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Use substitution to find the antiderivative of $\int \frac{dx}{25+4{x}^{2}}.$
$\frac{1}{10}\phantom{\rule{0.05em}{0ex}}{\text{tan}}^{\mathrm{-1}}\left(\frac{2x}{5}\right)+C$
Find the antiderivative of $\int \frac{1}{9+{x}^{2}}dx.$
Apply the formula with $a=3.$ Then,
Find the antiderivative of $\int \frac{dx}{16+{x}^{2}}.$
$\frac{1}{4}\phantom{\rule{0.05em}{0ex}}{\text{tan}}^{\mathrm{-1}}\left(\frac{x}{4}\right)+C$
Evaluate the definite integral ${\int}_{\sqrt{3}\text{/}3}^{\sqrt{3}}\frac{dx}{1+{x}^{2}}.$
Use the formula for the inverse tangent. We have
Evaluate the definite integral ${\int}_{0}^{2}\frac{dx}{4+{x}^{2}}}.$
$\frac{\pi}{8}$
In the following exercises, evaluate each integral in terms of an inverse trigonometric function.
$\int}_{0}^{\sqrt{3}\text{/}2}\frac{dx}{\sqrt{1-{x}^{2}}$
${\text{sin}}^{\mathrm{-1}}x{|}_{0}^{\sqrt{3}\text{/}2}=\frac{\pi}{3}$
$\int}_{\mathrm{-1}\text{/}2}^{1\text{/}2}\frac{dx}{\sqrt{1-{x}^{2}}$
$\int}_{\sqrt{3}}^{1}\frac{dx}{\sqrt{1+{x}^{2}}$
${\text{tan}}^{\mathrm{-1}}x{|}_{\sqrt{3}}^{1}=-\frac{\pi}{12}$
$\int}_{1\text{/}\sqrt{3}}^{\sqrt{3}}\frac{dx}{1+{x}^{2}$
$\int}_{1}^{\sqrt{2}}\frac{dx}{\left|x\right|\sqrt{{x}^{2}-1}$
${\text{sec}}^{\mathrm{-1}}x{|}_{1}^{\sqrt{2}}=\frac{\pi}{4}$
$\int}_{1}^{2\text{/}\sqrt{3}}\frac{dx}{\left|x\right|\sqrt{{x}^{2}-1}$
In the following exercises, find each indefinite integral, using appropriate substitutions.
$\int \frac{dx}{\sqrt{9-{x}^{2}}}$
${\text{sin}}^{\mathrm{-1}}\left(\frac{x}{3}\right)+C$
$\int \frac{dx}{\sqrt{1-16{x}^{2}}}$
$\int \frac{dx}{9+{x}^{2}}$
$\frac{1}{3}\phantom{\rule{0.05em}{0ex}}{\text{tan}}^{\mathrm{-1}}\left(\frac{x}{3}\right)+C$
$\int \frac{dx}{25+16{x}^{2}}$
$\int \frac{dx}{\left|x\right|\sqrt{{x}^{2}-9}}$
$\frac{1}{3}\phantom{\rule{0.05em}{0ex}}{\text{sec}}^{\mathrm{-1}}\left(\frac{x}{3}\right)+C$
$\int \frac{dx}{\left|x\right|\sqrt{4{x}^{2}-16}}$
Explain the relationship $\text{\u2212}{\text{cos}}^{\mathrm{-1}}t+C={\displaystyle \int \frac{dt}{\sqrt{1-{t}^{2}}}={\text{sin}}^{\mathrm{-1}}t+C.}$ Is it true, in general, that ${\text{cos}}^{\mathrm{-1}}t=\text{\u2212}{\text{sin}}^{\mathrm{-1}}t?$
$\text{cos}\left(\frac{\pi}{2}-\theta \right)=\text{sin}\phantom{\rule{0.1em}{0ex}}\theta .$ So, ${\text{sin}}^{\mathrm{-1}}t=\frac{\pi}{2}-{\text{cos}}^{\mathrm{-1}}t.$ They differ by a constant.
Explain the relationship ${\text{sec}}^{\mathrm{-1}}t+C={\displaystyle \int \frac{dt}{\left|t\right|\sqrt{{t}^{2}-1}}=\text{\u2212}{\text{csc}}^{\mathrm{-1}}t+C.}$ Is it true, in general, that ${\text{sec}}^{\mathrm{-1}}t=\text{\u2212}{\text{csc}}^{\mathrm{-1}}t?$
Explain what is wrong with the following integral: ${\int}_{1}^{2}\frac{dt}{\sqrt{1-{t}^{2}}}.$
$\sqrt{1-{t}^{2}}$ is not defined as a real number when $t>1.$
Explain what is wrong with the following integral: ${\int}_{\mathrm{-1}}^{1}\frac{dt}{\left|t\right|\sqrt{{t}^{2}-1}}.$
In the following exercises, solve for the antiderivative $\int f$ of f with $C=0,$ then use a calculator to graph f and the antiderivative over the given interval $\left[a,b\right].$ Identify a value of C such that adding C to the antiderivative recovers the definite integral $F\left(x\right)={\displaystyle {\int}_{a}^{x}f\left(t\right)dt.}$
[T] $\int \frac{1}{\sqrt{9-{x}^{2}}}dx$ over $\left[\mathrm{-3},3\right]$
The antiderivative is
${\text{sin}}^{\mathrm{-1}}\left(\frac{x}{3}\right)+C.$ Taking
$C=\frac{\pi}{2}$ recovers the definite integral.
[T] $\int \frac{9}{9+{x}^{2}}dx$ over $\left[\mathrm{-6},6\right]$
[T] $\int \frac{\text{cos}\phantom{\rule{0.1em}{0ex}}x}{4+{\text{sin}}^{2}x}dx$ over $\left[\mathrm{-6},6\right]$
The antiderivative is
$\frac{1}{2}\phantom{\rule{0.05em}{0ex}}{\text{tan}}^{\mathrm{-1}}\left(\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}x}{2}\right)+C.$ Taking
$C=\frac{1}{2}\phantom{\rule{0.05em}{0ex}}{\text{tan}}^{\mathrm{-1}}\left(\frac{\text{sin}\left(6\right)}{2}\right)$ recovers the definite integral.
[T] $\int \frac{{e}^{x}}{1+{e}^{2x}}dx$ over $\left[\mathrm{-6},6\right]$
In the following exercises, compute the antiderivative using appropriate substitutions.
$\int \frac{{\text{sin}}^{\mathrm{-1}}tdt}{\sqrt{1-{t}^{2}}}$
$\frac{1}{2}{\left({\text{sin}}^{\mathrm{-1}}t\right)}^{2}+C$
$\int \frac{dt}{{\text{sin}}^{\mathrm{-1}}t\sqrt{1-{t}^{2}}}$
$\int \frac{{\text{tan}}^{\mathrm{-1}}\left(2t\right)}{1+4{t}^{2}}dt$
$\frac{1}{4}{\left({\text{tan}}^{\mathrm{-1}}\left(2t\right)\right)}^{2}$
$\int \frac{t{\text{tan}}^{\mathrm{-1}}\left({t}^{2}\right)}{1+{t}^{4}}dt$
$\int \frac{{\text{sec}}^{\mathrm{-1}}\left(\frac{t}{2}\right)}{\left|t\right|\sqrt{{t}^{2}-4}}dt$
$\frac{1}{4}\left({\text{sec}}^{\mathrm{-1}}{\left(\frac{t}{2}\right)}^{2}\right)+C$
$\int \frac{t{\text{sec}}^{\mathrm{-1}}\left({t}^{2}\right)}{{t}^{2}\sqrt{{t}^{4}-1}}dt$
In the following exercises, use a calculator to graph the antiderivative $\int f$ with $C=0$ over the given interval $\left[a,b\right].$ Approximate a value of C , if possible, such that adding C to the antiderivative gives the same value as the definite integral $F\left(x\right)={\displaystyle {\int}_{a}^{x}f\left(t\right)dt.}$
[T] $\int \frac{1}{x\sqrt{{x}^{2}-4}}dx$ over $\left[2,6\right]$
The antiderivative is
$\frac{1}{2}\phantom{\rule{0.05em}{0ex}}{\text{sec}}^{\mathrm{-1}}\left(\frac{x}{2}\right)+C.$ Taking
$C=0$ recovers the definite integral over
$\left[2,6\right].$
[T] $\int \frac{1}{\left(2x+2\right)\sqrt{x}}dx$ over $\left[0,6\right]$
[T] $\int \frac{\left(\text{sin}\phantom{\rule{0.1em}{0ex}}x+x\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}x\right)}{1+{x}^{2}{\text{sin}}^{2}x}dx$ over $\left[\mathrm{-6},6\right]$
The general antiderivative is
${\text{tan}}^{\mathrm{-1}}\left(x\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}x\right)+C.$ Taking
$C=\text{\u2212}{\text{tan}}^{\mathrm{-1}}(6\phantom{\rule{0.1em}{0ex}}\text{sin}\left(6\right))$ recovers the definite integral.
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