3
$\begingroup$

How can i evaluate the following integral $$\int \frac{dx}{\sqrt{1+x}-\sqrt{1-x}}=?$$

This is taken from a definite integral where $x$ varies from $0$ to $1$.

My attempt:

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

  1. if i use $x=\sin^2\theta$ $$\int \frac{(\sqrt{1+\sin^2\theta}+\cos\theta)}{2\sin^2\theta}\sin2\theta\ d\theta=\int (\sqrt{1+\sin^2\theta}+\cos\theta)\cot\theta d\theta$$
  2. if i use $x=\tan^2\theta$ $$\int \frac{(\sec\theta-\sqrt{1-\tan^2\theta})}{2\tan^2\theta}2\tan\theta\sec^2\theta d\theta\ d\theta=\int \frac{(\sec\theta-\sqrt{1-\tan^2\theta})}{\sin\theta\cos\theta} d\theta$$

Should I use substitution $x=\sin^2\theta$ or $x=\tan^2\theta$?. I can't decide which substitution will work further. Please help me solve this integration.

Thanks

$\endgroup$
7
  • $\begingroup$ You cannot "solve" an integral. Integral is not a problem. $\endgroup$
    – Alexey
    Jul 1 '20 at 22:51
  • $\begingroup$ @Alexey sorry for my title i corrected it. you might want to answer now $\endgroup$
    – user766881
    Jul 1 '20 at 23:23
  • $\begingroup$ Just to be clear- you’re trying to evaluate $\int_{-1}^1 [\frac{dx}{\sqrt{1+x}-\sqrt{1-x}}]$. That’s undefined because the integrand is undefined at $x=0$. $\endgroup$ Jul 2 '20 at 2:57
  • $\begingroup$ @Alexey What is a problem? $\endgroup$ Jul 2 '20 at 3:08
  • $\begingroup$ @RadialArmSaw, something that can be solved :). Like a mathematical exercise. Equations can also be solved. Sums, products, integrals cannot be solved, they can be calculated/computed/evaluated. $\endgroup$
    – Alexey
    Jul 2 '20 at 8:54
7
$\begingroup$

You can split the integral into two parts $$\int \frac{(\sqrt{1+x}+\sqrt{1-x})}{2x} \, dx=\frac{1}{2}\left[\int \frac{\sqrt{1+x}}{x}\,dx+\int \frac{\sqrt{1-x}}{x}\,dx\right].$$ Solve these separately as follows: \begin{align*} \int \frac{\sqrt{1+x}}{x}\,dx & =\int \frac{t^2}{(t^2-1)}\,dt && (\text{ let } 1+x=t^2) \\ & =\int \frac{t^2-1+1}{(t^2-1)}\,dt\\ & =\int 1 \, dt+\int \frac{1}{(t^2-1)}\,dt\\ & =t+\frac{1}{2}\left[\int \frac{1}{(t-1)}\,dt-\int \frac{1}{(t+1)}\,dt\right]\\ &=t+\ln\frac{|t-1|}{|t+1|}+c\\ &=\sqrt{1+x}+\ln\frac{|\sqrt{1+x}-1|}{|\sqrt{1+x}+1|}+c\\ \end{align*} Observe that the second part is pretty much the same. If you use $x=-u$, then $$\int \frac{\sqrt{1-x}}{x}\, dx=\int \frac{\sqrt{1+u}}{u}\, du.$$ So you can write the answer without any further computation. $$\int \frac{\sqrt{1-x}}{x}\, dx=\sqrt{1\color{red}{-x}}+\ln\frac{|\sqrt{1\color{red}{-x}}-1|}{|\sqrt{1\color{red}{-x}}+1|}+c$$

$\endgroup$
2
$\begingroup$

$$\int \frac{dx}{\sqrt{1+x}-\sqrt{1-x}}=\int \frac{dx}{\sqrt{(\sqrt{1+x}-\sqrt{1-x})^2}}$$ $$=\int \frac{dx}{\sqrt{2-2\sqrt{1-x^2}}}$$ Let $x=\sin\theta\implies dx=\cos\theta d\theta$ $$=\int \frac{\cos\theta d\theta}{\sqrt{2-2\cos\theta}}$$ $$=\int \frac{\cos\theta d\theta}{\sqrt{4\sin^2\frac{\theta}{2}}}\quad \quad \left(\because \cos\theta=1-2\sin^2\frac{\theta}{2}\right)$$ $$=\int \frac{\left(1-2\sin^2\frac{\theta}{2}\right)d\theta}{2\sin\frac{\theta}{2}}$$ $$=\int \left(\frac12\csc\frac{\theta}{2}-\sin\frac{\theta}{2}\right)\ d\theta$$ $$=\ln \left|\tan\frac{\theta}{4}\right|+2\cos\frac{\theta}{2}+C$$

$\endgroup$
9
  • $\begingroup$ Why is $\sqrt{sin^2(\theta/2)} \ne |sin(\theta/2) |$? $\endgroup$
    – Koro
    Jul 1 '20 at 23:05
  • 2
    $\begingroup$ OP did mention the condition for $x$ so assuming $x$ varies from $0$ to $1$ and hence $\sin\frac{\theta}{2}, \sin\frac{\theta}{2}$ are positive hence $$\sqrt{4\sin^2\frac{\theta}{2}}=\left|2\sin\frac{\theta}{2}\right|=2\sin\frac{\theta}{2}\quad \quad \left(0<\theta<\frac{\pi}{2}\right)$$ $\endgroup$ Jul 1 '20 at 23:09
  • $\begingroup$ Indeed, $x$ is in between $-1$ and $1$ but how is it known that $x\ge 0$? $\endgroup$
    – Koro
    Jul 1 '20 at 23:11
  • 1
    $\begingroup$ for $\sqrt{1\pm x}$ to be defined we can conclude $-1<x<1$ $\endgroup$ Jul 1 '20 at 23:13
  • $\begingroup$ Ahhh, I got it. Either $0\le \theta \le \pi$ or $\pi\le \theta \le 2\pi$. In either case, $\theta/2$ is $\le \pi$ but $\ge 0$. Hence $\ sin$ is positive. $\endgroup$
    – Koro
    Jul 1 '20 at 23:14
0
$\begingroup$

With the change of variable $x=\sin 2t$, we have

$$\sqrt{1+x}-\sqrt{1-x}=\sqrt{\cos^2t+2\cos t\sin t+\sin^2t}-\sqrt{\cos^2t-2\cos t\sin t+\sin^2t}=2\sin t.$$

Then

$$\int\frac{dx}{\sqrt{1+x}-\sqrt{1-x}}=\int\frac{2\cos2t}{2\sin t}dt=\int\left(\frac1{\sin t}-2\sin t\right)dt \\=\text{arcoth}(\cos t)+2\cos t+C.$$


From the biquadratic equation

$$4\cos^2t\,(1-\cos^2t)=x^2$$ you draw $\cos t$ as a function of $x$.

$\endgroup$
0
$\begingroup$

$\begin{aligned} I &:= \int \frac{d x}{\sqrt{1+x}-\sqrt{1-x}}=\int \frac{\sqrt{1+x}+\sqrt{1-x}}{2 x} d x =\frac{1}{2}\left[ \underbrace{\int \frac{\sqrt{1+x}}{x}-d x}_{J}+\underbrace{\int \frac{\sqrt{1-x}}{x} d x}_{K}\right] \end{aligned}$ $$ \begin{aligned} J &=\int \frac{\sqrt{1+x}}{x} d x \\ &=\int \frac{1+x}{x \sqrt{1+x}} d x \\ &=\int \frac{1+x}{x} d (\sqrt{1+x}) \\ &=\int\left(\frac{1}{x}+1\right) d(\sqrt{1+x}) \\ &=\int \frac{d (\sqrt{1+x})}{(\sqrt{1+x})^{2}-1}+\sqrt{1+x} \\ &=\frac{1}{2} \ln \left|\frac{\sqrt{1+x}-1}{\sqrt{1+x}+1}\right|+\sqrt{1+x}+c_{1} \end{aligned} $$$$ K \stackrel{x\mapsto -x}{=} \sqrt{1-x}+\frac{1}{2} \ln \left|\frac{\sqrt{1-x}-1}{\sqrt{1-x}+1}\right|+c_{2} $$ Now we can conclude that $$ I=\sqrt{1-x}+\sqrt{1+x}+\frac{1}{2} \ln \left| \frac{(\sqrt{1+x}-1)(\sqrt{1-x}-1)}{(\sqrt{1+x}+1)(\sqrt{1-x}+1)}\right|+C $$

$\endgroup$

Your Answer

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