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Table of Integrals I WO Simplified.tex
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\documentclass[10pt]{article} \usepackage{amssymb,amsmath} \usepackage[hmargin=1cm,vmargin=0.5cm]{geometry} \begin{document} Table of Integrals \--- The Working Out (Simplified) \begin{align*} \hline\\ &I=\int\frac{1}{1+x^2}\:dx=\tan^{-1}x+C,\quad x\in\mathbb{R}.\\ \text{Solution:}\quad &\text{Since }1+\tan^2\theta=\sec^2\theta,\quad\text{let }x=\tan\theta,\;\text{where }\theta\in(-\tfrac{\pi}{2},\tfrac{\pi}{2}),\:x\in\mathbb{R}.\\ &dx=d\:(\tan\theta)=\sec^2\theta\:d\theta\quad\text{and}\quad 1+x^2=1+\tan^2\theta=\sec^2\theta\\ &I=\int\frac{1}{\sec^2\theta}\cdot\sec^2\theta\:d\theta=\int d\:\theta=\theta+C=\tan^{-1}x+C.\\ \\ \hline\\ &I=\int\frac{1}{1-x^2}\:dx=\frac{\:1\:}{2}\ln\left|\frac{1+x}{1-x}\right|+C,\quad x\neq\pm 1.\\ \text{Solution:}\quad &I=\frac{\:1\:}{2}\int\left(\frac{1}{1+x}+\frac{1}{1-x}\right)\:dx=\frac{\:1\:}{2}\left(\int\frac{1}{1+x}\:dx+\int\frac{1}{1-x}\:dx\right)\\ &=\frac{\:1\:}{2}\Big(\ln|1+x|-\ln|1-x|\Big)+C=\frac{\:1\:}{2}\ln\left|\frac{1+x}{1-x}\right|+C\\ &\text{\small (Note: $I$ is undefined when $x=\pm 1$. So any definite integrals must avoid having $\pm 1$ in the interval of integration.)}\\ \\ \hline\\ &I=\int\frac{1}{\sqrt{1-x^2}}\:dx=\sin^{-1}x+C,\quad -1
0\quad\text{for }\theta\in(-\tfrac{\pi}{2},\tfrac{\pi}{2}).\\ &I=\int\frac{1}{\cos\theta}\cdot\cos\theta\:d\:\theta=\int d\:\theta=\theta=\sin^{-1}x+C.\\ \\ \hline\\ &I=\int\frac{1}{\sqrt{x^2-1}}\:dx=\ln\left|x+\sqrt{x^2-1}\right|+C,\quad |x|>1\\ \text{Solution:}\quad &\text{Since }\sec^2\theta-1=\tan^2\theta,\quad\text{let }x=\sec\theta,\;\text{where }\theta\in(0,\tfrac{\pi}{2})\cup(\tfrac{\pi}{2},\pi),\:|x|>1.\\ &dx=d\:(\sec\theta)=\sec\theta\tan\theta\:d\theta\quad\text{and}\quad\sqrt{x^2-1}=\sqrt{\sec^2\theta-1}=|\tan\theta|.\\ &I=\int\frac{1}{|\tan\theta|}\cdot\sec\theta\tan\theta\:d\theta=\int\frac{1}{+\tan\theta}\cdot\sec\theta\tan\theta\:d\theta=\int\sec\theta\:d\theta\quad\text{for }\theta\in(0,\tfrac{\pi}{2}),\\ &\text{or }I=\int\frac{1}{-\tan\theta}\cdot\sec\theta\tan\theta\:d\theta=-\int\sec\theta\:d\theta\quad\text{for }\theta\in(\tfrac{\pi}{2},\pi).\\ &\text{Given }\int\sec\theta\:d\theta=\ln|\sec\theta+\tan\theta|+C,\\ &\text{For }\theta\in(0,\tfrac{\pi}{2}),\;\tan\theta=\sqrt{\sec^2\theta-1}=\sqrt{x^2-1},\quad I=\ln\left|x+\sqrt{x^2-1}\right|+C.\\ &\text{For }\theta\in(\tfrac{\pi}{2},\pi),\;\tan\theta=-\sqrt{\sec^2\theta-1}=-\sqrt{x^2-1},\quad I=-\ln\left|x-\sqrt{x^2-1}\right|+C\\ &=\ln\left|\frac{1}{x-\sqrt{x^2-1}}\right|+C=\ln\left|\frac{x+\sqrt{x^2-1}}{x^2-(x^2-1)}\right|+C=\ln\left|x+\sqrt{x^2-1}\right|+C.\\ &\text{Same result in both cases.}\quad\therefore I=\ln\left|x+\sqrt{x^2-1}\right|+C.\\ &\text{\small (Note: $\because|x|>1,\quad\therefore\left(\sqrt{x^2-1}\right)^2=\left|x^2-1\right|=x^2-1>0$.\quad)}\\ \\ \hline\\ \end{align*} \begin{align*} \hline\\ &I=\int\frac{1}{\sqrt{x^2+1}}\:dx=\ln\left|x+\sqrt{x^2+1}\right|+C,\quad x\in\mathbb{R}.\\ \text{Solution:}\quad &\text{Since }\tan^2\theta+1=\sec^2\theta,\quad\text{let }x=\tan\theta,\;\text{where }\theta\in(-\tfrac{\pi}{2},\tfrac{\pi}{2}),\:x\in\mathbb{R}.\\ &dx=d\:(\tan\theta)=\sec^2\theta\:d\theta\quad\text{and}\quad\sec\theta=\sqrt{\tan^2\theta+1}=\sqrt{x^2+1}>0\quad\text{for }\theta\in(-\tfrac{\pi}{2},\tfrac{\pi}{2}).\\ &I=\int\frac{1}{\sec\theta}\cdot\sec^2\theta\:d\theta=\int\sec\theta\:d\theta=\ln|\sec\theta+\tan\theta|+C=\ln\left|x+\sqrt{x^2+1}\right|+C.\\ \\ \hline\\ &I=\int\sqrt{1-x^2}\:dx=\frac{\:x\:}{2}\sqrt{1-x^2}+\frac{\:1\:}{2}\sin^{-1}x+C,\quad -1\leq x\leq 1\\ \text{Solution:}\quad &\text{Since }1-\sin^2\theta=\cos^2\theta,\quad\text{let }x=\sin\theta,\;\text{where }\theta\in[-\tfrac{\pi}{2},\tfrac{\pi}{2}],\:x\in[-1,1].\\ &dx=d\:(\sin\theta)=\cos\theta\:d\theta\quad\text{and}\quad\cos\theta=\sqrt{1-\sin^2\theta}=\sqrt{1-x^2}>0\quad\text{for }\theta\in[-\tfrac{\pi}{2},\tfrac{\pi}{2}].\\ &I=\int\cos\theta\cdot\cos\theta\:d\:\theta=\int\cos^2\theta\:d\:\theta= \int\frac{\cos(2\theta)+1}{2}d\:\theta\\ &=\frac{\:1\:}{4}\sin(2\theta)+\frac{\:\theta\:}{2}+C=\frac{\:1\:}{4}\cdot 2\sin\theta\cos\theta+\frac{\:\theta\:}{2}+C=\frac{\:x\:}{2}\sqrt{1-x^2}+\frac{\:1\:}{2}\sin^{-1}x+C.\\ \\ \hline\\ &I=\int\sqrt{x^2-1}\:dx=\frac{\:x\:}{2}\sqrt{x^2-1}-\frac{\:1\:}{2}\ln\left|x+\sqrt{x^2-1}\right|+C,\quad |x|\geq 1\\ \text{Solution:}\quad &\text{Since }\sec^2\theta-1=\tan^2\theta,\quad\text{let }x=\sec\theta,\;\text{where }\theta\in(0,\tfrac{\pi}{2})\cup(\tfrac{\pi}{2},\pi),\:|x|\geq 1.\\ &dx=d\:(\sec\theta)=\sec\theta\tan\theta\:d\theta\quad\text{and}\quad\sqrt{x^2-1}=\sqrt{\sec^2\theta-1}=|\tan\theta|.\\ &I=\int|\tan\theta|\cdot\sec\theta\tan\theta\:d\theta=\int+\tan\theta\cdot\sec\theta\tan\theta\:d\theta=\int\sec\theta\tan^2\theta\:d\theta\quad\text{for }\theta\in(0,\tfrac{\pi}{2}),\\ &\text{or }I=\int-\tan\theta\cdot\sec\theta\tan\theta\:d\theta=-\int\sec\theta\tan^2\theta\:d\theta\quad\text{for }\theta\in(\tfrac{\pi}{2},\pi).\\ \\ &\text{Let }I_2=\int\sec\theta\tan^2\theta\:d\theta=\int\sec\theta(\sec^2\theta-1)\:d\theta=\int\sec^3\theta\:d\theta-\int\sec\theta\:d\theta=\left[\int\sec\theta\:d(\tan\theta)\right]-\int\sec\theta\:d\theta\\ &=\left[\sec\theta\tan\theta-\int\tan\theta\:d(\sec\theta)\right]-\int\sec\theta\:d\theta =\left[\sec\theta\tan\theta-\int\tan\theta\cdot\sec\theta\tan\theta\:d\theta\right]-\int\sec\theta\:d\theta\\ &=\left[\sec\theta\tan\theta-\int\sec\theta\tan^2\theta\:d\theta\right]-\int\sec\theta\:d\theta =\left[\sec\theta\tan\theta-I_2\right]-\int\sec\theta\:d\theta =-I_2+\left(\sec\theta\tan\theta-\int\sec\theta\:d\theta\right).\\ &\because 2\:I_2=\left(\sec\theta\tan\theta-\int\sec\theta\:d\theta\right),\qquad \therefore I_2=\frac{\:1\:}{2}\left(\sec\theta\tan\theta-\ln|\sec\theta+\tan\theta|\right)+C.\\ \\ &\text{For }\theta\in(0,\tfrac{\pi}{2}),\;\tan\theta=\sqrt{\sec^2\theta-1}=\sqrt{x^2-1},\quad I=I_2=\frac{\:1\:}{2}\left(x\sqrt{x^2-1}-\ln\left|x+\sqrt{x^2-1}\right|\right)+C.\\ &\text{For }\theta\in(\tfrac{\pi}{2},\pi),\;\tan\theta=-\sqrt{\sec^2\theta-1}=-\sqrt{x^2-1},\quad I=-I_2=-\frac{\:1\:}{2}\left(-x\sqrt{x^2-1}-\ln\left|x-\sqrt{x^2-1}\right|\right)+C\\ &\qquad=\frac{\:1\:}{2}\left(x\sqrt{x^2-1}+\ln\left|x-\sqrt{x^2-1}\right|\right)+C =\frac{\:1\:}{2}\left(x\sqrt{x^2-1}-\ln\left|\frac{1}{x-\sqrt{x^2-1}}\right|\right)+C\\ &\qquad=\frac{\:1\:}{2}\left(x\sqrt{x^2-1}-\ln\left|\frac{x+\sqrt{x^2-1}}{x^2-(x^2-1)}\right|\right)+C =\frac{\:1\:}{2}\left(x\sqrt{x^2-1}-\ln\left|x+\sqrt{x^2-1}\right|\right)+C.\\ &\text{Same result in both cases.}\quad\therefore I=\frac{\:1\:}{2}\left(x\sqrt{x^2-1}-\ln\left|x+\sqrt{x^2-1}\right|\right)+C=\frac{\:x\:}{2}\sqrt{x^2-1}-\frac{\:1\:}{2}\ln\left|x+\sqrt{x^2-1}\right|+C.\\ &\text{\small (Note: $\because|x|>1,\quad\therefore\left(\sqrt{x^2-1}\right)^2=\left|x^2-1\right|=x^2-1>0$.\quad)}\\ \\ \hline\\ \end{align*} \begin{align*} \hline\\ &I=\int\sqrt{x^2+1}\:dx=\frac{\:x\:}{2}\sqrt{x^2+1}+\frac{\:1\:}{2}\ln\left|x+\sqrt{x^2+1}\right|+C,\quad x\in\mathbb{R}.\\ \text{Solution:}\quad &\text{Since }\tan^2\theta+1=\sec^2\theta,\quad\text{let }x=\tan\theta,\;\text{where }\theta\in(-\tfrac{\pi}{2},\tfrac{\pi}{2}),\:x\in\mathbb{R}.\\ &dx=d\:(\tan\theta)=\sec^2\theta\:d\theta\quad\text{and}\quad\sec\theta=\sqrt{\tan^2\theta+1}=\sqrt{x^2+1}>0\quad\text{for }\theta\in(-\tfrac{\pi}{2},\tfrac{\pi}{2}).\\ &I=\int\sec\theta\cdot\sec^2\theta\:d\theta=\int\sec\theta\cdot(\tan^2\theta+1)\:d\theta\\ &=\left[\int\sec\theta\tan^2\theta\:d\theta\right]+\int\sec\theta\:d\theta =\left[\frac{\:1\:}{2}\left(\sec\theta\tan\theta-\ln|\sec\theta+\tan\theta|\right)\right]+\ln|\sec\theta+\tan\theta|+C\\ &=\frac{\:1\:}{2}\sec\theta\tan\theta+\frac{\:1\:}{2}\ln|\sec\theta+\tan\theta|+C =\frac{\:1\:}{2}\tan\theta\sec\theta+\frac{\:1\:}{2}\ln|\tan\theta+\sec\theta|+C\\ &=\frac{\:x\:}{2}\sqrt{x^2+1}+\frac{\:1\:}{2}\ln|x+\sqrt{x^2+1}|+C.\\ &\text{\small (Note: From previous solutions: $I_2=\int\sec\theta\tan^2\theta\:d\theta=\frac{\:1\:}{2}\left(\sec\theta\tan\theta-\ln|\sec\theta+\tan\theta|\right)+C$.\quad)}\\ \\ \hline \\ \\ \\ \text{Find}\quad&\int\sqrt{a^2-b^2x^2}\:dx\:,\quad a>0\:\text,\quad b>0\:\text,\quad -\frac{\:a\:}{b}\leq x\leq\frac{\:a\:}{b}\\ &-1\leq\frac{\:b\:}{a}x\leq 1\:,\quad \text{Let }\theta=\sin^{-1}\frac{\:b\:}{a}x\quad\left(\text{where }-\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}\right)\:,\qquad\sin\theta=\frac{\:b\:}{a}x\:,\quad \cos\theta\:d\theta=\frac{\:b\:}{a}dx\\ &\sqrt{a^2-b^2x^2}=a\sqrt{1-\left(\frac{bx}{a}\right)^2}=a\sqrt{1-\sin^2\theta}=|a|\cos\theta\geq 0\quad\text{for }-\tfrac{\pi}{2}\leq\theta\leq\tfrac{\pi}{2}\:.\\ &\int\sqrt{a^2-b^2x^2}\:dx=\int a\cos\theta\cdot\frac{\:a\:}{b}\cos\theta\:d\theta=\frac{a^2}{b}\int\cos^2\theta\:d\theta\\ &=\frac{a^2}{b}\int\frac{1+\cos 2\theta}{2}\:d\theta =\frac{a^2}{2b}\left(\theta+\frac{\sin 2\theta}{2}\right)+C =\frac{a^2}{2b}\left(\sin^{-1}\frac{\:b\:}{a}x+\sin\theta\cos\theta\right)+C\\ &=\frac{a^2}{2b}\left(\sin^{-1}\frac{\:b\:}{a}x+\frac{\:b\:}{a}x\sqrt{1-\left(\frac{bx}{a}\right)^2}\right)+C =\frac{a^2}{2b}\sin^{-1}\frac{\:b\:}{a}x+\frac{\:x\:}{2}\sqrt{a^2-b^2x^2}+C\\ &=\frac{\:x\:}{2}\sqrt{a^2-b^2x^2}+\frac{a^2}{2b}\sin^{-1}\frac{bx}{a}+C\\ \end{align*} \end{document}