Category Archives: Calculus

Vectors

What is a vector?

A vector is a quantity that has both direction and magnitude. Examples of vectors include displacement, velocity, force, weight, momentum, etc. A quantity that has only magnitude is called a scalar. Scalars are really just numbers. Examples of scalars include distance, speed, mass, temperature, etc. Since a vector has both direction and magnitude, it can be visually represented by a directed arrow. For instance, if a particle moves from a point $A$ to another point $B$, its displacement (i.e. the shortest distance from $A$ to $B$) is denoted by $\overrightarrow{AB}$ and it is visually represented by a directed arrow as in Figure 1.

Figure 1. A vector

The direction at which the arrow is pointing is the direction of the vector $\overrightarrow{AB}$ and the length of the arrow is the magnitude of the vector $\overrightarrow{AB}$. When it is not necessary to specify the initial point and the terminal point, a vector is denoted by a lower case alphabet letter with arrow on top like $\vec v$ or in boldface like ${\bf v}$.

Two vectors are said to be the same or equivalent if they have the same direction and the magnitude regardless of where they are located. If vectors $\overrightarrow{AB}$ and $\overrightarrow{CD}$ are the same, we write $\overrightarrow{AB}=\overrightarrow{CD}$. Figure 2 shows two equivalent vectors $\overrightarrow{AB}$ and $\overrightarrow{CD}$.

Figure 2. Equivalent vectors

The equivalence of two vectors implies that a vector can be moved around maintaining its characters (direction and magnitude) so it stays as the same vector although its location has changed (meaning its initial and terminal points have changed). Moving a vector without changing direction and magnitude is called a parallel translation.

There are two types of operations on vectors. One is vector addition and the other is scalar multiplication. Vector addition is defined pictorially by using a parallelogram or a triangle. Figure 3 shows vector addition $$\overrightarrow{AB}+\overrightarrow{AC}=\overrightarrow{AD}$$ by using a parallelogram.

Figure 3. Vector addition by a parallelogram

Figure 4 shows vector addition $$\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}$$ by using a triangle. The two ways of adding two vectors are indeed equivalent. The only difference is how you locate two vectors to add them together.

Figure 4. Vector addition by a triangle

Scalar multiplication is a product between a scalar and a vector. While many people, even some (less careful) mathematicians, consider it as an operation, it is not an operation but an action. I am not going to talk about what an action is here. In case someone is curious, you can visit the Wikipedia page on Group Action here. In calculus level, distinction between an operation and an action is not really important at all. Let $c$ be a scalar and $v$ a vector. What the scalar multiplication $c{\bf v}$ does is, depends on the value of $c$, it can stretch (when $c>1$), shrink (when $0<c<1$), or reverse the direction (when $c=-1$) of vector ${\bf v}$ as illustrated in Figure 5.

Figure 5. Scalar multiplication

Using vector addition and scalar multiplication, one can define subtraction of a vector ${\bf v}$ from another vector ${\bf u}$: $${\bf u}-{\bf v}:={\bf u}+(-{\bf v})$$ See Figure 6.

Figure 6. Vector substraction

Earlier I mentioned that a vector can be moved around while preserving its direction and magnitude, and a parallel translation of a vector is still considered to be the same as the previous vector, although it is now at a different location. Among all those same vectors, we are particularly interested in vectors that are starting the origin $O$. Figure 7 shows an example of such a vector.

Figure 7. A position vector

A vectors whose initial point is the origin $O$ is called a position vector or a located vector. A position vector is determined only by its terminal point, thereby it can be identified with a point in space and conversely a point in space can be identified with a position vector. For example, if the terminal point of a position vector ${\bf v}$ is $(a_1,a_2,a_3)$, then we regard them the same i.e. ${\bf v}=(a_1,a_2,a_3)$. Why this is such a big deal? The directed arrow representation of a vector has a lot of limitations. The most severe limitation is that it can only be useful when we can see them, i.e. their usage is limited within 3-dimensions as our perception does not allow us to go beyond 3-dimensions. However, Einstein’s theory of relativity (which has also been confirmed by numerous experiments and observations) that our universe is actually 4-dimensional. But that’s the universe we observe right now. String theory tells us that the universe can have up to 26-dimensions. It does not have to go that far beyond though. There are many other places here down on earth including computer science, economics, etc. where the notions of vectors in higher dimensions are being used. Considering position vectors resolve the limitation. Furthermore, now that we identify vectors with points in space and points are represented by ordered $n$-tuples (ordered pairs, triples, quadruples, depending on the dimension of the space) which are algebraic objects, we can use the power of algebra to describe the properties of vectors.

Given the points $A(a_1,a_2,a_3)$ and $B(b_1,b_2,b_3)$, the vector ${\bf v}$ which is represented by the directed arrow $\overrightarrow{AB}$ is $${\bf v}=(b_1-a_1,b_2-a_2,b_3-a_3)$$

Example. Find the vector represented by the directed arrow with initial point $A(2,-3,4)$ and $B(-2,1,1)$.

Solution. ${\bf v}=(-2-2,1-(-3),1-4)=(-4,4,-3)$.

The length or magnitude of a vector ${\bf v}$ is denoted by $|{\bf v}|$. For a vector ${\bf v}=(v_1,v_2)$ in the plane, $|{\bf v}|$ is given by $$|{\bf v}|=\sqrt{v_1^2+v_2^2}$$ (It’s easy to see this from Figure 7 using the Pythagorean law.) Similarly, for a vector ${\bf v}=(v_1,v_2,v_3)$ in space, $$|{\bf v}|=\sqrt{v_1^2+v_2^2+v_3^3}$$ It follows from the definition that \begin{equation}\label{eq:length}|c{\bf v}|=|c||{\bf v}|\end{equation} where $c$ is a scalar.

Vector addition and scalar multiplication can be nicely defined algebraically without using parallelograms or triangles. Furthermore, these algebraic definitions apply to vectors in arbitrary $n$-dimensional space. For vectors ${\bf u}=(u_1,u_2,u_3)$, ${\bf v}=(v_1,v_2,v_3)$ and a scalar $c$, \begin{align*}{\bf u}+{\bf v}&:=(u_1+v_1,u_2+v_2,u_3+v+3)\\c{\bf u}&:=(cu_1,cu_2,cu_3)\end{align*}

Example. If ${\bf u}=(4,0,3)$ and ${\bf v}=(-2,1,5)$, find $|{\bf u}|$, ${\bf u}+{\bf v}$, ${\bf u}-{\bf v}$, $3{\bf v}$, $2{\bf u}+5{\bf v}$.

Solution. \begin{align*}|{\bf u}|&=\sqrt{4^2+0^2+3^3}=\sqrt{25}=5\\{\bf u}+{\bf v}&=(4+(-2),0+1,3+5)=(2,1,8)\\{\bf u}-{\bf v}&=(4-(-2),0-1,3-5)=(6,-1,-2)\\3{\bf v}&=(3(-2),3(1),3(5))=(-6,3,15)\\2{\bf u}+5{\bf v}&=(2(4),2(0),2(3))+(5(-2),5(1),5(5))=(8,0,6)+(-10,5,25)=(-2,5,31)\end{align*}

Theorem. Let ${\bf u}$, ${\bf v}$, and ${\bf w}$ be vectors in $n$-dimensional space and $c$ and $d$ are scalars. Then

  1. ${\bf u}+{\bf v}={\bf v}+{\bf u}$
  2. ${\bf u}+({\bf v}+{\bf w})=({\bf u}+{\bf v})+{\bf w})$
  3. ${\bf u}+{\bf 0}={\bf u}$, where ${\bf 0}=(0,0,\cdots,0)$
  4. ${\bf u}+(-{\bf u})={\bf 0}$
  5. $c({\bf u}+{\bf v})=c{\bf u}+c{\bf v}$
  6. $(c+d){\bf u}=c{\bf u}+d{\bf u}$
  7. $(cd){\bf u}=c(d{\bf u})$
  8. $1{\bf u}={\bf u}$

It turns out the the original definition of vectors as quantities that have both direction and magnitude is quite obsolete and that even the definition of vectors by ordered $n$-tuples is not adequate enough to address much needed a broader notion of vectors arising in modern physics and engineering. For this reason, in modern treatment of vectors we no longer define what an individual vector is but instead we define a vector space. Simply speaking, a set $V$ with addition $+$ and scalar multiplication $\cdot$ satisfying the properties 1-8 is called a vector space, and the elements of $V$ are called vectors. Under this broader notion of vectors, things that were previously inconceivable to become vectors are now considered vectors. For example, $V$ the set of all continuous real-valued functions on the closed interval $[0,1]$ with addition $+$ and scalar multiplication $\cdot$ are defined by: \begin{align*}(f+g)(x)&:=f(x)+g(x)\\(cf)(x)&:=cf(x)\end{align*} for $f,g\in V$ and a scalar $c$. Then it is straightforward to show that the properties 1-8 are satisfied and therefore, $(V,+,\cdot)$ is a vector space and we regard continuous real-valued functions on $[0,1]$ as vectors. In fact, in quantum mechanics wave functions are state vectors. Another example is signal processing where functions are regarded as vectors. We are not going to delve into vector spaces further here. It is a main topic of linear algebra. For those who are curious, more examples of vector spaces can be found here.

There are infinitely many vectors. So it is humanly impossible to check if a certain property regarding vectors holds for all vectors. However there a particular finite set of vectors, called a basis, that constitute the entire vectors. A vector ${\bf u}=(u_1,u_2,u_3)$ can be written as \begin{equation}\label{eq:lincomb}{\bf u}=u_1(1,0,0)+u_2(0,1,0)+u_3(0,0,1)\end{equation} So we see that any vector can be represented by the three vectors $${\bf i}=(1,0,0),\ {\bf j}=(0,1,0),\ {\bf k}=(0,0,1)$$ by applying vector addition and scalar multiplication finitely many times as in \eqref{eq:lincomb}. The expression on the right hand side of the identity in \eqref{eq:lincomb} is called a linear combination or a superposition of ${\bf i}$, ${\bf j}$, ${\bf k}$. The three vectors ${\bf i}$, ${\bf j}$, ${\bf k}$ are called the canonical or standard basis vectors.

Figure 8. The standard basis

The number of standard basis vectors determines the dimension of the space. The dimension of a space is not necessarily finite though we are considering only finite dimensional spaces here (actually only 2- or 3-dimensional spaces). The set $V$ of all continuous functions on $[0,1]$ is infinite dimensional. The set of all state vectors in a quantum mechanics system is, in general, an infinite dimensional space called a Hilbert space.

Example. If ${\bf u}={\bf i}+2{\bf j}-3{\bf k}$ and ${\bf v}=4{\bf i}+7{\bf k}$, express $2{\bf u}+3{\bf v}$ in terms of ${\bf i}$, ${\bf j}$, ${\bf k}$.

Solution. \begin{align*}2{\bf u}+3{\bf v}&=2({\bf i}+2{\bf j}-3{\bf k})+3(4{\bf i}+7{\bf k})\\&=2{\bf i}+4{\bf j}-6{\bf k}+12{\bf i}+21{\bf k}\\&=14{\bf i}+4{\bf j}+15{\bf k}\end{align*}

Often in geometry and physics, we are only interested in the direction of a vector. A unit vector is a vector with length 1. Any non-zero vector can be re-scaled to a unit vector with the same direction. All that’s required is dividing the given vector by its magnitude. If ${\bf u}\ne {\bf 0}$, then $$\hat{\bf u}:=\frac{{\bf u}}{|{\bf u}|}$$ is a unit vector which has the same direction as ${\bf u}$: Using \eqref{eq:length}, $$|\hat{\bf u}|=\left|\frac{{\bf u}}{|{\bf u}|}\right|=\frac{1}{|{\bf u}|}|{\bf u}|=1$$

Example. Find the unit vector in the direction of the vector $2{\bf i}-{\bf j}-2{\bf k}$.

Solution. The length of the vector is $\sqrt{2^2+(-1)^2+(-2)^2}=\sqrt{9}=3$. Hence the unit vector with the same direction is $$\frac{2}{3}{\bf i}-\frac{1}{3}{\bf j}-\frac{2}{3}{\bf k}$$

In physics, when several forces are acting on an object, the resultant force or the net force experienced by the object is the vector sum of these forces.

Example. A 100-lb weight hangs from two wires as shown in Figure 9. Find the tension forces ${\bf T}_1$ and ${\bf T}_2$ in both wires and their magnitudes.

Figure 9. The resultant force

Solution. We first express the tensions ${\bf T}_1$ and ${\bf T}_2$ in terms of their horizontal and vertical components (the vectors in green in Figure 10).

Figure 10. The resultant force

\begin{align}\label{eq:tension1}{\bf T}_1&=-|{\bf T}_1|\cos 50^\circ{\bf i}+|{\bf T}_1|\sin 50^\circ{\bf j}\\\label{eq:tension2}{\bf T}_2&=|{\bf T}_2|\cos 32^\circ{\bf i}+|{\bf T}_2|\sin 32^\circ{\bf j}\end{align} The net force ${\bf T}_1+{\bf T}_2$ of the tensions must counterbalance the weight ${\bf w}$ so that the mass stays hung as in the figure, i.e. $${\bf T}_1+{\bf T}_2=-{\bf w}=100{\bf j}$$ From equations \eqref{eq:tension1} and \eqref{eq:tension2}, we have $$(-|{\bf T}_1|\cos 50^\circ+|{\bf T}_2|\cos 32^\circ){\bf i}+(|{\bf T}_1|\sin 50^\circ+|{\bf T}_2|\sin 32^\circ){\bf j}=100{\bf j}$$ By comparing the components, we obtain the following equations: \begin{align*}-|{\bf T}_1|\cos 50^\circ+|{\bf T}_2|\cos 32^\circ&=0\\|{\bf T}_1|\sin 50^\circ+|{\bf T}_2|\sin 32^\circ&=100\end{align*} Solving these equations simultaneously we find \begin{align*}|{\bf T}_1|&=\frac{100}{\sin 50^\circ+\tan 32^\circ\cos 50^\circ}\approx 85.64\mathrm{lb}\\|{\bf T}_2|&=\frac{|{\bf T}_1|\cos 50^\circ}{\cos 32^\circ}\approx 64.91\mathrm{lb}\end{align*} Therefore, $${\bf T}_1\approx -55.05{\bf i}+65.60{\bf j},\ {\bf T}_2\approx 55.05{\bf i}+34.40{\bf j}$$

Examples in this note have been taken from [1].

References.

[1] Calculus, Early Transcendentals, James Stewart, 6th Edition, Thompson Brooks/Cole

Optimization Problems

In mathematics and also in applications, we often encounter problems that require to maximize or minimize the value of a certain quantity. The general procedure can be summarized as:

  1. Express the quantity to be maximized or minimized in terms of a single variable. The quantity may be described in terms of two variables however with given constraint it could be reduced to a single variable.
  2. Differentiate the function obtained in step 1 and set the derivative equal to 0.
  3. Solve the equation from step 2 to obtain critical values and determine whether they maximize or minimize the given quantity. Usually the first or second derivative test is a convenient tool for the required inspection.

Example. A farmer has 2400 ft of fencing and wants to fence off a rectangular field that borders a straight river. He needs no fence along the river. What are the dimensions of the field that has the largest area?

Solution. Let $x$ and $y$ denote the length and the width of the rectangular field. Suppose that the side along the river has the length $x$. Then the area is $A=xy$ and the required fencing in terms $x$ and $y$ is $x+2y=2400$. This fencing is a constraint and solve it for $y$ to obtain $y=1200-\frac{x}{2}$. Plugging this into $A$ for $y$, the area can be written as a function of a single variable $x$: $$A(x)=1200x-\frac{x^2}{2}$$ $A'(x)=1200-x$ and setting this equal to 0, we find $x=1200$. Since $A^{\prime\prime}(x)=-1<0$, by the second derivative test $x=1200$ gives rise to the absolute maximum of $A(x)$. The required dimensions are $1200\ \mbox{ft}\times 600\ \mbox{ft}$ where the side that borders the river is 1200 ft and the resulting largest area is 720,000 $\mbox{ft}^2$.

Example. A box with a square base and open top must have a volume of 32,000 $\mbox{cm}^3$. Find the dimensions of the box that minimize the amount of material used.

Solution. Let $x$ and $h$ be the length and the height of the box, respectively. Then $x^2h=32000$ and we want to minimize the surface area $A=x^2+4xh$. Solve the volume constraint for $h$ to obtain $h=\frac{32000}{x^2}$. Plugging this into $A$ for $h$, we write $A$ as a function of a single variable $x$: $$A(x)=x^2+\frac{128000}{x}$$ $A'(x)=2x-\frac{128000}{x^2}$ and setting it equalto 0, we find $x=40$. Since $A^{\prime\prime}(x)=2+\frac{256000}{x^3}>0$ for all $x>0$, $A(40)$ is the absolute minimum. Therefore the required dimensions are $40\ \mbox{cm}\times 40\ \mbox{cm}\times 20\ \mbox{cm}$.

Example. If 1200 $\mbox{cm}^2$ of material is available to make a box with a square base and an open top, find the largest possible volume of the box.

Solution. Let $x$ and $h$ be the length and the height of the box, respectively. Then $x^2+4xh=1200$ and we want to maximize $V=x^2h$. Solve the area for $h$ to obtain $h=\frac{1200-x^2}{4x}$. Plugging this into $V$, we write the volume as a function of a single variable $x$: $$V(x)=300x-\frac{1}{4}x^3$$ $V'(x)=300-\frac{3}{4}x^2$ and setting it equal to 0, we find $x=20$. Since $V'(x)$ is a quadratic polynomial with a negative leading coefficient, $V(2)=4000\ \mbox{cm}^3$ is the largest possible volume of the box.

Example. Find the point on the parabola $y^2=2x$ that is the closest to the point $(1,4)$.

Solution. Let $(x,y)$ denote a point on the parabola $y^2=2x$. The distance between $(x,y)$ and $(1,4)$ is $d=\sqrt{(x-1)^2+(y-4)^2}$ and we want to minimize this. Note minimizing $d$ is equivalent to minimizing $d^2=(x-1)^2+(y-4)^2$. Solve the equation of parabola for $x$ to obtain $x=\frac{y^2}{2}$. Plugging this into $d^2$, we can write it as a function of a single variable $y$: $$f(y)=\left(\frac{y^2}{2}-1\right)^2+(y-4)^2=\frac{y^4}{4}-8y+17$$ $f'(y)=y^3-8$ and setting it equal to 0, we find $y=2$. Since $f^{\prime\prime}(y)=3y^2>0$ for all $y\ne 0$, $(x,y)=(2,2)$ is the point on the parabola $y^2=2x$ that is the closest to (1,4)$.

The shortest distance from (1,4) to the parabola y^2=2x.

Remark. The above problem also can be solved using a simple geometric fact that the shortest path from $(1,4)$ to the parabola $y^2=2x$ would be normal to the tangent line (i.e. the path is perpendicular to the tangent line). Let $(a,b)$ be the point on the parabola that is closest to $(1,4)$. By implicit differentiation we find $\frac{dy}{dx}=\frac{1}{y}$ and so the normal line at $(a,b)$ has the slope $-b$. The equation of the normal line is then $y-4=-b(x-1)$. Since this line is passing through $(a,b)$, $b-a=-b(a-1)$ or $ab=4$. $(a,b)$ is also on the parabola so we have $b^2=2a$. Solve the two equations simultaneously to obtain $b=2$ and hence $a=2$. Therefore, $(a,b)=(2,2)$.

Improper Integrals

When we defined the definite integral $\int_a^b f(x)dx$, it was assumed that the limits $a$ and $b$ of the integral are finite and the integrand $f(x)$ is continuous on the closed interval $[a,b]$. Even if these assumptions are not satisfied, we can still consider a notion of integral extended from the definite integral (the Riemann integral). This extended integral is called the improper integral.

Infinite Limits

A definite integral, in which one or both limits of integration are infinite, is defined by the following: \begin{align*}\int_a^\infty f(x)dx&=\lim_{t\to\infty}\int_a^tf(x)dx\\\int_{-\infty}^b f(x)dx&=\lim_{t\to -\infty}\int_t^bf(x)dx\end{align*} The improper integrals are said to be convergent if the corresponding limit exists. Otherwise, divergent.

If both $\int_{-\infty}^a f(x)dx$ and $\int_a^\infty f(x)dx$ are convergent, we define $$\int_{-\infty}^\infty f(x)dx=\int_{-\infty}^a f(x)dx+\int_a^\infty f(x)dx$$

Examples:

  1. $\int_{-\infty}^0e^xdx=\lim_{t\to -\infty}\int_t^0 e^xdx=\lim_{t\to -\infty}(1-e^t)=1$.
  2. $\int_2^\infty\frac{dx}{x}=\lim_{t\to \infty}\int_2^t\frac{dx}{x}=\lim_{t\to \infty}(\ln t-\ln 2)=\infty$.
  3. $\int_{-\infty}^0\frac{1}{1+x^2}dx=\lim_{t\to -\infty}\int_t^0\frac{1}{1+x^2}dx=\lim_{t\to -\infty}(-\tan^{-1}t)=\frac{\pi}{2}$.
  4. $\int_0^\infty\frac{1}{1+x^2}dx=\lim_{t\to\infty}\int_0^t\frac{1}{1+x^2}dx=\lim_{t\to\infty}(\tan^{-1}t)=\frac{\pi}{2}$.
  5. $\int_{-\infty}^\infty\frac{1}{1+x^2}dx=\int_{-\infty}^0\frac{1}{1+x^2}dx+\int_0^\infty\frac{1}{1+x^2}dx=\frac{\pi}{2}+\frac{\pi}{2}=\pi$.

Remarks:

  1. $\int_{-\infty}^\infty f(x)dx$ can be also defined by the double limit $$\int_{-\infty}^\infty f(x)dx=\lim_{b\to\infty\\a\to -\infty}\int_a^b f(x)dx$$
  2. $\int_{-\infty}^\infty\frac{2x}{1+x^2}dx$ is divergent as $\int_{-\infty}^0\frac{2x}{1+x^2}dx=-\infty$ and $\int_0^\infty\frac{2x}{1+x^2}dx=\infty$. On the other hand, $$\lim_{a\to\infty}\int_{-a}^a\frac{2x}{1+x^2}dx=0$$ This is called the Cauchy principal value of the integral $\int_0^\infty\frac{1}{1+x^2}dx$ and is denoted by $$\mathrm{p.v.}\int_0^\infty\frac{1}{1+x^2}dx$$ The Cauchy principal value is a method of assigning values to certain ill-defined improper integrals. We will not, however, be considering the Cauchy principal value here.

Discontinuous Integrand

If $f(x)$ is continuous for all values of $x$ in the domain $a\leq x\leq b$ except $x=b$ or $x=a$, $\int_a^b f(x)dx$ is defined by \begin{equation}\label{eq:impropint}\int_a^b f(x)dx=\lim_{t\to b-}\int_a^t f(x)dx\end{equation} or \begin{equation}\label{eq:impropint2}\int_a^b f(x)dx=\lim_{t\to a+}\int_t^b f(x)dx\end{equation} provided the corresponding limit exists.

Examples:

  1. $\int_{-1}^0\frac{dx}{x^2}=\lim_{t\to 0-}\left(-\frac{1}{t}-1\right)=\infty$.
  2. $\int_0^a\frac{dx}{\sqrt{a^2-x^2}}=\lim_{t\to a-}\left(\sin^{-1}\frac{t}{a}\right)=\frac{\pi}{2}$.

When $f(x)$ is continuous for all values of $x$ in the domain $a\leq x\leq b$ except $x=c$ (where $a< c <b$), $\int_a^b f(x)dx$ is defined by $$\int_a^b f(x)dx=\int_a^c f(x)dx+\int_c^b f(x)dx$$ where the integrals in the RHS are evaluated in accordance with \eqref{eq:impropint} and \eqref{eq:impropint2}, respectively.

Example. Consider $\int_{-1}^1\frac{dx}{x^2}$.

The graph of y=1/x^2 on [-1,1]

Solution. Since the integrand is discontinous at $x=0$, we write the integral in two parts as $$\int_{-1}^1\frac{dx}{x^2}=\int_{-1}^0\frac{dx}{x^2}+\int_0^1\frac{dx}{x^2}=\infty+\infty=\infty$$

Remarks. If one mindlessly evaluates the integral as an ordinary definite integral, we obtain $$\int_{-1}^1\frac{dx}{x^2}=\left[-\frac{1}{x}\right]_{-1}^1=-2$$ However, this is nonsense because the integrand is always positive.

Integration of Rational Functions by Partial Fractions

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*}

Inverse Trigonometric Functions

$y=\sin x$ 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.

Figure 1. The graph of y=sin(x) on [-pi/2,pi/2]

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 $y=arccot x$) 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 $y=\sin^{-1} x$.

Figure 2. The graphs of y=sin(x) (in red), y=arcsin(x) (in blue) and y=x (in black).
Figure 3. The graph of y=arcsin(x)
Figure 4. The graph of y=arccos(x)
Figure 5. The graph of y=arctan(x) on [-20,20].
Figure 6. The graph of y=arccot(x) on [-20,20].

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

Figure 7. The graph of y=arcsec(x). The horizontal asymptote is y=pi/2.

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

The graph of y=arccsc(x).

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 $\frac{d}{dx}\sin^{-1}$, $\frac{d}{dx}\tan^{-1}x$, $\frac{d}{dx}\sec^{-1}$ 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*}