Chapter 6 Challenge Problems Solutions

SOLUTION In performing the substitution \(x + 1 = c \sec \theta,\) we attain the following bound changes.

• Lower Bound When \(x = 0,\) we have \(1 = c \sec \theta,\) or \(\sec \theta = 1/c.\)
• Upper Bound When \(x = 2,\) we have \(3 = c \sec \theta,\) or \(\sec \theta = 3/c.\)

But the range of secant is \((-\infty, -1] \cup [1, \infty).\) Thus, we must avoid \(1/c\) or \(3/c\) lying within \((-1, 1).\) For the lower bound \(x = 0,\) we have \[\abs{\frac{1}{c}} \lt 1 \iffArrow \abs{c} \gt 1 \pd\] Additionally, for the upper bound \(x = 2,\) \[\abs{\frac{3}{c}} \lt 1 \iffArrow \abs{c} \gt 3 \pd\] The intersection of the sets \(\{c \mid \abs c \gt 1\}\) and \(\{c \mid \abs c \gt 3\}\) is \[\boxed{(-\infty, -1) \cup (1, \infty)}\] In this case, we must antidifferentiate the integrand and return to working with \(x\)—not \(\theta.\)

SOLUTION The integral is \[ \ba \lim_{t \to 0^+} \int_t^1 x^{-q} \di x &= \lim_{t \to 0^+} \frac{x^{1 - q}}{1 - q} \intEval_t^1 \nl &= \frac{1}{1 - q} - \lim_{t \to 0^+} \frac{t^{1 - q}}{1 - q} \pd \ea \] If \(q \lt 1,\) then \(1 - q\) is positive and so \(t^{1 - q}/(1 - q) \to 0\) as \(t \to 0^+.\) Then the integral converges. But if \(q \gt 1,\) then \(1 - q\) is negative and so \(t^{1 - q}/(1 - q) \to \infty\) as \(t \to 0^+,\) meaning the integral diverges. And if \(q = 1,\) then the integral becomes \[ \ba \lim_{t \to 0^+} \ln x \intEval_t^1 &= 0 - \lim_{t \to 0^+} \ln t \nl &= \infty \pd \ea \] Accordingly, \(\int_0^1 \dd x/x^q\) converges for \(\boxed{q \lt 1}.\)

SOLUTION By factoring the numerator, the integral becomes \[ \int \frac{\tan x (\tan^2 x + 1)}{6 \tan^2 x - 2} \di x = \int \frac{\tan x \sec^2 x}{6 \tan^2 x - 2} \di x \pd \] Substituting \(u = 6 \tan^2 x - 2,\) we attain \(\dd u = 12 \tan x \sec^2 x \di x\) and so the integral becomes \[ \ba \int \frac{1}{12 u} \di u &= \tfrac{1}{12} \ln \abs u + C \nl &= \boxed{\tfrac{1}{12} \ln \abs{6 \tan^2 x - 2} + C} \ea \]

SOLUTION Multiplying the numerator and denominator each by \(\sec^6 x\) yields \[\int \frac{\sec^6 x}{\tan^6 x + 1} \di x \pd\] Since the power of secant is even, we can conserve an extra \(\sec^2 x\) factor, similar to the strategy in Section 6.2. Because \(\sec^2 x\) \(= \tan^2 x + 1,\) we have \[\int \frac{\sec^4 x \sec^2 x}{\tan^6 x + 1} \di x = \int \frac{(\tan^2 x + 1)^2 \sec^2 x}{\tan^6 x + 1} \di x \pd\] Substituting \(u = \tan x,\) we attain \(\dd u = \sec^2 x \di x\) and so the integral becomes \[ \ba \int \frac{(u^2 + 1)^2}{u^6 + 1} \di u &= \int \frac{(u^2 + 1)^2}{(u^2 + 1)(u^4 - u^2 + 1)} \di u \nl &= \int \frac{u^2 + 1}{u^4 - u^2 + 1} \di u \pd \ea \] Dividing the numerator and denominator each by \(u^2\) yields \[ \int \frac{1 + \dfrac{1}{\strut u^2}}{u^2 - 1 + \dfrac{\strut 1}{u^2}} \di u = \int \frac{1 + \dfrac{1}{\strut u^2}}{\par{u - \dfrac{1}{u}}^2 + 1} \di u \pd \] Letting \(v = u - 1/u,\) we get \(\dd v = (1 + 1/u^2) \di u\) and thus \[ \ba \int \frac{1}{v^2 + 1} \di v &= \atan v + C \nl &= \atan \par{u - \frac{1}{u}} + C \nl &= \atan \par{\tan x - \frac{1}{\tan x}} + C \nl &= \boxed{\atan \par{\tan x - \cot x} + C} \ea \]

SOLUTION While it appears menacing, the integrand can be drastically simplified. Note that \[ \ba \cos^4 x - \sin^4 x &= (\cos^2 x + \sin^2 x)(\cos^2 x - \sin^2 x) \nl &= \cos^2 x - \sin^2 x \nl &= \cos 2x \pd \ea \] Thus, the integral becomes \[\int \cos 2x e^{\sqrt x} \sec 2x \sqrt x \di x = \int \sqrt x \, e^{\sqrt x} \di x \pd\] Substituting \(t = \sqrt x,\) we get \(\dd t = 1/(2 \sqrt x) \di x\) \(= 1/2t \di x.\) Accordingly, \(\dd x = 2t \di t\) and so the integral becomes \[\int t e^t (2t) \di t = \int 2t^2 e^t \di t \pd\] Performing Integration by Parts (Section 6.1) with \[ \baat{2} u &= 2t^2 \lspace &\dd v &= e^t \di t \nl \dd u &= 4t \di t \lspace &v &= e^t \eaat \] yields \[2 t^2 e^t - \int 4t e^t \di t \pd\] We use Integration by Parts again with \[ \baat{2} u &= 4t \lspace &\dd v &= e^t \di t \nl \dd u &= 4 \di t \lspace &v &= e^t \cma \eaat \] attaining \[ \ba 2t^2 e^t - \par{4t e^t - \int 4 e^t \di t} &= 2t^2 e^t - \par{4t e^t - 4 e^t} + C \nl &= 2t^2 e^t - 4t e^t + 4e^t + C \nl &= \boxed{2x e^{\sqrt x} - 4 \sqrt x e^{\sqrt x} + 4 e^{\sqrt x} + C} \ea \]

SOLUTION We first substitute \(u = e^x\) to get \(\dd u = e^x \di x\) \(= u \di x.\) Accordingly, \(\dd x = \dd u/u,\) so the integral becomes \[ \int \frac{u^{-3}}{\sqrt{1 - u^2}} \par{\frac{1}{u}} \di u = \int \frac{1}{u^4 \sqrt{1 - u^2}} \di u \pd \] Now we perform the trigonometric substitution \(u = \sin \theta;\) then \(\dd u = \cos \theta \di \theta.\) Doing so yields \[ \ba \int \frac{1}{\sin^4 \theta \sqrt{1 - \sin^2 \theta}} (\cos \theta) \di \theta &= \int \frac{1}{\sin^4 \theta \abs{\cos \theta}} (\cos \theta) \di \theta \nl &= \int \csc^4 \theta \di \theta \nl &= \int (\cot^2 \theta + 1) \csc^2 \theta \di \theta \pd \ea \] Substituting \(v = \cot \theta,\) we attain \(\dd v = -\csc^2 \theta \di \theta.\) Consequently, we attain \[ \ba -\int \par{v^2 + 1} \di v &= -\tfrac{1}{3} v^3 - v + C \nl &= -\tfrac{1}{3} \cot^3 \theta - \cot \theta + C \pd \ea \] But since \(u = \sin \theta,\) we have \(\cot \theta = \sqrt{1 - u^2}/u.\) Also remembering the initial substitution \(u = e^x,\) we get \[-\frac{1}{3} \par{\frac{\sqrt{1 - e^{2x}}}{e^x}}^3 - \frac{\sqrt{1 - e^{2x}}}{e^x} + C = \boxed{-\frac{(1 - e^{2x})^{3/2}}{3e^{3x}} - \frac{\sqrt{1 - e^{2x}}}{e^x} + C} \]

SOLUTION Rewriting the integral as \(\int \cos^{n - 1} \theta \cos \theta \di \theta,\) we perform Integration by Parts with \[ \baat{2} u &= \cos^{n - 1} \theta \lspace &\dd v &= \cos \theta \di \theta \nl \dd u &= -(n - 1) \cos^{n - 2} \theta \sin \theta \di \theta \lspace &v &= \sin \theta \pd \eaat \] We then see \[ \ba \int \cos^n \theta \di \theta &= \sin \theta \cos^{n - 1} \theta + (n - 1) \int \sin^2 \theta \cos^{n - 2} \theta \di \theta \nl &= \sin \theta\cos^{n - 1} \theta + (n - 1) \int (1 - \cos^2 \theta) \cos^{n - 2} \theta \di \theta \nl &= \sin \theta\cos^{n - 1} \theta + (n - 1) \int \cos^{n - 2} \theta \di \theta - (n - 1)\int \cos^n \theta \di \theta \nl n \int \cos^n \theta \di \theta &= \sin \theta\cos^{n - 1} \theta + (n - 1) \int \cos^{n - 2} \theta \di \theta \nl \implies \int \cos^n \theta \di \theta &= \frac{1}{n} \sin \theta\cos^{n - 1} \theta + \frac{n - 1}{n} \int \cos^{n - 2} \theta \di \theta \cma \ea \] as requested.

SOLUTION Since \(\sec^2 x\) is easy to integrate, let's break up the integral as \[\int \sec^n x \di x = \int \sec^{n - 2} x \sec^2 x \di x\] and perform Integration by Parts. We choose \[ \baat{2} u &= \sec^{n - 2} x \lspace &\dd v &= \sec^2 x \di x \nl \dd u &= (n - 2) \sec^{n - 3} x (\sec x \tan x) \di x \lspace &v &= \tan x \pd \nl &= (n - 2) \sec^{n - 2} x \tan x \di x \eaat \] So we have \[ \ba \int \sec^n x \di x &= \sec^{n - 2} x \tan x - (n - 2) \int \sec^{n - 2} x \tan^2 x \di x \nl &= \sec^{n - 2} x \tan x - (n - 2) \int \sec^{n - 2} x (\sec^2 x - 1) \di x \nl &= \sec^{n - 2} x \tan x - (n - 2) \int \sec^n x \di x + (n - 2) \int \sec^{n - 2} x \di x \pd \ea \] Adding \(\int \sec^{n - 2} x \di x\) to both sides yields \[ \ba (n - 1) \int \sec^n x \di x &= \sec^{n - 2} x \tan x + (n - 2) \int \sec^{n - 2} x \di x \nl \implies \int \sec^n x \di x &= \frac{\sec^{n - 2} x \tan x}{n - 1} + \frac{n - 2}{n - 1} \int \sec^{n - 2} x \di x \cma \ea \] as requested.

SOLUTION By factoring the denominator, the integral becomes \[\int_c^{2c} \frac{1}{(x - 4)^2} \di x \pd\] The graph of \(y = 1/(x - 4)^2\) has a vertical asymptote at \(x = 4.\) For any \(a \lt 4 \lt b,\) observe that \[ \ba \int_4^b \frac{1}{(x - 4)^2} \di x &= \lim_{t \to 4^+} \int_t^b \frac{1}{(x - 4)^2} \di x \nl &= \lim_{t \to 4^+} \par{-\frac{1}{x - 4}} \intEval_t^b \nl &= \lim_{t \to 4^+} \par{\frac{1}{t - 4} - \frac{1}{b - 4}} \nl &= \infty \pd \ea \] Also, \[ \ba \int_a^4 \frac{1}{(x - 4)^2} \di x &= \lim_{t \to 4^-} \int_a^t \frac{1}{(x - 4)^2} \di x \nl &= \lim_{t \to 4^-} \par{-\frac{1}{x - 4}} \intEval_a^t \nl &= \lim_{t \to 4^-} \par{\frac{1}{a - 4} - \frac{1}{t - 4}} \nl &= \infty \pd \ea \] Thus, any integral whose bounds run through the vertical asymptote is divergent. Hence, the values of \(c\) must cross \(x = 4,\) meaning \(c \leq 4 \leq 2c.\) With \(4 \leq 2c,\) we have \(c \geq 2.\) Thus, we have \[ \ba \{c \mid c \geq 2 \text{ and } c \leq 4\} = \boxed{[2, 4]} \ea \]

SOLUTION Observe that \[\frac{1}{x^{2024} - x} = \frac{-1}{x} + \frac{x^{2022}}{x^{2023} - 1} \pd\] By substituting \(u = x^{2023} - 1,\) we get \(\dd u = 2023 x^{2022} \di x;\) thus, we see \[ \ba \int \frac{x^{2022}}{x^{2023} - 1} \di x &= \int \frac{1}{2023 u} \di u \nl &= \tfrac{1}{2023} \ln \abs u + C \nl &= \tfrac{1}{2023} \ln \abs{x^{2023} - 1} + C \pd \ea \] Hence, we have \[ \int \frac{1}{x^{2024} - x} \di x = \boxed{-\ln \abs x + \tfrac{1}{2023} \ln \abs{x^{2023} - 1} + C} \]

SOLUTION Note that \(y\) is simply a dummy variable; the convolutions of \(u\) and \(v\) are functions of \(x.\) Our goal is to show that \((u \circledast v)(x) = (v \circledast u)(x),\) or \[ \ba \underbrace{\int_{-\infty}^\infty u(y) v(x - y) \di y}_{(u \circledast v)(x)} &= \int_{-\infty}^\infty v(y) u(x - y) \di y \nl &= \int_{-\infty}^\infty u(x - y) v(y) \di y \pd \ea \] We use a change of variable \(t = x - y.\) Then \(y = x - t\) and \(\dd y = -\dd t.\) In addition, when \(y = -\infty,\) \(t = \infty;\) when \(y = \infty,\) \(t = -\infty.\) Hence, \[ \ba \underbrace{\int_{-\infty}^\infty u(y) v(x - y) \di y}_{(u \circledast v)(x)} &= \int_\infty^{-\infty} u(x - t) v(t) (-1) \di t \nl &= \int_{-\infty}^{\infty} u(x - t) v(t) \di t \nl &= \int_{-\infty}^\infty v(t) u(x - t) \di t \nl &= (v \circledast u)(x) \cma \ea \] where the last step is true because \(t\) is simply a dummy variable. Hence, \[ (u \circledast v)(x) = (v \circledast u)(x) \cma \] so convolution is commutative.

SOLUTION Let's perform Integration by Parts with \[ \baat{2} u &= x^n \lspace &\dd v &= e^x \di x \nl \dd u &= n x^{n - 1} \di x \lspace &v &= e^x \pd \eaat \] Doing so gives \[\int x^n e^x \di x = x^n e^x - \int n x^{n - 1} e^x \di x \pd\] Performing Integration by Parts again with \[ \baat{2} u &= n x^{n - 1} \lspace &\dd v &= e^x \di x \nl \dd u &= n (n - 1) x^{n - 2} \di x \lspace &v &= e^x \eaat \] shows \[\int x^n e^x \di x = x^n e^x - n x^{n - 1} e^x + \int n (n - 1) x^{n - 2} e^x \di x \pd\] Continuing this pattern, we have \[ \ba \int x^n e^x \di x &= \small x^n e^x - n x^{n - 1} e^x + n (n - 1) (n - 2) x^{n - 2} e^x - n (n - 1)(n - 2)(n - 3) x^{n - 3} e^x + \cdots \nl &= \boxed{\sum_{i = 0}^n (-1)^i \, x^{n - i} \, e^x \frac{n!}{(n - i)!}} \ea \] Note that \(0! = 1.\)

SOLUTION The integral to evaluate is \[ \ba V &= \int_0^L \frac{kQ}{L \sqrt{x^2 + d^2}} \di x \nl &= \frac{kQ}{L} \int_0^L \frac{1}{\sqrt{x^2 + d^2}} \di x \pd \ea \] We immediately substitute \(x = d \tan \theta,\) \(-\pi/2 \lt \theta \lt \pi/2;\) then \(\dd x = d \sec^2 \theta \di \theta.\) We now change the bounds.

• Lower Bound When \(x = 0,\) we have \(0 = d \tan \theta,\) so \(\theta = 0.\)
• Upper Bound When \(x = L,\) we have \(L = d \tan \theta,\) so \(\theta = \atan(L/d).\) For convenience, let \(\beta = \atan(L/d).\)

Since \(L\) and \(d\) are both positive, \(\beta = \atan(L/d)\) must be positive and less than \(\pi/2.\) Accordingly, secant is positive on \([0, \beta].\) Hence, we attain \[ \ba V &= \frac{kQ}{L} \int_0^\beta \frac{1}{\sqrt{d^2 \tan^2 \theta + d^2}} (d \sec^2 \theta) \di \theta \nl &= \frac{kQ}{L} \int_0^\beta \frac{d \sec^2 \theta}{d \abs{\sec \theta}} \di \theta \nl &= \frac{kQ}{L} \int_0^\beta \sec \theta \di \theta \nl &= \frac{kQ}{L} \ln \abs{\sec \theta + \tan \theta} \intEval_0^\beta \nl &= \frac{kQ}{L} \ln \abs{\sec \beta + \tan \beta} - 0 \pd \ea \] But we defined \(\tan \beta = L/d,\) so by the Pythagorean identity we have \[\sec \beta = \sqrt{\par{\frac{L}{d}}^2 + 1} = \sqrt{\frac{L^2 + d^2}{d^2}} = \frac{\sqrt{L^2 + d^2}}{d} \cma\] where the positive solution is appropriate. Substituting expressions yields \[ \ba V &= \frac{kQ}{L} \ln \abs{\frac{\sqrt{L^2 + d^2}}{d} + \frac{L}{d}} \nl &= \boxed{\frac{kQ}{L} \ln \abs{\frac{L + \sqrt{L^2 + d^2}}{d}}} \ea \]

SOLUTION

The vertical distance between the boat and lighthouse is \(20 \un{ft}.\) Let \(x\) be the horizontal distance between the boat and lighthouse. Then by the Pythagorean Theorem, the distance between the boat and lighthouse is \[d(x) = \sqrt{x^2 + 20^2} = \sqrt{x^2 + 400} \pd\] Once the angle of elevation becomes \(45 \degree,\) the horizontal distance is \(20 \un{ft}.\) Hence, the boat's journey is modeled by \(0 \leq x \leq 20,\) on which the average value of \(d\) (Section 5.6) is \[ \ba \avg d &= \frac{1}{20 - 0} \int_0^{20} d(x) \di x \nl &= \tfrac{1}{20} \int_0^{20} \sqrt{x^2 + 400} \di x \pd \ea \] To evaluate the integral, we substitute \(x = 20 \tan \theta;\) then \(\dd x = 20 \sec^2 \theta \di \theta.\) When \(x = 0,\) \(\theta = 0;\) when \(x = 20,\) \(\theta = \pi/4.\) We therefore have \[ \ba \avg d &= \tfrac{1}{20} \int_0^{\pi/4} \sqrt{400 \tan^2 \theta + 400} \, (20 \sec^2 \theta) \di \theta \nl &= \int_0^{\pi/4} \sqrt{400 \sec^2 \theta} \, \sec^2 \theta \di \theta \nl &= \int_0^{\pi/4} \abs{20 \sec \theta} \sec^2 \theta \di \theta \nl &= 20 \int_0^{\pi/4} \sec^3 \theta \di \theta \pd \ea \] The result of Example 6.2-12 yields \[ \ba \avg d &= 20 \par{\tfrac{1}{2} \sec \theta \tan \theta + \tfrac{1}{2} \ln \abs{\sec \theta + \tan \theta}} \intEval_0^{\pi/4} \nl &= 10 \par{\sqrt 2} (1) + 10 \ln \abs{\sqrt 2 + 1} \nl &= \boxed{10 \sqrt 2 + 10 \ln \par{\sqrt 2 + 1}} \approx 22.956 \pd \ea \] Since \(d\) is continuous on \([0, 20],\) the Mean Value Theorem for Integrals guarantees a value \(c \in [0, 20]\) such that \(d(c) = \avg d.\) Solving gives \[ \ba \sqrt{c^2 + 400} &= 10 \sqrt 2 + 10 \ln \par{\sqrt 2 + 1} \nl \implies c &= \boxed{\sqrt{\parbr{10 \sqrt 2 + 10 \ln \par{\sqrt 2 + 1}}^2 - 400}} \approx 11.268 \un{ft} \pd \ea \] Thus, after the boat has sailed \(11.268 \un{ft},\) it reaches its average distance \(\avg d\) to the lighthouse.

SOLUTION Note that \(\theta\) is not constant, so we cannot take \(\cos \theta\) out of the integral. Instead, \(\cos \theta\) represents a function of \(x \col\) \[\cos \theta = \frac{x + 5}{\sqrt{(x + 5)^2 + 3^2}} = \frac{x + 5}{\sqrt{(x + 5)^2 + 9}} \pd\] So the work is \[ \ba W &= \int_0^1 -T \par{\frac{x + 5}{\sqrt{(x + 5)^2 + 9}}} \di x \nl &= -50 \int_0^1 \frac{x + 5}{\sqrt{(x + 5)^2 + 9}} \di x \pd \ea \] Substituting \(x + 5 = 3 \tan \theta,\) we see \(x = 3 \tan \theta - 5\) and so \(\dd x = 3 \sec^2 \theta \di \theta.\) When \(x = 0,\) \(\theta = \atan(5/3);\) when \(x = 1,\) \(\theta = \atan 2.\) For organization, let \(\alpha = \atan(5/3)\) and \(\beta = \atan 2.\) Therefore, the work is \[ \ba W &= -50 \int_\alpha^\beta \frac{3 \tan \theta}{\sqrt{9 \tan^2 \theta + 9}} \, (3 \sec^2 \theta) \di \theta \nl &= -50 \int_\alpha^\beta \frac{3 \tan \theta}{\abs{3 \sec \theta}} (3 \sec^2 \theta) \di \theta \nl &= -150 \int_\alpha^\beta \sec \theta \tan \theta \di \theta \nl &= -150 \sec \theta \intEval_\alpha^\beta \nl &= 150 \sec \alpha - 150 \sec \beta \pd \ea \] Since \(\alpha = \atan(5/3),\) we have \(\tan \alpha = 5/3\) and so \[ \ba \sec^2 \alpha &= \tan^2 \alpha + 1 \nl &= \par{\tfrac{5}{3}}^2 + 1 \nl &= \tfrac{34}{9} \pd \ea \] Thus, \(\sec \alpha\) \(= \sqrt{34}/3.\) [We use the positive solution, as \(-\pi/2 \lt \underbrace{\alpha}_{\atan \tfrac{5}{3}} \lt \pi/2\) and so \(\sec \alpha \gt 0.\)] Likewise, because \(\beta = \atan 2,\) we have \(\tan \beta = 2\) and thus \[ \ba \sec^2 \beta &= \tan^2 \beta + 1 \nl &= (2)^2 + 1 \nl &= 5 \pd \ea \] Thus, \(\sec \beta\) \(= \sqrt 5.\) [We use the positive solution, as \(-\pi/2 \lt \underbrace{\beta}_{\atan 2} \lt \pi/2\) and so \(\sec \beta \gt 0.\)] Accordingly, the work is \[ \ba W &= 150 \par{\frac{\sqrt{34}}{3}} - 150 \par{\sqrt 5} \nl &= \boxed{50 \sqrt{34} - 150 \sqrt{5}} \nl &\approx -43.863 \un{ft-lb} \ea \] Because the work is negative, the box has lost \(43.863 \un{ft-lb}\) of energy after sliding \(1 \un{ft}.\)