삼각치환

We use the substitution $x=\tan u$, $\mathrm{d}x=\sec^2 u~\mathrm{d}u$.

$$\begin{aligned} \int\dfrac{\sqrt{1+x^2}^3}{x}\mathrm{d}x&=\int\frac{\sqrt{1+\tan^2 u}^3}{\tan u}\sec^2 u~\mathrm{d}u \\&=\int\frac{\sec^3u}{\tan u}\sec^2 u~\mathrm{d}u \\&=\int\frac{(\sec^2u)^2}{\tan u}\sec u~\mathrm{d}u \\&=\int\frac{(\tan^2u+1)^2}{\tan u}\sec u~\mathrm{d}u \\&=\int\frac{\tan^4u+2\tan^2 u + 1}{\tan u}\sec u~\mathrm{d}u \\&=\int \tan^3 u \sec u ~\mathrm{d}u + \int 2\sec u \tan u~\mathrm{d}u + \int \frac{\sec u}{\tan u}~\mathrm{d}u \\ \text{For the first integral, we use the substitution $w=\sec u$. The second is the antiderivative of $2\sec u$. The third we simplify as $\frac{\sec u}{\tan u} = \frac{1}{\cos u}\cdot \frac{\cos u}{\sin u} = \csc u$~.} \\ &=\int\bigg((\sec^2 u -1)\sec u \tan u\bigg)~\mathrm{d}u + 2\sec u + \log|\cot u - \csc u |+C\\ &=\int (w^2-1)~\mathrm{d}w+2\sec u + \log|\cot u - \csc u|+C\\ &=\frac{1}{3}w^3-w+2\sec u + \log|\cot u - \csc u|+C\\ &=\frac{1}{3}\sec^3u-\sec u+2\sec u + \log|\cot u - \csc u|+C \\ \smash{  } \qquad &=\frac{1}{3}\sec^3u+\sec u + \log|\cot u - \csc u|+C \\ \text{We began with the substitution $x=\tan u$. Then $\cot u = \frac{1}{x}$. To find $\csc u$ and $\sec u$, we draw a right triangle with angle $u$, opposite side $x$, and adjacent side $1$. The Pythagorean Theorem gives us the hypotenuse.} \\ &=\frac{1}{3}\sqrt{1+x^2}^3+\sqrt{1+x^2}+\log\left| \frac{1}{x}-\frac{\sqrt{1+x^2}}{x} \right|+C \\&=\frac{1}{3}\sqrt{1+x^2}(4+x^2)+\log\left|\frac{1-\sqrt{1+x^2}}{x} \right|+C \end{aligned}$$

Note that $f(x)$ is an even function, nonnegative over its entire domain.

(a) To find the area of $R$, we evaluate

$$\begin{aligned} \text{Area}&=\int _{-1/2}^{1/2} \dfrac{|x|}{\sqrt[4]{1-x^2}}~\mathrm{d}x = 2\int _{0}^{1/2} \dfrac{x}{\sqrt[4]{1-x^2}}~\mathrm{d}x \\ \text{We use the substitution $u=1-x^2$, $\mathrm{d}u=-2x~\mathrm{d}x$.} \\ &=-\int_{1}^{3/4} \frac{1}{u^{1/4}}~\mathrm{d}u\\ &=-\left[\frac{4}{3}u^{3/4}\right]_1^{3/4} = -\frac{4}{3}\left(\left(\frac{3}{4}\right)^{3/4}-1\right)\\ &=\frac{4}{3} - \sqrt[4]{\frac{4}{3}} \end{aligned}$$

(b) We slice the solid of rotation into circular disks of width $\mathrm{d}x$ and radius $\dfrac{|x|}{\sqrt[4]{1-x^2}}$.

$$\begin{aligned} \text{Volume}&=\int_{-1/2}^{1/2} \pi\left(\dfrac{|x|}{\sqrt[4]{1-x^2}}\right)^2~\mathrm{d}x\\ &=2\pi\int_{0}^{1/2} \dfrac{x^2}{\sqrt{1-x^2}}~\mathrm{d}x \\ \text{We use the substitution $x=\sin \theta$, $\mathrm{d}x = \cos\theta~\mathrm{d}\theta$, so $\sqrt{1-x^2} = \sqrt{1-\sin^2\theta}=\cos \theta$. Note $\sin 0 =0$ and $\sin\frac{\pi}{6}=\frac{1}{2}.$} \\ &=2\pi\int_{0}^{\pi/6} \frac{\sin^2 \theta}{\cos \theta}\cos \theta~\mathrm{d}\theta \\&=2\pi\int_{0}^{\pi/6} \sin^2 \theta~\mathrm{d}\theta \\&=\pi\int_{0}^{\pi/6}\big(1- \cos(2 \theta)\big)~\mathrm{d}\theta\\ &=\pi\left[\theta - \frac{1}{2}\sin(2\theta)\right]_0^{\pi/6}\\ &=\pi\left(\frac{\pi}{6} - \frac{1}{2}\cdot \frac{\sqrt{3}}{2}\right)\\ &=\frac{\pi^2}{6} - \frac{\sqrt{3}\pi}{4} \end{aligned}$$

Remember that for any value $X$, $$|X| = \left\{\begin{array}{rl} X&\text{if }X \ge 0\\ -X&\text{if }X \le 0 \end{array}\right.$$ So, $|X| \neq X$ precisely when $X<0$.

(a) The range of arcsine is $\big[-\frac{\pi}{2},\frac{\pi}{2}\big]$. So, since $u=\arcsin(x/a)$, $u$ is in the range $\big[-\frac{\pi}{2},\frac{\pi}{2}\big]$. Therefore $\cos u \geq 0$. Since $a$ is positive, $a\cos u \ge 0$, so $a\cos u = |a\cos u|$. That is, $$\sqrt{a^2-x^2}=|a\cos u|=a\cos u$$ all the time.

(b) The range of arctangent is $\big(-\frac{\pi}{2},\frac{\pi}{2}\big)$. So, since $u=\arctan(x/a)$, $u$ is in the range $\big(-\frac{\pi}{2},\frac{\pi}{2}\big)$. Therefore $\sec u = \frac{1}{\cos u} >0$. Since $a$ is positive, $a\sec u > 0$, so $a\sec u = |a\sec u|$.That is, $$\sqrt{a^2+x^2}=|a\sec u|=a\sec u$$ all the time.

(c) The range of arccosine is $\big[0,\pi \big]$. So, since $u=\mathop{\mathrm{arcsec}}(x/a) = \arccos(a/x)$, $u$ is in the range $\big[0,\pi\big]$. (Actually, it's in the range $[0,\frac{\pi}{2}) \cup (\frac{\pi}{2},\pi]$, since secant is undefined at $\pi/2$.) If $|a\tan u| \neq a\tan u$, then $\tan u <0$, which happens when $u$ is in the range $\big (\frac{\pi}{2},\pi)$. This is the same range over which $-1<\cos u <0$, and so $-1<\frac{a}{x}<0$. Since $\frac{a}{x}<0$, $a$ and $x$ have different signs, so $x<0$. Then since $-1<\frac{a}{x}$, also $x<-a$.

So, $$\sqrt{x^2-a^2} = |a\tan u| = -a\tan u \neq a\tan u$$ happens precisely when when $x< -a$.