Let us consider the integral $\int\frac{5x-3}{x^2-2x-3}dx$. $\frac{d}{dx}(x^2-2x-3)=2x-2$ so substitution is not an option. Noting $x^2-2x-3=(x+1)(x-3)$, let us assume instead that $$\frac{5x-3}{x^2-2x-3}=\frac{A}{x+1}+\frac{B}{x-3}$$ Then \begin{align*}5x-3&=A(x-3)+B(x+1)\\&=(A+B)x-3A+B\end{align*} Hence we obtain a system of linear equations $$\left\{\begin{aligned}A+B&=5\\-3A+B&=-3\end{aligned}\right.$$ Solving this system simultaneously we find $A=2$ and $B=3$.

Alternation: Let’s begin with $5x-3=A(x-3)+B(x+1)$. For $x=3$, we get $12=4B$ so $B=3$. For $x=-1$, we get $-8=-4A$ so $A=2$. This method certainly has a computational advantage when it works over getting a system of linear equations and solving it.

Now \begin{align*}\int\frac{5x-3}{x^2-2x-3}dx&=2\int\frac{dx}{x+1}+3\int\frac{dx}{x-3}\\&=2\ln|x+1|+3\ln|x-3|+C\end{align*}

General method of writing a rational function $\frac{f(x)}{g(x)}$ as a sum of partial fractions

1. Let $x-r$ be a linear factor of $g(x)$. Suppose that $(x-r)^m$ is the highest power of $x-r$ that divides $g(x)$ i.e. $x=r$ is a zero of $g(x)$ with multiplicity $m$. Then to this factor, assign the sum of the $m$ partial fractions $$\frac{A_1}{x-r}+\frac{A_2}{(x-r)^2}+\cdots+\frac{A_m}{(x-r)^m}$$
2. Let $x^2+px+q$ be a quadratic factor of $g(x)$ that cannot be factored further into linear factors with real coefficients. Suppose that $(x^2+px+q)^n$ is the highest power of this factor that divides $g(x)$. Then to this factor assign the sum of the $n$ partial fractions $$\frac{B_1x+C_1}{x^2+px+q}+\frac{B_2x+C_2}{(x^2+px+q)^2}+\cdots+\frac{B_nx+C_n}{(x^2+px+q)^n}$$

Example. Evaluate $\int\frac{x^2+4x+1}{(x-1)(x+1)(x+3)}dx$.

Solution. Let $$\frac{x^2+4x+1}{(x-1)(x+1)(x+3)}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{x+3}$$ Then $$A(x+1)(x+3)+B(x-1)(x+3)+C(x-1)(x+1)$$ For $x=1$, $8A=6$ so $A=\frac{3}{4}$. For $x=-1$, $-4B=-2$ so $B=\frac{1}{2}$. For $x=-3$, $8C=-2$ so $C=-\frac{1}{4}$. Hence, \begin{align*}\int\frac{x^2+4x+1}{(x-1)(x+1)(x+3)}dx&=\frac{3}{4}\int\frac{dx}{x-1}+\frac{1}{2}\int\frac{dx}{x+1}-\frac{1}{4}\int\frac{dx}{x+3}\\&=\frac{3}{4}\ln|x-1|+\frac{1}{2}\ln|x+1|-\frac{1}{4}\ln|x+3|+C\end{align*}

Example. Evaluate $\int\frac{6x+7}{(x+2)^2}dx$.

Solution. Let $\frac{6x+7}{(x+2)^2}=\frac{A}{x+2}+\frac{B}{(x+2)^2}$. Then $A=6$ and $B=-5$. Hence, \begin{align*}\int\frac{6x+7}{(x+2)^2}dx\\&=6\int\frac{dx}{x+2}-5\int\frac{dx}{(x+2)^2}\\&=6\ln|x+2|+\frac{5}{x+2}+C\end{align*}

Example. [Integrating an Improper Fraction] Evaluate $\int\frac{2x^3-4x^2-x-3}{x^2-2x-3}dx$.

Solution. By long division, divide $2x^3-4x^2-x-3$ by $x^2-2x-3$ to obtain quotient $2x$ and remainder $5x-3$. Thus $2x^3-4x^2-x-3=(x^2-2x-3)\cdot 2x+5x-3$ and $$\frac{2x^3-4x^2-x-3}{x^2-2x-3}=2x+\frac{5x-3}{x^2-2x-3}$$ $\frac{5x-3}{x^2-2x-3}$ is a proper fraction so we can apply the above method to write it as $$\frac{5x-3}{x^2-2x-3}=\frac{3}{x-3}+\frac{2}{x+1}$$ Hence, \begin{align*}\int\frac{2x^3-4x^2-x-3}{x^2-2x-3}dx&=\int 2xdx+3\int\frac{dx}{x-3}+2\int\frac{dx}{x+1}\\&=x^2+3\ln|x-3|+2\ln|x+1|+C\end{align*}

Remark. Instead of writing $\frac{5x-3}{x^2-2x-3}$ as $\frac{A}{x-3}+\frac{B}{x+1}$, let $$5x-3=A(x+1)+B(x-3)$$ $\frac{5x-3}{x+1}=A+\frac{B(x-3)}{x+1}$ and if $x=3$, $A=3$. $\frac{5x-3}{x-3}=\frac{A(x+1)}{x-3}+B$ and if $x=-1$, $B=2$. This is called Heaviside’s method and is easier to determine coefficients than the standard method we discussed above. However it can be useful only when the denominator has all linear factors.

Example. Evaluate $\int\frac{-2x+4}{(x^2+1)(x-1)^2}dx$.

Solution. Let $$\frac{-2x+4}{(x^2+1)(x-1)^2}=\frac{Ax+B}{x^2+1}+\frac{C}{x-1}+\frac{D}{(x-1)^2}$$ Then we obtain \begin{align*}-2x+4&=(Ax+B)(x-1)^2+C(x-1)(x^2+1)+D(x^2+1)\\&=(A+C)x^3+(-2A+B-C+D)x^2+(A-2B+C)x+(B-C+D)\end{align*} and hence the equations $A+C=0$, $-2A+B-C+D=0$, $A-2B+C=-2$, and $B-C+D=4$. Solve these equations simultaneously to obtain $A=2$, $B=1$, $C=-2$, and $D=1$. Therefore, \begin{align*}\int\frac{-2x+4}{(x^2+1)(x-1)^2}dx&=\int\frac{2x+1}{x^2+1}dx-2\int\frac{dx}{x-1}+\int\frac{dx}{(x-1)^2}\\&=\int\frac{2x}{x^2+1}dx+\int\frac{1}{x^2+1}dx-2\int{dx}{x-1}+\int\frac{dx}{(x-1)^2}\\&=\ln(x^2+1)+\tan^{-1}x-2\ln|x-1|-\frac{1}{x-1}+C\end{align*}

Example. Evaluate $\int\frac{dx}{x(x^2+1)^2}$.

Solution. Let $$\frac{1}{x(x^2+1)^2}=\frac{A}{x}+\frac{Bx+C}{x^2+1}+\frac{Dx+E}{(x^2+1)^2}$$ Then we have \begin{align*}1&=A(x^2+1)^2+(Bx+C)x(x^2+1)+(Dx+E)x\\&=(A+B)x^4+Cx^3+(2A+B+D)x^2+(C+E)x+A\end{align*} and comparing the coefficients we obtain the equations $A+B=0$, $C=0$, $2A+B+D=0$, $C+E=0$, and $A=1$. Solve these equations simultaneously to obtain $A=1$, $B=-1$, $C=0$, $D=-1$, and $E=0$. Therefore, \begin{align*}\int\frac{dx}{x(x^2+1)^2}&=\int\frac{dx}{x}-\int\frac{x}{x^2+1}dx-\int\frac{x}{(x^2+1)^2}dx\\&=\ln|x|-\frac{1}{2}\ln(x^2+1)+\frac{1}{2(x^2+1)}+C\end{align*}

$y=\sin x$ is not a one-to-one function so there cannot be an inverse function of $y=\sin x$. However we we restrict its domain to the closed interval $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ it becomes a one-to-one function as shown in Figure 1.

This means that we can consider $y=\sin^{-1}x$, the inverse function of $y=\sin x$. That is to say, $y=\sin^{-1}x$ is the value in $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ for which $x=\sin y$. $y=\sin^{-1}x$ is also denoted by $y=\arcsin x$. Similarly,

1. $y=\cos^{-1}x$ (or $y=\arccos x$) is the value in $[0,\pi]$ for which $x=\cos y$.
2. $y=\tan^{-1}x$ (or $y=\arctan x$) is the value in $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ for which $x=\tan y$.
3. $y=\cot^{-1}x$ (or $y=\mathrm{arccot} x$) is the value in $(0,\pi)$ for which $x=\cot y$.

Remark. In general, $y=\sin^{-1}x$ is an inverse relation of $y=\sin x$ which is multiple-valued. In order to consider differentiation, we require it to be single-valued and when $-\frac{\pi}{2}\leq\sin^{-1}x\leq\frac{\pi}{2}$ we call it the principal value of $\sin^{-1}x$ and denote it by $\mathrm{Sin}^{-1}x$. Throughout this note we will only consider principal values so we won’t be using the traditional notation like $y=\mathrm{Sin}^{-1}x$.

Recall that the graph of $y=f(x)$ and the graph of its inverse function $y=f^{-1}(x)$ are symmetric about the line $y=x$. Using this symmetry one can obtain the graph of an inverse trigonometric function. For example, Figure 2 shows the graph of $y=\sin x$ and the graph of $y=\sin^{-1} x$.

In addition, $y=\sec^{-1}x$ has domain $|x|\geq 1$ and range $\left[0,\frac{\pi}{2}\right)\cup\left(\frac{\pi}{2},\pi\right]$.

$y=\csc^{-1}x$ has domain $|x|\geq 1$ and range $\left[-\frac{\pi}{2},0\right)\cup\left(0,\frac{\pi}{2}\right]$.

Inverse Functions and Their Derivatives

Let $y=f^{-1}(x)$ denote the inverse function of $f(x)$. Then \begin{equation}\label{eq:invfn}x=f(y)\end{equation} Differentiate \eqref{eq:invfn} with respect to $x$. \begin{equation}\label{eq:dinvfn}1=f'(y)\frac{dy}{dx}\end{equation} Solving \eqref{eq:dinvfn} for $\frac{dy}{dx}$ we have \begin{equation}\label{eq:dinvfn2}\frac{dy}{dx}=\frac{1}{f'(f^{-1}(x))}\end{equation}

Example. Let $f(x)=\ln x$. Knowing $f'(x)=\frac{1}{x}$ find the derivative of $f^{-1}(x)=e^x$.

Solution. Using \eqref{eq:dinvfn2} $$\frac{d}{dx}e^x=\frac{1}{\frac{1}{e^x}}=e^x$$

Although knowing the formula \eqref{eq:dinvfn2} is convenient, you can always find the derivative of an inverse function by following the same process of deriving \eqref{eq:dinvfn2}. It’s not actually anymore difficult or complicated than using \eqref{eq:dinvfn2}.

The Derivative of Inverse Trigonometric Functions

Let $f(x)=\sin x$. Then $f^{-1}(x)=\sin^{-1}x$. Using \eqref{eq:dinvfn2} \begin{align*}\frac{d}{dx}\sin^{-1}x&=\frac{1}{\cos(\sin^{-1}x)}\\&=\frac{1}{\sqrt{1-\sin^2(\sin^{-1}x)}}\\&=\frac{1}{\sqrt{1-x^2}}\end{align*} where $|x|<1$. The reason the sign in front of $\sqrt{}$ is positive is that $-\frac{\pi}{2}<\sin^{-1}x<\frac{\pi}{2}$ so $\cos(\sin^{-1}x)> 0$.

Alternative derivation: Let $y=\sin^{-1}x$. Then $x=\sin y$. By implicit differentiation $$1=\cos y\frac{dy}{dx}$$ Hence $$\frac{dy}{dx}=\frac{1}{\cos y}=\frac{1}{\cos(\sin^{-1}x)}=\frac{1}{\sqrt{1-x^2}}$$

If $u$ is a functions of $x$, then $$\frac{d}{dx}\sin^{-1}u=\frac{1}{\sqrt{1-u^2}}\frac{du}{dx},\ |u|<1$$ Similarly we obtain the rest of derivative formulas. \begin{align*}\frac{d}{dx}\cos^{-1}u&=-\frac{1}{\sqrt{1-u^2}}\frac{du}{dx},\ |u|<1\\\frac{d}{dx}\tan^{-1}x&=\frac{1}{1+u^2}\frac{du}{dx}\\\frac{d}{dx}\cot^{-1}x&=-\frac{1}{1+u^2}\frac{du}{dx}\\\frac{d}{dx}\sec^{-1}u&=\frac{1}{|u|\sqrt{u^2-1}}\frac{du}{dx},\ |u|>1\\\frac{d}{dx}\csc^{-1}x&=-\frac{1}{|u|\sqrt{u^2-1}}\frac{du}{dx},\ |u|>1\end{align*} In turns out we don’t really have to calculate all these formulas. We just need to calculate for example $\frac{d}{dx}\sin^{-1}x$, $\frac{d}{dx}\tan^{-1}x$, $\frac{d}{dx}\sec^{-1}x$ the rest can be obtained by inverse function-inverse cofunction identities \begin{align*}\cos^{-1}x&=\frac{\pi}{2}-\sin^{-1}x\\\cot^{-1}x&=\frac{\pi}{2}-\tan^{-1}x\\\csc^{-1}x&=\frac{\pi}{2}-\sec^{-1}x\end{align*}In case you haven’t seen this identities before they can be easily obtained from cofunction identities. For example sine and cosine are cofunctions of each other as you learned in trigonometry, namely $$\sin\left(\frac{\pi}{2}-x\right)=\cos x,\ \cos\left(\frac{\pi}{2}-x\right)=\sin x$$ Let $$\cos\left(\frac{\pi}{2}-y\right)=\sin y=x$$ Then $$\frac{\pi}{2}-y=\cos^{-1}x$$ i.e. the first identity above $$\cos^{-1}x=\frac{\pi}{2}-\sin^{-1}x$$

Integration Formulas

For any constant $a\ne 0$,

1. $\int\frac{du}{\sqrt{a^2-u^2}}=\sin^{-1}\left(\frac{u}{a}\right)+C$, valid for $u^2<a^2$
2. $\int\frac{du}{a^2+u^2}=\frac{1}{a}\tan^{-1}\left(\frac{u}{a}\right)+C$
3. $\int\frac{du}{u\sqrt{u^2-a^2}}=\frac{1}{a}\sec^{-1}\left|\frac{u}{a}\right|+C$, valid for $|u|>a>0$

Example. \begin{align*}\int_{\sqrt{2}/2}^{\sqrt{3}/2}\frac{dx}{\sqrt{1-x^2}}&=\left.\sin^{-1}x\right|_{\sqrt{2}/2}^{\sqrt{3}/2}\\&=\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)-\sin^{-1}\left(\frac{1}{\sqrt{2}}\right)\\&=\frac{\pi}{3}-\frac{\pi}{4}=\frac{\pi}{12}\end{align*}

Example. \begin{align*}\int\frac{dx}{\sqrt{3-4x^2}}&=\frac{1}{2}\int\frac{du}{\sqrt{3-u^2}} (u=2x)\\&=\frac{1}{2}\sin^{-1}\left(\frac{u}{\sqrt{3}}\right)+C\\&=\frac{1}{2}\sin^{-1}\left(\frac{2x}{\sqrt{3}}\right)+C\end{align*}

Example. \begin{align*}\int\frac{dx}{4x-x^2}&=\int\frac{dx}{4-(x-2)^2}\\&=\int\frac{du}{4-u^2} (u=x-2)\\&=\sin^{-1}\left(\frac{u}{2}\right)+C\\&=\sin^{-1}\left(\frac{x-2}{2}\right)+C\end{align*}

Example. \begin{align*}\int\frac{dx}{4x^2+4x+2}&=\int\frac{dx}{4\left(x+\frac{1}{2}\right)^2+1}\\&=\frac{1}{2}\int\frac{du}{u^2+1} (u=2x+1)\\&=\frac{1}{2}\tan^{-1}u+C\\&=\frac{1}{2}\tan^{-1}(2x+1)+C\end{align*}

Example. \begin{align*}\int\frac{dx}{e^{2x}-6}&=\int\frac{du}{u\sqrt{u^2-6}}\ (u=e^x)\\&=\frac{1}{\sqrt{6}}\sec^{-1}\left(\frac{e^x}{\sqrt{6}}\right)+C\end{align*} where $e^x>\sqrt{6}$ or equivalently $x>\ln\sqrt{6}\approx 0.8959$.

Integrating Inverses of Functions

Let $y=f^{-1}(x)$. Then $x=f(y)$ and $dx=f'(y)dy$. So with integration by parts, we have \begin{equation}\begin{aligned}\int f^{-1}(x)dx&=\int yf'(y)dy\\&=yf(y)-\int f(y)dy\\&=xf^{-1}(x)-\int f(y)dy\end{aligned}\label{eq:intinvfn}\end{equation} Using \eqref{eq:dinvfn2} we can also rewrite \eqref{eq:intinvfn} as \begin{equation}\label{eq:intinvfn2}\int f^{-1}(x)dx=xf^{-1}(x)-\int\frac{x}{f'(f^{-1}(x))}dx\end{equation}

Using \eqref{eq:intinvfn}\begin{align*}\int\cos^{-1}xdx&=x\cos^{-1}x-\sin y+C\\&=x\cos^{-1}x-\sin(\cos^{-1}x)+C\end{align*}Since $x=\cos y$, $$\sin(\cos^{-1}x)=\sin y=\sqrt{1-x^2}$$ (Recall that $0\leq\cos^{-1}x\leq\pi$ so $\sin^{-1}x\geq 0$.) Hence, \begin{equation}\label{eq:intinvcos}\int\cos^{-1}xdx=x\cos^{-1}x-\sqrt{1-x^2}+C\end{equation} Of course one can obtain \eqref{eq:intinvcos} using \eqref{eq:intinvfn2} though the required calculation is a bit longer. The integral \eqref{eq:intinvcos} can be also found without using \eqref{eq:intinvfn} or \eqref{eq:intinvfn2}. Using integration by parts\begin{align*}\int\cos^{-1}xdx&=x\cos^{-1}x+\int\frac{x}{\sqrt{1-x^2}}dx\\&=x\cos^{-1}x-\sqrt{1-x^2}+C\end{align*} The rest of the integrals of inverse trigonometric functions are given by\begin{align*}\int\sin^{-1}xdx&=x\sin^{-1}x+\sqrt{1-x^2}+C\\\int\tan^{-1}xdx&=x\tan^{-1}x-\frac{1}{2}\ln(1+x^2)+C\\\int\cot^{-1}xdx&=x\cot^{-1}x+\frac{1}{2}\ln(1+x^2)+C\\\int\sec^{-1}xdx&=x\sec^{-1}x-\frac{x}{|x|}\ln|x+\sqrt{x^2-1}|+C\\\int\csc^{-1}xdx&=x\csc^{-1}x+\frac{x}{|x|}\ln|x+\sqrt{x^2-1}|+C\end{align*}

When the integrand contains $\sqrt{a^2+x^2}$, $\sqrt{a^2-x^2}$ or $\sqrt{x^2-a^2}$ and the integral cannot be evaluated by a simple substitution, the following trigonometric substitutions may come in handy:

1. For $\sqrt{a^2+x^2}$, use $x=a\tan\theta$ with $-\frac{\pi}{2}<\theta<\frac{\pi}{2}$. Accordingly $dx=a\sec^2\theta d\theta$ and $a^2+x^2=a^2\sec^2\theta$.
2. For $\sqrt{a^2-x^2}$, use $x=a\sin\theta$ with $-\frac{\pi}{2}\leq x\leq\frac{\pi}{2}$. Accordingly $dx=a\cos\theta d\theta$ and $a^2-x^2=a^2\cos^2\theta$. Note: You can also use the substitution $x=a\cos\theta$ with $0\leq\theta\leq\pi$. Accordingly $dx=-a\sin\theta d\theta$ and $a^2-x^2=a^2\sin^2\theta$.
3. For $\sqrt{x^2-a^2}$, use $x=a\sec\theta$ with $0\leq\theta<\frac{\pi}{2}$ or $\pi\leq\theta<\frac{3\pi}{2}$. Accordingly $dx=a\sec\theta\tan\theta d\theta$ and $x^2-a^2=a^2\tan^2\theta$.

Example. Evaluate $\int\frac{dx}{\sqrt{4+x^2}}$.

Solution. Let $x=2\tan\theta$. Then $dx=2\sec^2\theta d\theta$ and $4+x^2=4\sec^2\theta$. So \begin{align*}\int\frac{dx}{\sqrt{4+x^2}}&=\int\frac{2\sec^2\theta d\theta}{2|\sec\theta|}\\&=\int\sec\theta d\theta\ \left(\sec\theta>0\ \mbox{for}\ -\frac{\pi}{2}<\theta<\frac{\pi}{2}\right)\\&=\ln|\sec\theta+\tan\theta|+C’\end{align*} SInce $\tan\theta=\frac{x}{2}$ one may consider the triangle

Hence $$\int\frac{dx}{\sqrt{4+x^2}}=\ln|\sqrt{4+x^2}+x|+C$$ where $C=C’-\ln 2$.

Example. Evaluate $\int\frac{x^2}{\sqrt{9-x^2}}dx$.

Solution. Let $x=3\sin\theta$. Then $dx=3\cos\theta d\theta$ and $9-x^2=9\cos^2\theta$. \begin{align*}\int\frac{x^2}{\sqrt{9-x^2}}dx&=\int\frac{9\sin^2\theta(3\cos\theta d\theta)}{3|\cos\theta|}\\&=9\int\sin^2\theta d\theta\ \left(\cos\theta>0\ \mbox{for}\ -\frac{\pi}{2}<\theta<\frac{\pi}{2}\right)\\&=9\int\frac{1-\cos 2\theta}{2}d\theta\\&=\frac{9}{2}(\theta-\sin\theta\cos\theta)+C\\&=\frac{9}{2}\left(\sin^{-1}\frac{x}{3}-\frac{x}{3}\frac{\sqrt{9-x^2}}{3}\right)+C\\&=\frac{9}{2}\sin^{-1}\frac{x}{3}-\frac{x}{2}\sqrt{9-x^2}+C\end{align*}

Example. Evaluate $\int\frac{dx}{\sqrt{25x^2-4}}$.

Solution. $\int\frac{dx}{\sqrt{25x^2-4}}=\frac{1}{5}\int\frac{dx}{\sqrt{x^2-\left(\frac{2}{5}\right)^2}}$. Let $x=\frac{2}{5}\sec\theta$. Then $dx=\frac{2}{5}\sec\theta\tan\theta d\theta$ and $x^2-\left(\frac{2}{5}\right)^2=\frac{4}{25}\tan^2\theta$. \begin{align*}\int\frac{dx}{\sqrt{25x^2-4}}&=\frac{1}{5}\int\frac{dx}{\sqrt{x^2-\left(\frac{2}{5}\right)^2}}\\&=\frac{1}{5}\int\frac{\frac{2}{5}\sec\theta\tan\theta d\theta}{5\left(\frac{2}{5}\right)\tan\theta}\ \left(\tan\theta>0\ \mbox{for}\ 0<\theta<\frac{\pi}{2}\right)\\&=\frac{1}{5}\int\sec\theta d\theta\\&=\frac{1}{5}\ln|\sec\theta +\tan\theta|+C’\\&=\frac{1}{5}\ln|5x+\sqrt{25x^2-4}|+C\end{align*} where $C=C’-\frac{\ln 2}{5}$.

Example. Evaluate $\int\sqrt{1+x^2}dx$.

Solution. Let $x=\tan\theta$. Then $dx=\sec^2\theta d\theta$ and $1+x^2=\sec^2\theta$. \begin{align*}\int\sqrt{1+x^2}dx&=\int\sec^3\theta d\theta\\&=\frac{1}{2}\sec\theta\tan\theta+\frac{1}{2}\ln|\sec\theta+\tan\theta|+C\\&=\frac{1}{2}x\sqrt{1+x^2}+\frac{1}{2}\ln|x+\sqrt{1+x^2}|+C\end{align*} The integration $$\int\sec^3\theta d\theta=\frac{1}{2}\sec\theta\tan\theta+\frac{1}{2}\ln|\sec\theta+\tan\theta|+C$$ was discussed here.

Example. Evaluate $\int\frac{\sqrt{1+x^2}}{x}dx$.

Solution. Let $x=\tan\theta$. Then $dx=\sec^2\theta d\theta$ and $1+x^2=\sec^2\theta$. \begin{align*}\int\frac{\sqrt{1+x^2}}{x}dx&=\int\frac{\sec^3\theta}{\tan\theta}d\theta\\&=\int\frac{\sec\theta\sec^2\theta}{\tan\theta}d\theta\\&=\int\frac{\sec\theta(1+\tan^2\theta)}{\tan\theta}d\theta\\&=\int\frac{\sec\theta}{\tan\theta}d\theta+\int\sec\theta\tan\theta d\theta\\&=\int\csc\theta d\theta+\int\sec\theta\tan\theta d\theta\\&=-\ln|\csc\theta+\cot\theta|+\sec\theta+C\\&=-\ln\left|\frac{\sqrt{1+x^2}}{x}+\frac{1}{x}\right|+\sqrt{1+x^2}+C\end{align*}

Example. Find the area enclosed by the ellipse $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$

Solution. The ellipse in the first quadrant is given by the function $y=\frac{b}{a}\sqrt{a^2-x^2}$, $0\leq x\leq a$. So the area is \begin{align*}A&=4\int_0^a\frac{b}{a}\sqrt{a^2-x^2}dx\\&=4\frac{b}{a}\int_0^{\frac{\pi}{2}}a\cos\theta(a\cos\theta d\theta)\ (x=a\sin\theta$)\\&=4ab\int_0^{\frac{\pi}{2}}\cos^2\theta d\theta\\&=4ab\int_0^{\frac{\pi}{2}}\frac{1+\cos 2\theta}{2}d\theta\\&=\pi ab\end{align*}

If two functions $f(x)$ and $g(x)$ both approach zero as $x\to a$, the function $\frac{f(x)}{g(x)}$ is said to assume the indeterminate form $\frac{0}{0}$ at $x=a$. Regardless of the term, $\frac{f(x)}{g(x)}$ may approach a limit as $x$ approaches $a$. The following theorem named after the French mathematician G. F. A. de L’Hôpital (1661-1704) is useful for the process of determining this limit if it exists.

Theorem (L’Hôpital’s Rule). If the functions $f(x)$ and $g(x)$ are continuous in an interval containing $a$ and if $f'(x)$ and $g'(x)$ exist such that $g'(x)\ne 0$ in this interval (except possibly at $x=a$), then when $\lim_{x\to a}f(x)=\lim_{x\to 0}g(x)=0$ (or equivalently $f(a)=g(a)=0$), we have $$\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)}$$ provided the limit on the right exists.

Example. Evaluate $\lim_{x\to 0}\frac{\tan x}{x}$.

Solution. The limit is an indeterminate form $\frac{0}{0}$. By applying L’Hôpital’s Rule, we obtain $$\lim_{x\to 0}\frac{\tan x}{x}=\lim_{x\to 0}\frac{\sec^2x}{1}=1$$

The Indeterminate form $\frac{\infty}{\infty}$

If $f(x)\to\infty$ and  $g(x)\to \infty$ as $x\to a$ (or $x\to\pm\infty$), the function $\frac{f(x)}{g(x)}$ is said to assume the indeterminate form $\frac{\infty}{\infty}$ at $x=a$ (or at $x=\pm\infty$). The limit of $\frac{\infty}{\infty}$ as $x\to a$ (or $x\to\pm\infty$) may still be found by L’Hôpital’s Rule if it exists.

Example. $$\lim_{x\to\infty}\frac{x^2}{e^x}=\lim_{x\to\infty}\frac{2x}{e^2}=\lim_{x\to\infty}\frac{2}{e^x}=0$$

Example. Evaluate $\lim_{x\to 0+}\frac{e^{-\frac{1}{x}}}{x}$.

Solution. Applying L’Hôpital’s Rule, we have $$\lim_{x\to 0+}\frac{e^{-\frac{1}{x}}}{x}=\lim_{x\to 0+}\frac{e^{-\frac{1}{x}}}{x^2}$$ While L’Hôpital’s rule can still be applied, we would accomplish nothing by doing so. This example shows that L’Hôpital’s rule may not necessarily leads to a desirable result. For the above example, the limit can be found by $$\lim_{x\to 0+}\frac{e^{-\frac{1}{x}}}{x}=\lim_{x\to 0+}\frac{\frac{1}{x}}{e^{\frac{1}{x}}}=\lim_{z\to\infty}\frac{z}{e^z}=\lim_{z\to\infty}\frac{1}{e^z}=0$$

The indeterminate form $0\cdot\infty$

If $f(x)\to 0$ and $g(x)\to\infty$ as $x\to a$ (or $x\to\pm\infty$), the function $f(x)g(x)$ is said to assume the indeterminate form $0\cdot\infty$ at $x=a$ (or at $x=\pm\infty$). If the limit of $f(x)g(x)$ as $x\to a$ (or $x\to\pm\infty$) exists, it may be found by writing $f(x)g(x)$ as $$\frac{f(x)}{\frac{1}{g(x)}}\ \mbox{or}\ \frac{g(x)}{\frac{1}{f(x)}}$$ and applying L’Hôpital’s rule.

Example.

1. $\lim_{x\to\infty}xe^{-x}=\lim_{x\to\infty}\frac{x}{e^x}=\lim_{x\to\infty}\frac{1}{e^x}=0$.
2. $\lim_{x\to 0}\sin 3x\cot 2x=\lim_{x\to 0}\frac{\sin 3x}{\tan 2x}=\lim_{x\to 0}\frac{3\cos 3x}{2\sec^22x}=\frac{3}{2}$.

The Indeterminate Form $\infty-\infty$

If $f(x)\to\infty$ and $g(x)\to\infty$ as $x\to a$ (or $x\to\pm\infty$), the difference $f(x)-g(x)$ is said to assume the indeterminate form $\infty-\infty$  at $x=a$ (or at $x=\pm\infty$). If the limit of $f(x)-g(x)$ as $x\to a$ (or as $x\to\pm\infty$) exists, it may be found by transforming the difference into a fraction by algebraic means and applying L’Hôpital’s rule.

Example. $\lim_{x\to 0}(\csc x-\cot x)=\lim_{x\to 0}\frac{1-\cos x}{\sin x}=\lim_{x\to 0}\frac{\sin x}{\cos x}=0$

Example. Evaluate $\lim_{x\to\infty}(x-\ln x)$.

Solution. While $x-\ln x$ assumes the indeterminate form $\infty-\infty$ at $x=\infty$, there is no algebraic means to transform it to a fraction. An indeterminate of the form $\infty-\infty$ such as the one in consideration may be evaluated by finding the limit of its exponential. Let $y=x-\ln x$. Then $$e^y=e^{x-\ln x}=\frac{e^x}{e^{\ln x}}=\frac{e^x}{x}$$ Hence, $$\lim_{x\to\infty}e^y=\lim_{x\to\infty}\frac{e^x}{x}=\lim_{x\to\infty}\frac{e^x}{1}=\infty$$ Since $y\to\infty$ when $e^y\to\infty$, $$\lim_{x\to\infty}(x-\ln x)=\infty$$

The Indeterminate Forms $0^0$, $\infty^0$, $1^\infty$

If $f(x)\to 0$ and $g(x)\to 0$, or $f(x)\to\infty$ and $g(x)\to 0$, or $f(x)\to 1$ and $g(x)\to\infty$ as $x\to a$ (or $x\to\pm\infty$), $f(x)^{g(x)}$ is said to assume the indeterminate form $0^0$, $\infty^0$, or $1^\infty$, respectively at $x=a$ (or at $x=\pm\infty$). If the limit of $f(x)^{g(x)}$ exists as $x\to a$ (or $x\to\pm\infty$), it may be found by denoting $f(x)^{g(x)}$ by $y$ and investigating the limit approached by the logarithm $$\ln y=g(x)\ln f(x)$$ If $\lim_{x\to a}\ln y=k$, then $\lim_{x\to a}y=e^k$.

Example. Evaluate $\lim_{x\to 0}x^x$.

Solution. $x^x$ assumes the indeterminate form $0^0$ at $x=0$. Let $y=x^x$. Then $\ln y=x\ln x$ and $$\lim_{x\to 0}\ln y=\lim_{x\to 0}\frac{\ln x}{\frac{1}{x}}=\lim_{x\to 0}\frac{\frac{1}{x}}{-\frac{1}{x^2}}=\lim_{x\to 0}(-x)=0$$ Hence $\lim_{x\to 0}y=\lim_{x\to 0}x^x=e^0=1$.

Example. Evaluate $\lim_{x\to 0}(1-\sin x)^{\frac{1}{x}}$.

Solution. The function assumes the indeterminate form $1^\infty$ at $x=0$. Let $y=(1-\sin x)^{\frac{1}{x}}$. Then $\ln y=\frac{\ln(1-\sin x)}{x}$ and $$\lim_{x\to 0}\ln y=\lim_{x\to 0}\frac{\ln(1-\sin x)}{x}=\lim_{x\to 0}\frac{\frac{-\cos x}{1-\sin x}}{1}=-1$$ Hence, $\lim_{x\to 0}(1-\sin x)^{\frac{1}{x}}=e^{-1}=\frac{1}{e}$.

Example. Use L’Hôpital’s rule to show the limit $$\lim_{x\to 0}(1+x)^{\frac{1}{x}}=e$$

Proof. Let $y=(1+x)^{\frac{1}{x}}$. Then $y$ assumes the indeterminate form $1^\infty$ at $x=0$. $\ln y=\frac{\ln(1+x)}{x}$ and $$\lim_{x\to 0}\ln y=\lim_{x\to 0}\frac{\ln(1+x)}{x}=\lim_{x\to 0}\frac{1}{1+x}=1$$ Hence, $$\lim_{x\to 0}(1+x)^{\frac{1}{x}}=e$$

Suppose that $f$ is differentiable infinitely many times on an open interval containing $a$. Then $$\sum_{n=0}^\infty\frac{f^{(n)}(a)}{n!}(x-a)^n=f(a)+\frac{f'(a)}{1!}(x-a)+\frac{f^{\prime\prime}(a)}{2!}(x-a)^2+\frac{f^{\prime\prime\prime}(a)}{3!}(x-a)^3+\cdots$$ is called the Taylor series of $f$ centered at $a$. A Taylor series centered at 0 is called a Maclaurin series.

Example. Find the Maclaurin series of $f(x)=e^x$ and its radius of convergence $R$.

Solution. The Maclaurin series is $\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}x^n=\sum_{n=0}^\infty\frac{x^n}{n!}$. By the ratio test $$\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\lim_{n\to\infty}\frac{|x|}{n+1}=0<1$$ for all $x$ so the series converges for all $x$ i.e. the radius of convergence is $R=\infty$.

The following theorem tells when a function $f(x)$ can be represented by its Taylor series. Recall that $f(x)=T_n(x)+R_n(x)$ where $T_n(x)$ is the $n$-th degree Taylor polynomial of $f$ at $a$ and $R_n(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1}$ where $a<\xi<x$ or $x<\xi<a$.

Theorem. $f=\sum_{n=0}^\infty\frac{f^{(n)}(a)}{n!}(x-a)^n$ for $|x-a|<R$ if and only if $\lim_{n\to\infty}R_n(x)=0$ for $|x-a|<R$.

Since $\sum_{n=0}^\infty\frac{x^n}{n!}$ converges for all $x$, \begin{equation}\label{eq:explim}\lim_{n\to\infty}\frac{x^n}{n!}=0\end{equation} Now we show that \begin{equation}\label{eq:expx}e^x=\sum_{n=0}^\infty\frac{x^n}{n!}\end{equation} for all $x$. $R_n(x)=\frac{e^\xi}{(n+1)!}x^{n+1}$ where $0<\xi<x$ or $x<\xi<0$. If $0<\xi<x$ then $0<R_n(x)<\frac{e^x}{(n+1)!}x^{n+1}\to 0$ as $n\to\infty$ by \eqref{eq:explim}. If $x<\xi<0$ then $0\leq |R_n(x)|<\frac{|x|^{n+1}}{(n+1)!}\to 0$ as $n\to\infty$ again by \eqref{eq:explim}. This completes the proof of \eqref{eq:expx}. For $x=1$ we have a definition of the Euler number $e$ in terms of a series as \begin{equation}\label{eq:eulernum}e=\sum_{n=0}^\infty\frac{1}{n!}=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\cdots\end{equation}

Example. Find the Taylor series of $f(x)=e^x$ at $a=2$.

Solution. By the same manner, one can show that $$e^x=\sum_{n=0}^\infty\frac{e^2}{n!}(x-2)^n$$ for all $x$.

Example. Find the Maclaurin series of $\sin x$ and show that it represents $\sin x$ for all $x$.

Solution. The Maclaurin series is $\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)!}$ and by the ratio test one can show that it converges for all $x$. Since $|f^{(n+1)}(\xi)|\leq 1$, $0\leq |R_n(x)|\leq\frac{|x|^{n+1}}{(n+1)!}\to 0$ as $n\to\infty$. Hence, \begin{equation}\label{eq:sinx}\sin x=\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)!}\end{equation} for all $x$.

Similarly one can also show that \begin{equation}\label{eq:cosx}\cos x=\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n)!}\end{equation} for all $x$.

Euler’s Formula

If $x$ is replaced by $ix$ in \eqref{eq:expx}, we obtain \begin{align*}e^{ix}&=\sum_{n=0}^\infty\frac{(ix)^n}{n!}\\&=\sum_{n=0}^\infty\frac{x^{2n}}{(2n)!}+i\sum_{n=0}^\infty\frac{x^{2n+1}}{(2n+1)!}\\&=\cos x+i\sin x\end{align*}\begin{equation}\label{eq:eulerformula}e^{ix}=\cos x+i\sin x\end{equation} is called the Euler’s Formula and it represents the point on the unit circle centered at the origin corresponding to the angle $x$. The formula in \eqref{eq:eulerformula} comes in handy in so many places of mathematics from trigonometry, calculus, differential equations to abstract algebra, topology, geometry to name a few. It is also a useful tool in physics on many occasions. For example, one can derive the sine sum and the cosine sum formulas using \eqref{eq:eulerformula}: $e^{i(\theta_1+\theta_2)}=\cos(\theta_1+\theta_2)+i\sin(\theta_1+\theta_2)$.  On the other hand, \begin{align*}e^{i(\theta_1+\theta_2)}&=e^{i\theta_1}e^{i\theta_2}\\&=(\cos\theta_1+i\sin\theta_1)(\cos\theta_2+i\sin\theta_2)\\&=(\cos\theta_1\cos\theta_2-\sin\theta_1\sin\theta_2)+i(\cos\theta_1\sin\theta_2+\sin\theta_1\cos\theta_2)\end{align*} Hence we obtain the sine and the cosine sum formulas \begin{align*}\sin(\theta_1+\theta_2)&=\sin\theta_1\cos\theta_2+\cos\theta_1\sin\theta_2\\\cos(\theta_1+\theta_2)&=\cos\theta_1\cos\theta_2-\sin\theta_1\sin\theta_2\end{align*} When $\theta=\pi$, we obtain the so-called the Euler’s identity \begin{equation}\label{eq:euleridentity}e^{\pi i}+1=0\end{equation} which contains the most fundamental constants of mathematics $\pi$, $e$, $i$, 0, and 1.

The Binomial Series

Let us find the Maclaurin series of $f(x)=(1+x)^k$ where $k$ is any real number. First, we find $$f^{(n)}(x)=k(k-1)\cdots(k-n+1)(1+x)^{k-n}$$ and so $$f^{(n)}(0)=k(k-1)\cdots(k-n+1)$$ Hence the Maclaurin series is given by $$\sum_{n=0}^\infty\frac{k(k-1)\cdots(k-n+1)}{n!}x^n$$ The coefficients $\frac{k(k-1)\cdots(k-n+1)}{n!}$, $n=0,1,2,\cdots$ are denoted by ${}_k\mathrm{C}_n$ or $\begin{pmatrix}k\\n\end{pmatrix}$ and are called binomial coefficients. $$\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\left|\frac{k-n}{n+1}\right||x|=\lim_{n\to\infty} |x|$$ So by the ratio test if $|x|<1$ then the series converges and if $|x|>1$ then it diverges. Furthermore it can be shown (we omit the proof) that \begin{equation}\label{eq:binomialseries}(1+x)^k=\sum_{n=0}^\infty\frac{k(k-1)\cdots(k-n+1)}{n!}x^n\end{equation} for $|x|<1$. The series in \eqref{eq:binomialseries} is called the binomial series.

Example. FInd the Maclaurin series of $f(x)=\frac{1}{\sqrt{4-x}}$.

Solution. \begin{align*}\frac{1}{\sqrt{4-x}}&=\frac{1}{2}\frac{1}{\sqrt{1-\frac{x}{4}}}\\&=\frac{1}{2}\left(1-\frac{x}{4}\right)^{-\frac{1}{2}}\\&=\frac{1}{2}\sum_{n=0}^\infty\begin{pmatrix}-\frac{1}{2}\\n\end{pmatrix}\left(-\frac{x}{4}\right)^n\\&=\frac{1}{2}\left\{1+\left(-\frac{1}{2}\right)\left(-\frac{x}{4}\right)+\frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2!}\left(-\frac{x}{4}\right)^2+\frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)}{3!}\left(-\frac{x}{4}\right)^3+\cdots\\+\frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)\cdots\left(-\frac{2n-1}{2}\right)}{n!}\left(-\frac{x}{4}\right)^n+\cdots\right\}\\&=\frac{1}{2}\left\{1+\frac{x}{8}+\frac{1\cdot 3}{2!8^2}x^2+\frac{1\cdot\cdot 5}{3!8^3}x^3+\cdots+\frac{1\cdot 3\cdot 5\cdots(2n-1)}{n!8^n}x^n+\cdots\right\}\end{align*} This series converges if $\left|-\frac{x}{4}\right|<1$ i.e. $|x|<4$. The radius of convergence is 4.

Working with Taylor Series

Example. Use Taylor series to evaluate $\lim_{x\to 0}\frac{x^2+2\cos x-2}{3x^4}$.

Solution. \begin{align*}\lim_{x\to 0}\frac{x^2+2\cos x-2}{3x^4}&=\lim_{x\to 0}\frac{x^2+2\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots\right)-2}{3x^4}\\&=\lim_{x\to 0}\frac{2\left(\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots\right)}{3x^4}\\&=\frac{2}{4!3}=\frac{1}{36}\end{align*}

Remark. The above limit can be also calculated using the L’Hôpital’s rule as it is an indeterminate form of type $\frac{0}{0}$.

Example. Approximate $\int_0^1 e^{-x^2}dx$ with an error no greater than $5\times 10^{-4}$.

Solution. \begin{align*}\int_0^1 e^{-x^2}dx&=\int_0^1\left(1-x^2+\frac{x^4}{2!}-\frac{x^6}{3!}+\cdots+\frac{(-1)^nx^{2n}}{n!}+\cdots\right)dx\\&=\left[\left(x-\frac{x^3}{3}+\frac{x^5}{2!5}-\frac{x^7}{3!7}+\cdots+\frac{(-1)^nx^{2n+1}}{n!(2n+1)}+\cdots\right)\right]_0^1\\&=1-\frac{1}{3}+\frac{1}{2!5}-\frac{1}{3!7}+\cdots+\frac{(-1)^n}{n!(2n+1)}+\cdots\end{align*} Recall that for an alternating series $\sum_{n=0}^\infty(-1)^n a_n$, $|R_n|\leq a_{n+1}$. For $n=4$, $|R_4|\leq a_5=\frac{1}{5!11}=7.6\times 10^{-3}>5\times 10^{-4}$. For $n=5$, $|R_5|\leq a_6=\frac{1}{6!13}=1.07\times 10^{-4}<5\times 10^{-4}$. Thus the error is less than $5\times 10^{-4}$ if $n\geq 5$. The approximation with $n=5$ is given by \begin{align*}\int_0^1 e^{-x^2}dx&\approx 1-\frac{1}{3}+\frac{1}{2!5}-\frac{1}{3!7}+\frac{1}{5!11}\\&\approx 0.747\end{align*}