Chapter 5, Answers | Math 140 - Calculus and Analytic Geometry 1

Chapter 5

\(\frac{x^9}{9}-x+C \).
\( -2e^{-2x} + 6x – 3\tan 2x + C \).
\( \frac{1}{2}(\cosh 2x – \cos 2x) +C \).
\( \pi \tan 6x +C \).
\( x\left (1-\frac{9}{4}\sqrt[3]{x}+\frac{9}{5}x \sqrt[3]{x}-\frac{1}{2}x\right ) +C \).
\(\frac{1}{\pi}\left (-2\cos \pi x + \cot \pi x\right ) +C \).
\(-x+4\arctan x +\arcsin x+C \).
\(x- 2\tan \frac{x}{2} +\frac{1}{\pi }\left (\sin \pi x – \cos \pi x\right ) +C \).
\(-\frac{2}{3}\csc 3x+C\).
\(x^x (\ln x +1)+2^{(x^2+1)} x \ln 2\).
\(\frac{1}{2\sqrt{1-x}} + \sec^{-1}\sqrt{x}\).
\(\frac{x (6 \ln x+3)}{4\ln 2}\).
\(\frac{x e^{x^2} (2 x^4+3 x^2+2)}{(1+x^2)\,\sqrt{1+x^2}}\).
(a)–(c) False.

\(\frac{2}{3}\sqrt{x^3}+x+\frac{1}{3}\).
\(\frac{1}{2}\tan 2x- x-3\).
\(2\sin^{-1}x – \frac{4\pi}{3}\).
\(x+3\tan^{-1}x + \pi + \frac{1}{\sqrt{3}}\).
\(\frac{2}{\pi}\sin \pi x -\frac{\sqrt 2}{\pi}\).
\(\frac{x}{2e^2}-2x+\frac{1}{4}e^{-2x}-\frac{3}{4e^2}+3\).
\(\frac{x}{\pi}-x -\frac{1}{\pi^2}\cos \pi x -\frac{3}{2\pi} +\frac{5}{2}\).
\(ex+\frac{x}{e}-\ln x+e^2+e\).
\(-\frac{x}{\pi}-\frac{3}{2}\sin 2x – \cos 2x -\frac{1}{2}\).
\(\sin 2 x – \cos 2 x -1\).

\(\frac{1}{2}\left (1+\sqrt[3]{x^2}\right )^3+C\).
\(\frac{1}{\pi}\ln \sec \pi x +C\).
\(\frac{1}{9}\arctan^3 3x+C\).
\(-\frac{1}{3}\sqrt{\cos^3 e^{2x}}+C\).
\(\frac{1}{12}\ln^4x^3 + C\).
\(2\tan \sqrt{x} + C\).
\( -2e^{\cos(x/2)}+C\).
\(\frac{1}{\pi}e^{\sin^2\pi x}+C\).
\(\frac{1}{4}\arcsin^2(2x)+C\).
\(\frac{1}{8}\ln(1+4x^2)+ \frac{1}{4}\arctan^2(2x)+C\).
\(\frac{1}{2}\arctan e^{2x}+C\).
\(\frac{1}{3}\arcsin x^3 + \frac{1}{12}\arcsin^4 x^3+C\).
\(e^x-e^{-x} + C\).
\(w-\frac{\sqrt{2}}{\pi}\cos \pi x + C\).
\(-\frac{1}{4\pi}(\tan \pi x + \cot \pi x)+ C\) or \(-\frac{1}{2\pi}\csc 2\pi x+C\).

(a) \(\frac{1}{4}x^4+x^3+x^2 + C\); (b) \(\frac{1}{4}x^2 (x + 2)^2 + D\).
\(-\frac{32}{3\ln^2x^3} +C\).
\(x-\frac{1}{8}\sin 4x + \frac{1}{16}\sin 4x + C\).
\(\frac{\ln^4 x}{4\ln^3x} + C\).
\(\frac{1}{\ln 2}2^{\sin^2x} + C\).
\(\frac{1}{8}\arcsin^2 4x + C\).
\(\frac{3}{16}\left (1+\sqrt[3]{x^4}\right )^4\).
\(-\frac{1}{2\sqrt{t(1-t)}} \).
\(\frac{e^{\sqrt{z}}\left (2 (z^2+1) + (z^2\sqrt{z} – 2 z^2 + \sqrt{z}\right ) \ln z }{x\sqrt{(1+z^2)^3}}\).
\(-\frac{2u}{(u^2+1)^2}\).
\(\sin^t(\pi t) (\pi t \cot\pi t+\ln \sin \pi t) \).
\(2^{2+u^2}\ln 2\).
\(\frac{1}{8}\frac{1}{ \sqrt{1+x}}\frac{1}{ \sqrt{1+\sqrt{1+x}}}\frac{1}{\sqrt{1+\sqrt{1+\sqrt{1+x}}}}\).
\(\frac{1}{2\pi}\frac{1}{\cos^2\pi x}+C\), \(\frac{1}{2\pi}\sec^2\pi x+C\), \(\frac{1}{2\pi} \tan^2\pi x+C\). Why are these three answers compatible?
\(-\frac{1}{2\pi}\cos^2\pi x + C\); \(\frac{1}{2\pi}\sin^2\pi x + C\); \(-\frac{1}{4\pi}\cos 2\pi x + C\). Why are these three answers compatible?
\(\frac{1}{2\pi}\csc^2 \pi x+C\), \(-\frac{1}{2\pi}\cot^2 \pi x+C\), \(-\frac{1}{2\pi \sin^2\pi x}+ C\).

\(\sin x^2\).
\(-\ln x\).
\(2x \,\sqrt{1+x^8} – 2\sqrt{1 + 16x^4}\).
\(\frac{x\,\sqrt{x}}{2(1+x)} – \tan^4x\).
\(e^{-e^{2x}+x}- \frac{x^{-x}(\ln x + 1)}{2\sqrt{x\ln x}}\).
\(\frac{\sin x}{x} – \frac{x}{2\arcsin(x/2)\sqrt{4-x^2}}\).
\(-|\sin x|\sin x – \sqrt{1-x^2}\).
\(\frac{1}{2\sqrt{\tan x}}\).
(a) \(f(x)= -\frac{3}{\sqrt{1-9x^2}}\), \(c=-\frac{\pi}{3}\); (b) \(f(x)= \frac{1}{3x}\), \(c=-\frac{2}{3}\).
(a) \(f(x)= 2\sin x\), \(c\) can be any value \(c=\frac{(2n+1)\pi}{2}\), \(k \in \mathbb{Z}\); (b) \(f(x)= \frac{\arctan \sqrt{x}}{(1+x)\,\sqrt{x}}\), \(c=\frac{1}{3}\).

(b) \(e^{3/4}-1\).
(b) \(\frac{\pi^2}{9}\).
(b) \(\frac{12}{5}\).
(b) \(\frac{17\pi}{96}\).
(b) \(\frac{1}{\sqrt{2}\pi}\).
(b) \(\frac{33}{4}\).
(b) \(6 \).
(b) \(\pi\).
\(f(x) = \frac{1}{x^2}\) is not continuous on \([-1,2]\).
\([2,4]\) is not in the domain of \(f(x)= \frac{1}{\sqrt{1-x^2}}\).
\(2^x\) is not an antiderivative of \(2^x\).
The antiderivative to use is \(F(x) =\ln |x|\).
(a) \(0 \); (b) \(\frac{8}{3} \).
(a) \(-2 \); (b) \(\frac{2}{3\pi}(6\sqrt{3}+\pi) \).
(a) \(-\frac{1}{8}+\frac{e^4}{2}-e(2+\ln 2) \); (b) \(\frac{1}{8}(1-16e+4e^4+8e \ln 2 ) \).
(a) \(-\frac{3}{2}+2\ln 2 \); (b) \(\frac{1}{2}\).
\(\frac{8}{3}\).
\(\frac{4\sqrt{2}}{3}\).
\(\sqrt{3}-\sqrt{2}\).
\(\frac{4}{7}\).
\(\frac{242}{5}\).
\(\sqrt{3}-\frac{\pi}{3}\).
Area of a semicircle with \(r=2\): \(A=2\pi\).

\(\frac{1}{12}(2 + x^4)^3+C \).
\(\frac{1}{3}\tan^3t+C\).
\(-\frac{7}{64\pi}\).
\(\frac{1}{2}\sin e^{2w}+C \).
\(\frac{15}{8}\).
\(-\frac{\pi^2}{24}\).
\(\frac{\pi^3}{24}\left (\frac{1}{8}+\frac{1}{27}\right )\).
\(\frac{e-1}{\pi} \).
\(-\frac{1}{3\pi}\cot^3x+C \).
\( \frac{1}{\ln 2} 2^{\sqrt{t}+1} + e^{e^t} +C\).
\(\frac{1}{\pi} \left (\sin \pi x -\cos \pi x\right ) +C\)
\( \frac{3}{2}\ln(1+x^2)+4x – 2\arctan x + C\).

\(\frac{3}{8}(x^2+1)\sqrt[3]{x^2+1} + C\).
\( \frac{1}{2}\arctan x^2+ C\).
\( -\frac{1}{3}\ln \left [\cos e^{3x}\right ]+ C\).
\(2 \sec\sqrt{t}+ C\).
\( \frac{1}{4}\arctan^2 2t + C\).
\( -\frac{1}{\pi}\arcsin e^{\cos \pi x} + C\).
\(-\frac{1}{\pi}\arctan (\cos^2\pi x) +C\).
\( \frac{1}{3}\arcsin\left (\ln x^3\right )+ C\).
\(\frac{1}{2}(x^3+1)^(2/3)+C\).
\(\frac{1}{2}(e^{2x}+e{-2x}) + C \)
\(\frac{1}{3}\ln|\cos e^{3x} + C\)
\(-\frac{3}{2}\cot^2\sqrt[3]{x} + C\)
\( \frac{1}{3}\sqrt{ \ln^2 ( x^3)} + C\)
\(\frac{1}{2\pi}\arctan (\sin^2\pi x) +C\).
\(\frac{175}{64}\).
\(\frac{1}{24}\pi(4+\pi) \).
\( \frac{1}{12}\).
\( \frac{\pi^3}{72}\).
\(\frac{1}{6}(3\sqrt{3}-1) \).
\(\left (\sqrt{e} -1\right )\sqrt[4]{e} \).

\(\frac{2}{3}\ln|3x+4| +C\)
\(6 (x + 1) – 7 \ln |x+1|+C\)
\(2 x^3 – 3 x^2 + 9 x – 11\ln|x+1|+C\).
\(x+\frac{1}{2}e^2x+C\).
\(\frac{2}{3}\sin^3\theta -\ln|\sin \theta+2|+C\).
\(x^3-x +\frac{1}{2} \ln(x^2+1)+2 tan^{-1} x + C\).
\(3 x^2-\frac{3}{2} \ln(x^2+1)-2 tan^{-1} x + C\).
\(\ln^2 x – 6 \ln x + \frac{13}{2} \ln|2 \ln x + 3| + C\).

\(\frac{2}{27} \sqrt{3 x-1} (6 x-5)\).
\(\frac{4}{15}\,\sqrt{(\sqrt{x}+2)^3} \,(3\sqrt{x}-4)+ C\).
\(-\frac{6 x^2+1}{48 \left(2 x^2+1\right)^3}+C\).
\(\frac{1}{15} \sqrt{\left(2 x-1\right)^3}\left (3x+1\right)+C\).
\(\frac{3}{2} \left(\sqrt[3]{x^2} -2\sqrt[3]{x} +2 \ln \left(\sqrt[3]{x}+1\right)\right)+C\).
\(- x^2 -9 x-27 \ln |x-3| +C\).
\(\frac{1}{2}(\sin 2x + \cos 2x)+C\).
\(-\frac{4}{5}\sqrt{1+\sqrt{x}}\left (x-3\sqrt{x}+6\right )+C.\)

\(\arcsin \frac{x}{2}+C\).
\(\frac{1}{4} \arctan 2 x+C\).
\(\frac{\sqrt{2}}{2}\text{arcsec}\frac{x}{\sqrt{2}}+C\).
\(\frac{1}{\sqrt{6}}\arctan \frac{\sqrt{2}x}{\sqrt{3}} +C\).
\(\arctan (x+3)+C\).
\(\arcsin \frac{x-3}{2}+C\).
\(\frac{1}{\sqrt{2}}\arcsin \frac{\sqrt{2}}{\sqrt{7}}
(x-1)+C\).
\(\frac{1}{3}\text{arcsec}\frac{x+3}{3}+C\)
\(\frac{1}{2}\left (\ln(x^2+4 x+6)-\sqrt{2}\arctan\frac{x+2}{\sqrt{2}}\right )+C\).
Use Theorem 1 in 5.1.1.