07. 여러 가지 함수Transcendental Functions

Solution.

(1) Since $\sin(\pi/3) = \sqrt{3}/2$ and $\pi/3 \in [-\pi/2, \pi/2]$, we get $\arcsin\left(\dfrac{\sqrt{3}}{2}\right) = \dfrac{\pi}{3}$.

(2) Since $\cos(2\pi/3) = -1/2$ and $2\pi/3 \in [0,\pi]$, we get $\arccos\left(-\dfrac{1}{2}\right) = \dfrac{2\pi}{3}$.

(3) Since $\csc(\pi/4) = 1/\sin(\pi/4) = \sqrt{2}$ and $\pi/4$ lies in the range $[-\pi/2,0)\cup(0,\pi/2]$ of $\csc^{-1}$, we get $\csc^{-1}\sqrt{2} = \dfrac{\pi}{4}$.

Solution.

(1) Since $\sin^{-1}\left(-\dfrac{\sqrt{3}}{2}\right) = -\dfrac{\pi}{3}$, $$\cot\left(\sin^{-1}\left(-\dfrac{\sqrt{3}}{2}\right)\right) = \cot\left(-\dfrac{\pi}{3}\right) = \frac{\cos(-\pi/3)}{\sin(-\pi/3)} = \frac{1/2}{-\sqrt{3}/2} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3}.$$

(2) The function $\sec^{-1}x$ has range $[0,\pi/2)\cup(\pi/2,\pi]$. Since $\sec^{-1}x = \cos^{-1}(1/x)$ and $1/x \to 0-$ as $x \to -\infty$, $$\lim_{x \to -\infty}\sec^{-1}x = \lim_{x\to-\infty}\cos^{-1}\frac1x = \cos^{-1}(0) = \frac{\pi}{2}.$$

Note. Read literally, the turn from $C$ toward $A$ is exactly $180°$ (see the solution below). However, the intent of the original (Thomas) appears to be the turn needed to head toward the destination $B$, in which case the turn angle is $\sin^{-1}\dfrac{12}{180}+\tan^{-1}\dfrac{12}{240-\sqrt{180^2-12^2}}\approx 15.06°$.

Solution. Let $F$ be the foot of the perpendicular from $C$ to line $AB$. Then triangle $AFC$ is a right triangle with $\angle F = 90°$, $\overline{FC} = 12$, and $\overline{AC} = 180$. (Since the angle between segments $AB$ and $AC$ is acute, $F$ lies on segment $AB$.) Hence the angle between the two straight paths, $\angle BAC$, satisfies $$\sin(\angle BAC) = \frac{\overline{FC}}{\overline{AC}} = \frac{12}{180} = \frac{1}{15},$$ so $$\angle BAC = \sin^{-1}\frac{1}{15} \approx 0.0667\,\text{rad} \approx 3.82°.$$

Since the plane flew straight from $A$ to $C$ without changing direction, its heading at $C$ is the direction of the vector $\overrightarrow{AC}$. To reach point $A$, it must change its heading to the direction of $\overrightarrow{CA} = -\overrightarrow{AC}$, which is exactly opposite its current heading. Hence the angle through which the plane must turn at $C$ to reach $A$ is $$180°.$$

Solution.

(1) $\dfrac{d}{dx}\arcsin(x^2) = \dfrac{1}{\sqrt{1-(x^2)^2}}\cdot 2x = \dfrac{2x}{\sqrt{1-x^4}}$.

(2) $\dfrac{d}{dx}\arctan(5x^4) = \dfrac{1}{1+(5x^4)^2}\cdot 20x^3 = \dfrac{20x^3}{1+25x^8}$.

Solution. With $u = \sqrt{x^2-1}$, $$\frac{d}{dx}\tan^{-1}\sqrt{x^2-1} = \frac{1}{1+u^2}\cdot\frac{x}{\sqrt{x^2-1}} = \frac{1}{1+(x^2-1)}\cdot\frac{x}{\sqrt{x^2-1}} = \frac{1}{x^2}\cdot\frac{x}{\sqrt{x^2-1}} = \frac{1}{x\sqrt{x^2-1}},$$ and since $x \gt 1$, $$\frac{d}{dx}\csc^{-1}x = -\frac{1}{|x|\sqrt{x^2-1}} = -\frac{1}{x\sqrt{x^2-1}}.$$ Hence $$y' = \frac{1}{x\sqrt{x^2-1}} - \frac{1}{x\sqrt{x^2-1}} = 0.$$ (In fact, $\tan^{-1}\sqrt{x^2-1} + \csc^{-1}x$ is constant for $x \gt 1$, equal to $\pi/2$: for instance, at $x=\sqrt{2}$, $\tan^{-1}1 + \csc^{-1}\sqrt{2} = \pi/4+\pi/4 = \pi/2$.)

Solution. Differentiating $y\cos^{-1}(xy) = -\dfrac{3\sqrt{2}}{4}\pi$ implicitly with respect to $x$, $$y'\cos^{-1}(xy) + y\cdot\left(-\frac{1}{\sqrt{1-(xy)^2}}\right)(y+xy') = 0.$$ At $P(1/2,-\sqrt{2})$, $xy = -\dfrac{\sqrt{2}}{2}$, so $$\cos^{-1}(xy) = \cos^{-1}\left(-\frac{\sqrt{2}}{2}\right) = \frac{3\pi}{4}, \qquad 1-(xy)^2 = \frac{1}{2}, \qquad \sqrt{1-(xy)^2} = \frac{\sqrt{2}}{2}.$$ Substituting, $$\frac{3\pi}{4}y' - \frac{y(y+xy')}{\sqrt{2}/2} = 0,$$ and plugging in $y=-\sqrt{2}$, $x=1/2$, $$\frac{3\pi}{4}y' + \sqrt{2}\cdot\frac{-\sqrt{2}+\frac{1}{2}y'}{\sqrt{2}/2} = 0 \implies \frac{3\pi}{4}y' + \left(-2\sqrt{2}+y'\right) = 0 \implies \left(\frac{3\pi}{4}+1\right)y' = 2\sqrt{2}.$$ Hence $$y' = \frac{2\sqrt{2}}{\frac{3\pi+4}{4}} = \frac{8\sqrt{2}}{3\pi+4},$$ so the slope of the tangent line at $P$ is $\dfrac{8\sqrt{2}}{4+3\pi}$.

Solution.

(1) With $u=5x$, $u\to0$ as $x\to0$, so $$\lim_{x\to0}\frac{\sin^{-1}5x}{x} = 5\lim_{u\to0}\frac{\sin^{-1}u}{u} = 5\cdot1 = 5,$$ using $\lim_{u\to0}\sin^{-1}u/u = 1$, since this is the derivative of $\sin^{-1}u$ at $u=0$.

(2) With $t=2/x$, $t\to0$ as $x\to\infty$, so $$\lim_{x\to\infty}x\tan^{-1}\frac{2}{x} = 2\lim_{t\to0}\frac{\tan^{-1}t}{t} = 2\cdot1 = 2.$$

(3) As $x\to0+$, $\tan^{-1}\sqrt{x} \sim \sqrt{x}$, so $(\tan^{-1}\sqrt{x})^2 \sim x$, and $x\sqrt{x+1}\sim x$. Hence $$\lim_{x\to0+}\frac{(\tan^{-1}\sqrt{x})^2}{x\sqrt{x+1}} = \left(\lim_{x\to0+}\left(\frac{\tan^{-1}\sqrt{x}}{\sqrt{x}}\right)^2\right)\cdot\frac{1}{\sqrt{0+1}} = 1^2\cdot1 = 1.$$

Solution. Let $\alpha, \beta$ be the angles of elevation from $C(x,0)$ to $A$ and $B$, respectively, measured from the positive $x$-axis: $$\alpha = \tan^{-1}\frac{1}{x}, \qquad \beta = \tan^{-1}\frac{5}{x},$$ so $\angle ACB = \beta - \alpha$. Let $f(x) = \tan^{-1}\dfrac{5}{x} - \tan^{-1}\dfrac{1}{x}$ for $x\gt0$. Then $$f'(x) = \frac{1}{1+(5/x)^2}\cdot\left(-\frac{5}{x^2}\right) - \frac{1}{1+(1/x)^2}\cdot\left(-\frac{1}{x^2}\right) = -\frac{5}{x^2+25} + \frac{1}{x^2+1}.$$ Setting $f'(x)=0$, $$\frac{1}{x^2+1} = \frac{5}{x^2+25} \implies x^2+25 = 5(x^2+1) \implies 4x^2 = 20 \implies x^2=5,$$ so $x=\sqrt{5}$ since $x\gt0$. As $x\to0+$ and $x\to\infty$, $f(x)\to0$, and $f'$ changes sign from positive to negative near $x=\sqrt5$, so $x=\sqrt5$ gives the maximum of $f$. Hence the value of $x$ that maximizes $\angle ACB$ is $$x=\sqrt{5}.$$

Solution. Since $y=\sec^{-1}x$ satisfies $y=0$ at $x=1$, the region bounded by the curve, the $x$-axis, and the line $x=2$ lies over $1\le x\le2$. By the disk method, $$V = \pi\int_1^2(\sec^{-1}x)^2\,dx.$$ Substituting $x=\sec t$ $(0\le t\le\pi/3)$, so $dx=\sec t\tan t\,dt$ and $\sec^{-1}x=t$, $$V = \pi\int_0^{\pi/3}t^2\sec t\tan t\,dt.$$ Integrating by parts with $u=t^2$, $dv=\sec t\tan t\,dt$, $v=\sec t$, $$\int_0^{\pi/3}t^2\sec t\tan t\,dt = \Bigl[t^2\sec t\Bigr]_0^{\pi/3} - \int_0^{\pi/3}2t\sec t\,dt = \frac{2\pi^2}{9} - 2\int_0^{\pi/3}t\sec t\,dt.$$ However, $\int t\sec t\,dt$ has no elementary closed form (it requires special functions such as the dilogarithm), so this volume cannot be expressed in closed form and must be evaluated numerically. Numerical integration (e.g., Simpson's Rule) gives $$\int_1^2(\sec^{-1}x)^2\,dx \approx 0.6563,$$ so $$V \approx \pi\times0.6563 \approx 2.0618.$$

Solution. With $A(a,0)$, $B(0,b)$ and $a^2+b^2=9$ ($a,b\gt0$), revolving segment $AB$ about the $y$-axis produces a cone of base radius $a$ and height $b$, so its volume is $$V = \frac{1}{3}\pi a^2 b.$$ Substituting $a^2=9-b^2$, $$V(b) = \frac{\pi}{3}(9-b^2)b = \frac{\pi}{3}(9b-b^3) \qquad (0\lt b\lt3),$$ so $$V'(b) = \frac{\pi}{3}(9-3b^2) = 0 \implies b^2=3 \implies b=\sqrt{3}.$$ Since $V''(b) = \frac{\pi}{3}(-6b) \lt 0$ for $b\gt0$, $V$ attains its maximum at $b=\sqrt3$, where $a^2=9-3=6$, so $a=\sqrt6$. In the right triangle $OAB$ (right angle at $O$), $$\tan(\angle OBA) = \frac{\overline{OA}}{\overline{OB}} = \frac{a}{b} = \frac{\sqrt6}{\sqrt3} = \sqrt2,$$ so when the volume of the cone is maximized, $$\angle OBA = \tan^{-1}\sqrt2 \; (\approx 54.7°).$$

Solution.

(1) $$\int_{\sqrt2/2}^{\sqrt3/2}\frac{dx}{\sqrt{1-x^2}} = \Bigl[\sin^{-1}x\Bigr]_{\sqrt2/2}^{\sqrt3/2} = \sin^{-1}\frac{\sqrt3}{2} - \sin^{-1}\frac{\sqrt2}{2} = \frac{\pi}{3}-\frac{\pi}{4} = \frac{\pi}{12}.$$

(2) $$\int\frac{dx}{\sqrt{3-4x^2}} = \int\frac{dx}{2\sqrt{(\sqrt3/2)^2-x^2}} = \frac{1}{2}\sin^{-1}\frac{2x}{\sqrt3}+C.$$

(3) Substituting $u=e^x$, so $du=e^x\,dx=u\,dx$ and $dx=du/u$, $$\int\frac{dx}{\sqrt{e^{2x}-6}} = \int\frac{du}{u\sqrt{u^2-6}} = \frac{1}{\sqrt6}\sec^{-1}\frac{u}{\sqrt6}+C = \frac{1}{\sqrt6}\sec^{-1}\frac{e^x}{\sqrt6}+C.$$ If $\sec\theta = u/\sqrt6$, the opposite side of the corresponding right triangle is $\sqrt{u^2-6}$, so $\tan\theta = \sqrt{u^2-6}/\sqrt6$, which represents the same angle as $\theta = \tan^{-1}\bigl(\sqrt{u^2-6}/\sqrt6\bigr)$. Hence we may also write $$\int\frac{dx}{\sqrt{e^{2x}-6}} = \frac{1}{\sqrt6}\tan^{-1}\frac{\sqrt{e^{2x}-6}}{\sqrt6}+C.$$

Solution.

(1) Since $4x-x^2 = 4-(x-2)^2$, $$\int\frac{dx}{\sqrt{4x-x^2}} = \int\frac{dx}{\sqrt{4-(x-2)^2}} = \sin^{-1}\left(\frac{x-2}{2}\right)+C = \sin^{-1}\left(\frac{x}{2}-1\right)+C.$$

(2) Since $4x^2+4x+2 = 4\left(x+\dfrac12\right)^2+1$, with $w=x+1/2$, $$\int\frac{dx}{4x^2+4x+2} = \int\frac{dw}{4w^2+1} = \int\frac{dw}{1+(2w)^2} = \frac{1}{2}\tan^{-1}(2w)+C = \frac{1}{2}\tan^{-1}(2x+1)+C.$$

(3) Since $\dfrac{d}{dt}(t^2-6t+10) = 2t-6$ and $t-2 = \dfrac12(2t-6)+1$, $$\int\frac{t-2}{t^2-6t+10}\,dt = \frac12\int\frac{2t-6}{t^2-6t+10}\,dt + \int\frac{dt}{(t-3)^2+1} = \frac12\ln(t^2-6t+10) + \tan^{-1}(t-3)+C.$$

(4) Substituting $u=\sin^{-1}x$, so $du=dx/\sqrt{1-x^2}$, $$\int\frac{e^{\sin^{-1}x}}{\sqrt{1-x^2}}\,dx = \int e^u\,du = e^u+C = e^{\sin^{-1}x}+C.$$

(5) Substituting $u=\tan^{-1}\sqrt{x}$, $$\frac{du}{dx} = \frac{1}{1+x}\cdot\frac{1}{2\sqrt{x}} = \frac{1}{2\sqrt{x}(x+1)},$$ so $\dfrac{dx}{\sqrt{x}(x+1)} = 2\,du$. Hence $$\int\frac{dx}{\sqrt{x}(x+1)\bigl((\tan^{-1}\sqrt{x})^2+9\bigr)} = \int\frac{2\,du}{u^2+9} = \frac{2}{3}\tan^{-1}\frac{u}{3}+C = \frac{2}{3}\tan^{-1}\left(\frac{\tan^{-1}\sqrt{x}}{3}\right)+C.$$

(6) Integrating by parts with $u=\tan^{-1}x$, $dv=dx/x^2$, $v=-1/x$, $$\int\frac{\tan^{-1}x}{x^2}\,dx = -\frac{\tan^{-1}x}{x} + \int\frac{dx}{x(1+x^2)},$$ and by the partial fraction decomposition $\dfrac{1}{x(1+x^2)} = \dfrac1x - \dfrac{x}{1+x^2}$, $$\int\frac{dx}{x(1+x^2)} = \ln|x| - \frac12\ln(x^2+1)+C.$$ Hence $$\int\frac{\tan^{-1}x}{x^2}\,dx = \ln x - \frac12\ln(x^2+1) - \frac{\tan^{-1}x}{x}+C.$$

(7) Integrating by parts with $u=(\sin^{-1}x)^2$, $dv=dx$, $v=x$, $$\int(\sin^{-1}x)^2\,dx = x(\sin^{-1}x)^2 - 2\int\frac{x\sin^{-1}x}{\sqrt{1-x^2}}\,dx.$$ Integrating by parts again with $p=\sin^{-1}x$, $dq = \dfrac{x}{\sqrt{1-x^2}}\,dx$, $q=-\sqrt{1-x^2}$, $$\int\frac{x\sin^{-1}x}{\sqrt{1-x^2}}\,dx = -\sqrt{1-x^2}\sin^{-1}x + \int dx = -\sqrt{1-x^2}\sin^{-1}x + x+C.$$ Hence $$\int(\sin^{-1}x)^2\,dx = x(\sin^{-1}x)^2 + 2\sqrt{1-x^2}\sin^{-1}x - 2x+C.$$

(8) Integrating by parts with $u=\ln(1+x^2)$, $dv=dx$, $v=x$, $$\int\ln(1+x^2)\,dx = x\ln(1+x^2) - \int\frac{2x^2}{1+x^2}\,dx,$$ and since $\dfrac{2x^2}{1+x^2} = 2-\dfrac{2}{1+x^2}$, $$\int\frac{2x^2}{1+x^2}\,dx = 2x-2\tan^{-1}x+C.$$ Hence $$\int\ln(1+x^2)\,dx = x\ln(1+x^2) - 2x + 2\tan^{-1}x+C.$$

Solution.

(1) $\sinh(\ln2) = \dfrac{e^{\ln2}-e^{-\ln2}}{2} = \dfrac{2-1/2}{2} = \dfrac34$.

(2) $\dfrac{d}{dx}\left(6\sinh\dfrac{x}{3}\right) = 6\cosh\dfrac{x}{3}\cdot\dfrac13 = 2\cosh\dfrac{x}{3}$.

(3) $\dfrac{d}{dt}\left(\tanh\sqrt{1+t^2}\right) = \operatorname{sech}^2\!\sqrt{1+t^2}\cdot\dfrac{t}{\sqrt{1+t^2}} = \dfrac{t\operatorname{sech}^2\sqrt{1+t^2}}{\sqrt{1+t^2}}$.

(4) $\displaystyle\int\coth(5x)\,dx = \frac15\ln|\sinh5x|+C$.

(5) Since $\sinh^2x = \dfrac{\cosh2x-1}{2}$, $$\int_0^1\sinh^2x\,dx = \int_0^1\frac{\cosh2x-1}{2}\,dx = \left[\frac{\sinh2x}{4}-\frac{x}{2}\right]_0^1 = \frac{\sinh2}{4}-\frac12.$$

(6) Since $4e^x\sinh x = 4e^x\cdot\dfrac{e^x-e^{-x}}{2} = 2e^{2x}-2$, $$\int_0^{\ln2}4e^x\sinh x\,dx = \int_0^{\ln2}(2e^{2x}-2)\,dx = \Bigl[e^{2x}-2x\Bigr]_0^{\ln2} = (4-2\ln2)-1 = 3-2\ln2 = 3-\ln4.$$

(7) The derivative of $\operatorname{csch}^{-1}x$ is $\dfrac{d}{dx}\operatorname{csch}^{-1}x = -\dfrac{1}{|x|\sqrt{1+x^2}}$ for $x\ne0$. With $x=2^\theta\gt0$ and $\dfrac{d}{d\theta}2^\theta = 2^\theta\ln2$, $$\frac{d}{d\theta}\operatorname{csch}^{-1}2^\theta = -\frac{1}{2^\theta\sqrt{1+2^{2\theta}}}\cdot2^\theta\ln2 = -\frac{\ln2}{\sqrt{2^{2\theta}+1}}.$$

(8) Since $3+4x^2 = 4\left(x^2+\dfrac34\right)$, $$\int_0^1\frac{2\,dx}{\sqrt{3+4x^2}} = \int_0^1\frac{dx}{\sqrt{x^2+(\sqrt3/2)^2}} = \left[\sinh^{-1}\frac{2x}{\sqrt3}\right]_0^1 = \sinh^{-1}\frac{2}{\sqrt3}.$$

Solution. Since $\cosh^2x-\sinh^2x=1$, $$\sinh^2x = \cosh^2x-1 = \left(\frac{17}{15}\right)^2-1 = \frac{289-225}{225} = \frac{64}{225},$$ and since $x\gt0$, $\sinh x\gt0$, so $$\sinh x = \frac{8}{15}.$$ Also, $$\tanh x = \frac{\sinh x}{\cosh x} = \frac{8/15}{17/15} = \frac{8}{17}.$$

Solution.

(1) $2\cosh(\ln x) = 2\cdot\dfrac{e^{\ln x}+e^{-\ln x}}{2} = e^{\ln x}+e^{-\ln x} = x+\dfrac1x$.

(2) By the identity $\cosh u+\sinh u = e^u$, $\cosh(5x)+\sinh(5x) = e^{5x}$.

(3) Since $\sinh x+\cosh x = e^x$, $(\sinh x+\cosh x)^4 = (e^x)^4 = e^{4x}$.

Solution.

(1) Since $\sinh^{-1}x=\ln\bigl(x+\sqrt{x^2+1}\bigr)$, $$\sinh^{-1}\frac{5}{12} = \ln\left(\frac{5}{12}+\sqrt{\frac{25}{144}+1}\right) = \ln\left(\frac{5}{12}+\frac{13}{12}\right) = \ln\frac32.$$ Since $\sinh^{-1}$ is an odd function, $$\sinh^{-1}\left(-\frac{5}{12}\right) = -\sinh^{-1}\frac{5}{12} = -\ln\frac32 = \ln\frac23.$$

(2) We have $\operatorname{csch}^{-1}x = \ln\left(\dfrac1x+\sqrt{\dfrac1{x^2}+1}\right)$ for $x\ne0$. With $x=-1/\sqrt3$, $1/x=-\sqrt3$ and $1/x^2=3$, so $$\operatorname{csch}^{-1}\left(-\frac{1}{\sqrt3}\right) = \ln\left(-\sqrt3+\sqrt{3+1}\right) = \ln(2-\sqrt3).$$

(3) Since $$\frac{d}{dv}\bigl(\ln\sinh v\bigr) = \coth v, \qquad \frac{d}{dv}\left(-\frac12\coth^2v\right) = -\coth v\cdot(-\operatorname{csch}^2v) = \coth v\operatorname{csch}^2v,$$ we get $$\frac{d}{dv}\left(\ln\sinh v-\frac12\coth^2v\right) = \coth v(1+\operatorname{csch}^2v).$$ Using the identity $1+\operatorname{csch}^2v=\coth^2v$, $$\frac{d}{dv}\left(\ln\sinh v-\frac12\coth^2v\right) = \coth v\cdot\coth^2v = \coth^3v.$$

(4) Let $y=\sinh^{-1}(\tan x)$, so $\sinh y=\tan x$ and $\cosh y = \sqrt{1+\sinh^2y} = \sqrt{1+\tan^2x} = \sqrt{\sec^2x} = |\sec x|$. Differentiating both sides with respect to $x$, $\cosh y\cdot\dfrac{dy}{dx} = \sec^2x$, so $$\frac{d}{dx}\sinh^{-1}(\tan x) = \frac{\sec^2x}{\sqrt{1+\tan^2x}} = \frac{\sec^2x}{|\sec x|} = |\sec x|.$$ (In particular $|\sec x| = \sec x$ for $-\pi/2 \lt x \lt \pi/2$; this function is the inverse of the Gudermannian function.)

(5) Substituting $u=\sqrt t$, so $du = dt/(2\sqrt t)$ and $dt/\sqrt t = 2\,du$, $$\int\frac{\operatorname{sech}\sqrt t\,\tanh\sqrt t}{\sqrt t}\,dt = \int2\operatorname{sech}u\tanh u\,du = -2\operatorname{sech}u+C = -2\operatorname{sech}\sqrt t+C.$$

(6) Substituting $u=\sqrt x$, so $dx/\sqrt x = 2\,du$, with $u=1$ at $x=1$ and $u=\sqrt2$ at $x=2$, $$\int_1^2\frac{8\cosh\sqrt x}{\sqrt x}\,dx = \int_1^{\sqrt2}8\cosh u\cdot2\,du = 16\Bigl[\sinh u\Bigr]_1^{\sqrt2} = 16\bigl(\sinh\sqrt2-\sinh1\bigr).$$

(7) $$\int_0^{2\sqrt3}\frac{dx}{\sqrt{4+x^2}} = \left[\sinh^{-1}\frac{x}{2}\right]_0^{2\sqrt3} = \sinh^{-1}\sqrt3 - \sinh^{-1}0 = \sinh^{-1}\sqrt3 = \ln\bigl(\sqrt3+2\bigr).$$

(8) Substituting $u=\sin x$, so $du=\cos x\,dx$: at $x=0$, $u=0$, and at $x=\pi$, $u=\sin\pi=0$ as well, so both endpoints give $u=0$. Hence $$\int_0^{\pi}\frac{\cos x\,dx}{\sqrt{1+\sin^2x}} = \int_0^0\frac{du}{\sqrt{1+u^2}} = 0.$$ (Using the antiderivative $\sinh^{-1}(\sin x)$ gives the same result: $\sinh^{-1}(\sin\pi)-\sinh^{-1}(\sin0) = 0-0=0$.)

Proof.

(1) Multiplying numerator and denominator of $\operatorname{sech}x = 1/\cosh x$ by $\cosh x$, $$\int\operatorname{sech}x\,dx = \int\frac{\cosh x}{\cosh^2x}\,dx = \int\frac{\cosh x}{1+\sinh^2x}\,dx.$$ Substituting $u=\sinh x$, so $du=\cosh x\,dx$, $$\int\operatorname{sech}x\,dx = \int\frac{du}{1+u^2} = \tan^{-1}u+C = \tan^{-1}(\sinh x)+C.$$ On the other hand, since $1-\tanh^2x = \operatorname{sech}^2x$ and $\operatorname{sech}x \gt 0$, $$\frac{d}{dx}\sin^{-1}(\tanh x) = \frac{\operatorname{sech}^2x}{\sqrt{1-\tanh^2x}} = \frac{\operatorname{sech}^2x}{\operatorname{sech}x} = \operatorname{sech}x,$$ so $\sin^{-1}(\tanh x)$ is also an antiderivative of $\operatorname{sech}x$. Since $\tan^{-1}(\sinh x)$ and $\sin^{-1}(\tanh x)$ have the same derivative, they differ by a constant; at $x=0$, $\tan^{-1}(\sinh0)=\tan^{-1}0=0$ and $\sin^{-1}(\tanh0)=\sin^{-1}0=0$, so that constant is $0$. Hence $$\int\operatorname{sech}x\,dx = \tan^{-1}(\sinh x)+C = \sin^{-1}(\tanh x)+C.$$

(2) Let $u=\operatorname{sech}^{-1}x$, $dv=x\,dx$, so for $0\lt x\lt1$, $du=-\dfrac{1}{x\sqrt{1-x^2}}\,dx$ and $v=\dfrac{x^2}{2}$. Then $$\int x\operatorname{sech}^{-1}x\,dx = \frac{x^2}{2}\operatorname{sech}^{-1}x - \int\frac{x^2}{2}\left(-\frac{1}{x\sqrt{1-x^2}}\right)dx = \frac{x^2}{2}\operatorname{sech}^{-1}x + \frac12\int\frac{x}{\sqrt{1-x^2}}\,dx.$$ Since $\displaystyle\int\frac{x}{\sqrt{1-x^2}}\,dx = -\sqrt{1-x^2}+C$, $$\int x\operatorname{sech}^{-1}x\,dx = \frac{x^2}{2}\operatorname{sech}^{-1}x - \frac12\sqrt{1-x^2}+C.$$

Solution. Since $y=\dfrac12\cosh2x$, $y'=\sinh2x$, so $$1+(y')^2 = 1+\sinh^22x = \cosh^22x.$$ Hence the arc length is $$L = \int_0^{\ln\sqrt5}\sqrt{1+(y')^2}\,dx = \int_0^{\ln\sqrt5}\cosh2x\,dx = \left[\frac{\sinh2x}{2}\right]_0^{\ln\sqrt5} = \frac{\sinh(2\ln\sqrt5)}{2} = \frac{\sinh(\ln5)}{2}.$$ Since $\sinh(\ln5) = \dfrac{e^{\ln5}-e^{-\ln5}}{2} = \dfrac{5-1/5}{2} = \dfrac{24/5}{2} = \dfrac{12}{5}$, $$L = \frac12\cdot\frac{12}{5} = \frac65.$$

Solution. Since $\cosh x\ge\sinh x\ge0$ for $x\ge0$, the volume of the solid obtained by revolving about the $x$-axis the region bounded by $y=\cosh x$, $y=\sinh x$, the $y$-axis, and the line $x=2$ is, by the washer method, $$V = \pi\int_0^2\bigl[(\cosh x)^2-(\sinh x)^2\bigr]\,dx.$$ By the identity $\cosh^2x-\sinh^2x=1$, the integrand is the constant function $1$, so $$V = \pi\int_0^2 1\,dx = 2\pi.$$