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
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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?