Chapter 8

T

$\sin x-x\cos x+C$

$-x^2\cos x+2x\sin x+2\cos x+C$

$1/2e^{x^2}+C$

$-\frac{1}{2}x{e}^{-2x}-\frac{e^{-2x}}{4}+C$

$1/5e^{2x}\left(\sin x+2\cos x\right)+C$

$\frac{1}{10}{e}^{5x}\left(\sin \left(5x\right)+\cos \left(5x\right)\right)+C$

$\sqrt{1-x^2}+x{\sin }^{-1}\left(x\right)+C$

$\frac{1}{2}{x}^2{\tan }^{-1}\left(x\right)-\frac{x}{2}+\frac{1}{2}{\tan }^{-1}\left(x\right)+C$

$\frac{1}{2}{x}^2\ln \left|x\right|-\frac{x^2}{4}+C$

$-\frac{x^2}{4}+\frac{1}{2}{x}^2\ln \left|x-1\right|-\frac{x}{2}-\frac{1}{2}\ln \left|x-1\right|+C$

$\frac{1}{3}{x}^3\ln \left|x\right|-\frac{x^3}{9}+C$

$2x+x{\left(\ln \left|x+1\right|\right)}^2+{\left(\ln \left|x+1\right|\right)}^2-2x\ln \left|x+1\right|-2\ln \left|x+1\right|+2+C$

$\ln \left|\sin x\right|-x\cot x+C$

$\frac{1}{3}{\left(x^2-2\right)}^{3/2}+C$

$x\sec x-\ln \left|\sec x+\tan x\right|+C$

$x\sinh x-\cosh x+C$

$x{\sinh }^{-1}x-\sqrt{x^2+1}+C$

$1/2x\left(\sin \left(\ln x\right)-\cos \left(\ln x\right)\right)+C$

$\frac{1}{2}x\ln \left|x\right|-\frac{x}{2}+C$

$1/2x^2+C$

$\pi$

$0$

$1/2$

$\frac{3}{4e^2}-\frac{5}{4e^4}$

$\frac{1}{5}\left(e^{\pi }+e^{-\pi }\right)$

F

F

$\frac{1}{4}{\sin }^{4}x+C$

$\frac{3}{8}x+\frac{1}{4}\sin 2x+\frac{1}{32}\sin 4x+C$

$\frac{1}{6}{\cos }^{6}x-\frac{1}{4}{\cos }^{4}x+C$

$\frac{1}{2}{\cos }^{2}x-\ln \left|\cos x\right|+C$

$\left(\frac{2}{7}{\cos }^{3}x-\frac{2}{3}\cos x\right)\sqrt{\cos x}+C$

$\frac{1}{2}\left(\frac{1}{4}\sin \left(4x\right)-\frac{1}{10}\sin \left(10x\right)\right)+C$

$\frac{1}{2}\left(\sin \left(x\right)+\frac{1}{3}\sin \left(3x\right)\right)+C$

$\tan x-x+C$

$\frac{{\tan }^6\left(x\right)}{6}+\frac{{\tan }^{4}x}{4}+C$

$\frac{{\sec }^5\left(x\right)}{5}-\frac{{\sec }^{3}x}{3}+C$

$\frac{1}{3}{\tan }^{3}x-\tan x+x+C$

$\frac{1}{2}\left(\sec x\tan x-\ln \left|\sec x+\tan x\right|\right)+C$

$\ln \left|\csc x-\cot x\right|+C$

$-\frac{1}{2}{\cot }^{2}x+\ln \left|\csc x\right|+C$

$\frac{2}{5}$

$32/315$

$2/3$

$16/15$

backwards

$\frac{1}{2}\left(x\sqrt{x^2+1}+\ln \left|\sqrt{x^2+1}+x\right|\right)+C$

$x\sqrt{x^2+1/4}+\frac{1}{4}\ln \left|2\sqrt{x^2+1/4}+2x\right|+C=\frac{1}{2}x\sqrt{4x^2+1}+\frac{1}{4}\ln \left|\sqrt{4x^2+1}+2x\right|+C$

$4\left(\frac{1}{2}x\sqrt{x^2-1/16}-\frac{1}{32}\ln \left|4x+4\sqrt{x^2-1/16}\right|\right)+C=\frac{1}{2}x\sqrt{16x^2-1}-\frac{1}{8}\ln \left|4x+\sqrt{16x^2-1}\right|+C$

$3{\sin }^{-1}\left(\frac{x}{\sqrt{7}}\right)+C$ (Trig. Subst. is not needed)

$\sqrt{x^2-11}-\sqrt{11}{\sec }^{-1}\left(x/\sqrt{11}\right)+C$

$-\frac{1}{\sqrt{x^2+9}}+C$ (Trig. Subst. is not needed)

$\frac{1}{18}\frac{x+2}{x^2+4x+13}+\frac{1}{54}{\tan }^{-1}\left(\frac{x+2}{3}\right)+C$

$\frac{1}{7}\left(-\frac{\sqrt{5-x^2}}{x}-{\sin }^{-1}\left(x/\sqrt{5}\right)\right)+C$

$\pi /2$

$2\sqrt{2}+2\ln \left(1+\sqrt{2}\right)$

$9{\sin }^{-1}\left(1/3\right)+2\sqrt{2}$ Note: the new bounds of integration are . The final answer comes with recognizing that ${\sin }^{-1}\left(-1/3\right)=-{\sin }^{-1}\left(1/3\right)$ and that $\cos \left({\sin }^{-1}\left(1/3\right)\right)=\cos \left({\sin }^{-1}\left(-1/3\right)\right)=2\sqrt{2}/3$.

rational

$\frac{A}{x}+\frac{B}{x-3}$

$\frac{A}{x-\sqrt{7}}+\frac{B}{x+\sqrt{7}}$

$3\ln \left|x-2\right|+4\ln \left|x+5\right|+C$

$\frac{1}{3}\left(\ln \left|x+2\right|-\ln \left|x-2\right|\right)+C$

$-\frac{4}{x+8}-3\ln \left|x+8\right|+C$

$-\ln \left|2x-3\right|+5\ln \left|x-1\right|+2\ln \left|x+3\right|+C$

$x+\ln \left|x-1\right|-\ln \left|x+2\right|+C$

$2x+C$

$\frac{1}{x}+\frac{1}{2}\ln \left|\frac{x-1}{x+1}\right|+C$

$\ln \left|3x^2+5x-1\right|+2\ln \left|x+1\right|+C$

$\ln \left|x\right|-\frac{1}{2}\ln \left(x^2+1\right)-{\tan }^{-1}x-\frac{1}{2\left(x^2+1\right)}+C$

$\frac{1}{2}\left(3\ln \left|x^2+2x+17\right|-4\ln \left|x-7\right|+{\tan }^{-1}\left(\frac{x+1}{4}\right)\right)+C$

$-\frac{1}{4}\ln \left(x^2+3\right)+\frac{1}{4}\ln \left(x^2+1\right)+C=\frac{1}{4}\ln \frac{x^2+1}{x^2+3}+C$

$3\left(\ln \left|x^2-2x+11\right|+\ln \left|x-9\right|\right)+3\sqrt{\frac{2}{5}}{\tan }^{-1}\left(\frac{x-1}{\sqrt{10}}\right)+C$

$\frac{1}{32}\ln \left|x-2\right|-\frac{1}{32}\ln \left|x+2\right|-\frac{1}{16}{\tan }^{-1}\left(x/2\right)+C$

$\ln \left(2000/243\right)\approx 2.108$

$-\pi /4+{\tan }^{-1}3-\ln \left(11/9\right)\approx 0.263$

$x{\sin }^{-1}x+\sqrt{1-x^2}+C$

$18\ln \left|x-2\right|-9\ln \left|x-1\right|-5\ln \left|x-3\right|+C$

${\displaystyle \frac{x}{25\sqrt{x^2+25}}}+C$

$2\ln \left|x-1\right|-\ln \left|x\right|-{\displaystyle \frac{1}{x-1}}-{\displaystyle \frac{1}{{\left(x-1\right)}^2}}+C$

${\displaystyle \frac{1}{2}}{e}^{x^2}\left(x^2-1\right)+C$

${\displaystyle \frac{1}{13}}{e}^{2x}\left(2\sin 3x-3\cos 3x\right)+C$

$-\sqrt{4-x^2}+C$

$2{\tan }^{-1}\sqrt{x}+C$

${\displaystyle \frac{1}{27}}\left[6x\sin 3x-\left(9x^2-2\right)\cos 3x\right]+C$

${\displaystyle \frac{2}{3}}{\left(1+e^x\right)}^{3/2}+C$

${\displaystyle \frac{1}{3}}{\tan }^{3}x+C$

$-{\displaystyle \frac{1}{4}}{\left(8-x^3\right)}^{4/3}+C$

${\displaystyle \frac{1}{10}}{\left(3-2x\right)}^{5/2}-{\displaystyle \frac{1}{2}}{\left(3-2x\right)}^{3/2}+C$

${\displaystyle \frac{2}{5}}{x}^{5/2}-{\displaystyle \frac{8}{3}}{x}^{3/2}+6x^{1/2}+C$

${\displaystyle \frac{11}{2}}\ln \left|x+5\right|-{\displaystyle \frac{15}{2}}\ln \left|x+7\right|+C$

$e^{\tan x}+C$

$-{\displaystyle \frac{1}{5}}{\cot }^{5}x+{\displaystyle \frac{1}{3}}{\cot }^{3}x-\cot x-x+C$

${\displaystyle \frac{1}{3}}{x}^3-{\displaystyle \frac{1}{4}}\tanh 4x+C$

$3{\sin }^{-1}\left({\displaystyle \frac{x+5}{6}}\right)+C$

${\displaystyle \frac{1}{3}}{\sec }^{3}x-\sec x+C$

$-2{\sin }^{-1}\left({\displaystyle \frac{2x}{3}}\right)-{\displaystyle \frac{1}{x}}\sqrt{9-4x^2}+C$

$-\ln x+{\displaystyle \frac{4}{\sqrt[4]{x}}}+4\ln \left|1-\sqrt[4]{x}\right|+C$

${\displaystyle \frac{-x}{2\left(25+x^2\right)}}+{\displaystyle \frac{1}{10}}{\tan }^{-1}\left({\displaystyle \frac{x}{5}}\right)+C$

${\displaystyle \frac{1}{4}}{x}^4-2x^2+4\ln \left|x\right|+C$

${\displaystyle \frac{3}{64}}{\left(2x+3\right)}^{8/3}-{\displaystyle \frac{9}{20}}{\left(2x+3\right)}^{5/3}+{\displaystyle \frac{27}{16}}{\left(2x+3\right)}^{2/3}+C$

$-{\displaystyle \frac{1}{7}}\cos 7x+C$

$\ln \left|1+\tan \frac{\theta }{2}\right|-\ln \left|1-\tan \frac{\theta }{2}\right|+C$

The interval of integration is finite, and the integrand is continuous on that interval.

converges; could also state $\le 10$.

$p>1$

$e^5/2$

$1/3$

$1/\ln 2$

diverges

$1$

diverges

diverges

$2\sqrt{3}$

diverges

diverges

$1$

$0$

$-1/4$

$-1$

diverges

diverges; Limit Comparison Test with $1/x$.

diverges; Limit Comparison Test with $1/x$.

converges; Direct Comparison Test with $e^{-x}$.

converges; Direct Comparison Test with $1/\left(x^2-1\right)$.

converges; Direct Comparison Test with $1/e^x$.

F

They are superseded by the Trapezoidal Rule; it takes an equal amount of work and is generally more accurate.

Trapezoidal Rule: $0.9006$

Simpson’s Rule: $0.90452$

Trapezoidal Rule: $13.9604$

Simpson’s Rule: $13.9066$

Trapezoidal Rule: $1.1703$

Simpson’s Rule: $1.1873$

Trapezoidal Rule: $1.0803$

Simpson’s Rule: $1.077$

Let $f\left(x\right)=a{\left(x-x_1\right)}^2+b\left(x-x_1\right)+c$, so that $f\left(x_1\right)=c=y_1$, $f\left(x_1+\mathrm{\Delta }x\right)=a\mathrm{\Delta }{x}^2+b\mathrm{\Delta }x+c=y_2$, and $f\left(x_1+2\mathrm{\Delta }x\right)=4a\mathrm{\Delta }{x}^2+2b\mathrm{\Delta }x+c=y3$. Therefore, $a=\frac{y_1-2y_2+y_3}{2{\left(\mathrm{\Delta }x\right)}^2}$ and $b=\frac{4y_2-y_3-3y_1}{2\mathrm{\Delta }x}$, and ${\displaystyle {\int }_{x_1}^{x_1+2\mathrm{\Delta }x}}a{\left(x-x_1\right)}^2+b\left(x-x_1\right)+cdx={\displaystyle \frac{a{\left(2\mathrm{\Delta }x\right)}^3}{3}}+{\displaystyle \frac{b{\left(2\mathrm{\Delta }x\right)}^2}{2}}+c\left(2\mathrm{\Delta }x\right)={\displaystyle \frac{4\left(y_1-2y_2+y_3\right)\mathrm{\Delta }x}{3}}+\left(4y_2-y_3-3y_1\right)\mathrm{\Delta }x+2y_1\mathrm{\Delta }x={\displaystyle \frac{\mathrm{\Delta }x}{3}}\left(4y_1-8y_2+4y_3+12y_2-3y_3-9y_1+6y_1\right)={\displaystyle \frac{\mathrm{\Delta }x}{3}}\left(y_1+4y_2+y_3\right)$.