sqrt{1-x^2} \right )}+2} \\ &\!\!\!\!\!\overset{x = \sin u}{=\! =\! =\! =\! =\!} \int_{-\pi/2}^{\pi/2} \frac{\cos u}{\sqrt{2\left ( 1+\cos u \right )}+2}\, \mathrm{d}u\\ &=\int_{-\pi/2}^{\pi/2} \frac{\cos u}{\sqrt{4 \cos^2 \frac{u}{2}}+2} \, \mathrm{d}u \\ &= \int_{-\pi/2}^{\pi/2} \frac{\cos u}{2 \cos \frac{u}{2} + 2} \, \mathrm{d} u \\ &= \frac{1}{2} \int_{-\pi/2}^{\pi/2} \frac{\cos u}{1+ \cos \frac{u}{2}} \, \mathrm{d}u \\ &= \frac{1}{4} \int_{-\pi/2}^{\pi/2} \frac{\cos u}{\cos^2 \frac{u}{4}} \, \mathrm{d}u \\ &= {\require{cancel}\cancelto{0}{\frac{1}{4} \left [ 4 \tan \frac{u}{4} \cos u \right ]_{-\pi/2}^{\pi/2}}} + \int_{-\pi/2}^{\pi/2} \tan \frac{u}{4} \sin u \, \mathrm{d}u \end{align*}

The last integral can be dealt as follows:

\begin{align*} \int_{-\pi/2}^{\pi/2} \tan \frac{u}{4} \sin u \, \mathrm{d}u \; &\overset{y=u/4}{=\! =\! =\! =\!} \; 4 \int_{-\pi/8}^{\pi/8} \tan y \sin 4y \, \mathrm{d}y\\ &=4 \int_{-\pi/8}^{\pi/8} \tan y \left ( 4 \sin y \cos^3 y - 4 \sin^3 y \cos y \right )\, \mathrm{d}y \\ &=4 \int_{-\pi/8}^{\pi/8} \left ( 4 \sin^2 y \cos^2 y - 4 \sin^4 y \right )\, \mathrm{d}y \\ &= 16 \int_{-\pi/8}^{\pi/8} \left ( \sin y \cos y \right )^2 \, \mathrm{d} y - 16 \int_{-\pi/8}^{\pi/8} \sin^4 y \, \mathrm{d}y \\ &= \left (\frac{\pi}{2} - 1 \right ) + \left (-1 + 4 \sqrt{2} - \frac{3\pi}{2} \right ) \\ &= 4\sqrt{2} - \pi - 2 \end{align*}

Basically as in @RE60K's answer. The results arise in view of the identities:

$$\sin^4 x = \frac{1}{8} \left ( 3 + \cos 4x - 4 \cos 2x \right )$$

and

\begin{align*} \int_{-\pi/8}^{\pi/8} \sin^2 x \cos^2 x \, \mathrm{d}x &= \int_{-\pi/8}^{\pi/8} \left ( \sin x \cos x \right )^2 \, \mathrm{d} x\\ &=\int_{-\pi/8}^{\pi/8} \left ( \frac{\sin 2x}{2} \right )^2 \, \mathrm{d}x\\ &= \frac{1}{4} \int_{-\pi/8}^{\pi/8} \sin^2 2x \, \mathrm{d}x\\ &=\frac{1}{8}\int_{-\pi/8}^{\pi/8} \left ( 1- \cos 4x \right ) \, \mathrm{d}x \\ &= \frac{1}{8} \left ( \frac{\pi}{4} - \frac{1}{2} \right ) \\ &= \frac{\pi}{32} - \frac{1}{16} \end{align*}

I know I'm 9 years late but I wanted to present the method of squares.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.