Check to see whether the integral can be evaluated easily by using another method. In some cases, it is more convenient to use an alternative method.
Substitute
$x=a\phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}\theta $ and
$dx=a\phantom{\rule{0.1em}{0ex}}{\text{sec}}^{2}\theta d\theta .$ This substitution yields
$\sqrt{{a}^{2}+{x}^{2}}=\sqrt{{a}^{2}+{(a\phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}\theta )}^{2}}=\sqrt{{a}^{2}(1+{\text{tan}}^{2}\theta )}=\sqrt{{a}^{2}{\text{sec}}^{2}\theta}=\left|a\phantom{\rule{0.1em}{0ex}}\text{sec}\phantom{\rule{0.1em}{0ex}}\theta \right|=a\phantom{\rule{0.1em}{0ex}}\text{sec}\phantom{\rule{0.1em}{0ex}}\theta .$ (Since
$-\frac{\pi}{2}<\theta <\frac{\pi}{2}$ and
$\text{sec}\phantom{\rule{0.1em}{0ex}}\theta >0$ over this interval,
$\left|a\phantom{\rule{0.1em}{0ex}}\text{sec}\phantom{\rule{0.1em}{0ex}}\theta \right|=a\phantom{\rule{0.1em}{0ex}}\text{sec}\phantom{\rule{0.1em}{0ex}}\theta .)$
Simplify the expression.
Evaluate the integral using techniques from the section on trigonometric integrals.
Use the reference triangle from
[link] to rewrite the result in terms of
$x.$ You may also need to use some trigonometric identities and the relationship
$\theta ={\text{tan}}^{\mathrm{-1}}\left(\frac{x}{a}\right).$ (
Note : The reference triangle is based on the assumption that
$x>0;$ however, the trigonometric ratios produced from the reference triangle are the same as the ratios for which
$x\le 0.)$
Integrating an expression involving
$\sqrt{{a}^{2}+{x}^{2}}$
Evaluate
$\int \frac{dx}{\sqrt{1+{x}^{2}}}$ and check the solution by differentiating.
Begin with the substitution
$x=\text{tan}\phantom{\rule{0.1em}{0ex}}\theta $ and
$dx={\text{sec}}^{2}\theta d\theta .$ Since
$\text{tan}\phantom{\rule{0.1em}{0ex}}\theta =x,$ draw the reference triangle in the following figure.
Since
$\sqrt{1+{x}^{2}}+x>0$ for all values of
$x,$ we could rewrite
$\text{ln}\left|\sqrt{1+{x}^{2}}+x\right|+C=\text{ln}\left(\sqrt{1+{x}^{2}}+x\right)+C,$ if desired.
Evaluating
$\int \frac{dx}{\sqrt{1+{x}^{2}}}$ Using a different substitution
Use the substitution
$x=\text{sinh}\phantom{\rule{0.1em}{0ex}}\theta $ to evaluate
$\int \frac{dx}{\sqrt{1+{x}^{2}}}.$
Because
$\text{sinh}\phantom{\rule{0.1em}{0ex}}\theta $ has a range of all real numbers, and
$1+{\text{sinh}}^{2}\theta ={\text{cosh}}^{2}\theta ,$ we may also use the substitution
$x=\text{sinh}\phantom{\rule{0.1em}{0ex}}\theta $ to evaluate this integral. In this case,
$dx=\text{cosh}\phantom{\rule{0.1em}{0ex}}\theta d\theta .$ Consequently,
To evaluate this integral, use the substitution
$x=\frac{1}{2}\text{tan}\phantom{\rule{0.1em}{0ex}}\theta $ and
$dx=\frac{1}{2}{\text{sec}}^{2}\theta d\theta .$ We also need to change the limits of integration. If
$x=0,$ then
$\theta =0$ and if
$x=\frac{1}{2},$ then
$\theta =\frac{\pi}{4}.$ Thus,
Rewrite
${{\displaystyle \int}}^{\text{}}{x}^{3}\sqrt{{x}^{2}+4}\phantom{\rule{0.1em}{0ex}}dx$ by using a substitution involving
$\text{tan}\phantom{\rule{0.1em}{0ex}}\theta .$
The domain of the expression
$\sqrt{{x}^{2}-{a}^{2}}$ is
$\left(\text{\u2212}\infty ,\text{\u2212}a\right]\cup \left[a,\text{+}\infty \right).$ Thus, either
$x<\text{\u2212}a$ or
$x>a.$ Hence,
$\frac{x}{a}\le -1$ or
$\frac{x}{a}\ge 1.$ Since these intervals correspond to the range of
$\text{sec}\phantom{\rule{0.1em}{0ex}}\theta $ on the set
$\left[0,\frac{\pi}{2}\right)\cup \left(\frac{\pi}{2},\pi \right],$ it makes sense to use the substitution
$\text{sec}\phantom{\rule{0.1em}{0ex}}\theta =\frac{x}{a}$ or, equivalently,
$x=a\phantom{\rule{0.1em}{0ex}}\text{sec}\phantom{\rule{0.1em}{0ex}}\theta ,$ where
$0\le \theta <\frac{\pi}{2}$ or
$\frac{\pi}{2}<\theta \le \pi .$ The corresponding substitution for
$dx$ is
$dx=a\phantom{\rule{0.1em}{0ex}}\text{sec}\phantom{\rule{0.1em}{0ex}}\theta \phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}\theta d\theta .$ The procedure for using this substitution is outlined in the following problem-solving strategy.
Questions & Answers
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
In this morden time nanotechnology used in many field .
1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc
2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc
3- Atomobile -MEMS, Coating on car etc.
and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change .
maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
how did you get the value of 2000N.What calculations are needed to arrive at it
Leaves accumulate on the forest floor at a rate of 2 g/cm2/yr and also decompose at a rate of 90% per year. Write a differential equation governing the number of grams of leaf litter per square centimeter of forest floor, assuming at time 0 there is no leaf litter on the ground. Does this amount approach a steady value? What is that value?
You have a cup of coffee at temperature 70°C, which you let cool 10 minutes before you pour in the same amount of milk at 1°C as in the preceding problem. How does the temperature compare to the previous cup after 10 minutes?