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
where we get a research paper on Nano chemistry....?
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest.
Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.?
How this robot is carried to required site of body cell.?
what will be the carrier material and how can be detected that correct delivery of drug is done
Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
Adin
why?
Adin
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
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?