Integral Play
Each of the following indefinite integrals can be found by using a suitable substitution or integration by parts. In some cases, algebraic simplification may also be necessary (e.g., long division, rationalizing terms, or a trigonometric identity).
 \( \displaystyle \int \cos^{3}(x) ~ dx \)
 \( \displaystyle \int \dfrac{a^x}{b^x} ~ dx \), with \( a, b \gt 0 \) and \( a \neq b \)
 \( \displaystyle \int \dfrac{1}{\sqrt{x + 2}+ \sqrt{x  2}} ~ dx \)
 \( \displaystyle \int \dfrac{1}{1 + \sin(x)} ~ dx \)
 \( \displaystyle \int e^{e^x + x} ~ dx \)
 \( \displaystyle \int \dfrac{3x^2 + 2x + 1}{x + 1} ~ dx \)
 \( \displaystyle \int \ln\big( \cos(x) \big) \tan(x) ~ dx \)
 \( \displaystyle \int x^{p  1} \ln(x) ~ dx \)
 \( \displaystyle \int \dfrac{1}{1  e^{x}} ~ dx \)
 \( \displaystyle \int \sec^{3}(x) ~ dx \)
 \( \displaystyle \int \cos{\big( \ln(x) \big)} ~ dx \)
 \( \displaystyle \int \dfrac{x}{1 + \sin(x)} ~ dx \)
PreNewtonian Laws
Long before Newton developed his laws of motion, a simple rule for computing distance in terms of velocity and time was used for a body falling straight downwards under the force of gravity.
If an object's initial velocity at time \( t = 0 \) is \( v(0) = v_0 > 0 \), and its final velocity at time \( t = T \) is \( v(T) = v_f > v_0 \), then \( s \), the distance fallen after \( T \) seconds, was calculated as \[ s = \dfrac{v_0 + v_f}{2} T \]
 Suppose that \( v'(t) = g \), where \( g \) is a constant value. Make a sketch of \( v(t) \) versus \( t \) for \( 0 \leq t \leq T \) which illustrates that if \( v'(t) = g \), then the rule \( s = \dfrac{v_0 + v_f}{2} \) is accurate. Interpret the distance fallen, \( s \), as an area \( A \) using your sketch for \( v(t) \).
 Make a second sketch to illustrate that if \( v'(t) \) increases over \( 0 \lt t \lt T \), then this preNewtonian rule will overestimate the true value of \( s \) for the same value of \( v_0 \) and \( v_f \).
A Balancing Act
Consider the integral function \( F_a(x) = \displaystyle \int_{a}^{x} \left( 1  \left \lvert t  1 \right \rvert \right) ~ dt \). For each of the following values of \( a \), determine the values of \( x \) for which the function \( F_a(x) \) is zero.
 \( a = 1 \)
 \( a = 1 \)
 \( a = 0 \)
 \( a = c \) for some constant \( c \gt 1 \) (Hint: a careful sketch of the integrand will be helpful.)
No Effort Needed!
 Use a suitable substitution to show that \[ \displaystyle c \int_{a}^{b} f(cx) ~ dx = \int_{ca}^{cb} f(x) ~ dx \] for any real constant \( c \).

The area \( A_c \) enclosed by the circle \( x^2 + y^2 = 1 \) is equal to \( \pi \). As an integral, this can be expressed by noting that the portion of the circle in the first quadrant can be modelled using the function \( y = \sqrt{1  x^2} \), giving the total area as \[ A_c = 4 \int_{0}^{1} \sqrt{1  x^2} ~ dx = \pi \] since the circle is symmetric about both axes (as shown in the diagram.)
In a similar manner, the area enclosed by the ellipse \( \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1 \) can be expressed as \[ A_e = 4b \int_{0}^{a} \sqrt{1  \dfrac{x^2}{a^2}} ~ dx \] by noting that the portion of the ellipse in the first quadrant can be modelled using the function \( y = b\sqrt{1  \dfrac{x^2}{a^2}} \) and observing that the ellipse is symmetric about both axes.
Without directly evaluating the integral, find the area of \( A_e \).
Unexpectedly...
Use a suitable substitution to prove each of the following results.
Hint: Remember that the variable in a definite integral is just a “dummy”, i.e., \( \displaystyle \int_{a}^{b} f(x) ~ dx = \int_{a}^{b} f(t) ~ dt \).
 If \( f(t) \) is continuous on \( [0, 1] \), then \( \displaystyle \int_{0}^{1} f(1  t) ~ dt = \int_{0}^{1} f(t) ~ dt \).
 If \( f(t) \) is continuous on \( \left[0, \frac{\pi}{2} \right] \), then \( \displaystyle \int_{0}^{\tfrac{\pi}{2}} f\big(\cos{(t)}\big) ~ dt = \int_{0}^{\tfrac{\pi}{2}} f\big(\sin{(t)}\big) ~ dt \).
 If \( F(x) = \displaystyle \int_{e}^{x} \dfrac{1}{\ln{(t)}} ~ dt \) and \( G(x) = \displaystyle \int_{1}^{x} \dfrac{e^u}{u} du \), then \( G(x) = F(e^x) \) for all \( x \geq 1 \).
Reduction, but not in value...
Repeated use of integration by parts often simplifies integrals such as \( \displaystyle \int x^n e^{x} ~ dx \), which depend on an integer \( n \). This process can sometimes be expressed more concisely by a reduction formula. For example, if \( \displaystyle I_n = \int x^n e^{x} ~ dx \), then it can be shown that \( I_n = x^n e^{x} + n I_{n1} \).
 Find a simple reduction formula for \( \displaystyle I_n = \int \big( \ln(x) \big)^n ~ dx \), \( n \geq 1 \). Use it to evaluate \( \displaystyle \int_{1}^{e} \big( \ln(x) \big)^3 ~ dx \).

Suppose that \( C_n = \displaystyle \int_{0}^{\tfrac{\pi}{2}} \cos^n(x) ~ dx \) for \( n = 0, 1, 2, \ldots \).
Show that \( C_n = \left( \dfrac{n  1}{n} \right) C_{n  2} \) for \( n \geq 2 \), and hence evaluate \( \displaystyle \int_{0}^{\tfrac{\pi}{2}} \cos^4(x) ~ dx \).
“Right” On!
For \( f(x) = \sin^{2}(x) \) on the interval \( 0 \leq x \leq \pi \), the Riemann sums \( M_3 \), \( L_4 \), and \( R_6 \) are each equal to \( \frac{\pi}{2} \).
 Write down the general form of the right Riemann sum \( R_{2n} \) \( (n \geq 1) \) for \( f(x) = \sin^2(x) \) on \( [0, \pi] \).
 Show that \( \displaystyle \sum_{k = 1}^{2n} \cos{\left( \frac{k\pi}{n} \right)} = 0 \).
 Use the result from part b) to show that \( R_{2n} = \frac{\pi}{2} \) for any \( n \geq 1 \).
 What is the value of \( \displaystyle \int_{0}^{\pi} f(x) ~ dx \)?
Gleaning Info on Integral Functions
 Consider the integral function \( F(x) = \displaystyle \int_{0}^{x} \dfrac{\sin{(t)}}{t + 1} ~ dt \).
 Explain why the integrand \( f(t) = \dfrac{\sin{(t)}}{t + 1} \) satisfies the inequality \[ \dfrac{1}{t + 1} \leq f(t) \leq \dfrac{1}{t + 1} \] for all \( t \geq 0 \).
 Carefully sketch the graphs of \( y = \dfrac{1}{t + 1} \) and \( y = \dfrac{1}{t + 1} \) on the same axes for \( 0 \leq t \leq 4\pi \) along with the graph \( y = f(t) \). Indicate any points where \( f(t) = 0 \), \( f(t) = \dfrac{1}{t + 1} \), or \( f(t) = \dfrac{1}{t + 1} \).
 Use your graph to give an intuitive explanation of why \( F(2\pi) \gt 0 \) and \( F(4\pi) \gt 0 \).
 Use the result from part iii) to give an intuitive explanation of why \( F(x) \gt 0 \) for all \( x \gt 0 \).
 Many integral functions cannot be expressed in terms of elementary functions such as those we study in calculus. Consider, for example,
\[ F(x) = \int_{0}^{x} t^2 e^{t^2} ~ dt \]
 Show that \( F \) passes through \( (0, 0) \) and that it is an odd function, i.e., \( F(x) = F(x) \).
 Find any critical points of \( F(x) \) and any points of inflection.
 Show that \( 0 \leq F(1) \leq \frac{1}{3} \). (Hint: Compare \( F(1) \) to a simpler integral.)
 Make a qualitative sketch of \( y = F(x) \).
When and Why: Trignometric Substitutions
Many of the integrals of interest which can occur in applications will contain expressions of the form
 \( \sqrt{a^2  x^2} \) or
 \( \sqrt{a^2 + x^2} \).
which can yield integrable forms when trigonometric substitutions are performed. (Note that we may assume \( a \gt 0 \) since \( a^2 = (a)^2 \).)
Substituting \( x = a\sin(\theta) \) into the first form gives
\begin{align*}
\sqrt{a^2  x^2} &= \sqrt{a^2  a^2\sin^2(\theta)} \\
&= \left\lvert a \right\rvert \sqrt{1  \sin^2(\theta)} \\
&= \left\lvert a \right\rvert \sqrt{\cos^2(\theta)} \\
&= \left\lvert a \right\rvert \left\lvert \cos(\theta) \right\rvert \\
&= a \left\lvert \cos(\theta) \right\rvert & \text{since }a \gt 0
\end{align*}
If we restrict \( \theta \) to be between \( \frac{\pi}{2} \) and \( \frac{\pi}{2} \), then \( \cos(\theta) \geq 0 \), and hence \[ \sqrt{a^2  x^2} = a \left\lvert \cos(\theta) \right\rvert = a \cos(\theta) \]
Similarly, substituting \( x = a\tan(\theta) \) into the second form, where \( \frac{\pi}{2} \lt \theta \lt \frac{\pi}{2} \), yields
\begin{align*}
\sqrt{a^2 + x^2} &= \sqrt{a^2 + a^2\tan^2(\theta)} \\
&= \left\lvert a \right\rvert \sqrt{1 + \tan^2(\theta)} \\
&= \left\lvert a \right\rvert \sqrt{\sec^2(\theta)} \\
&= \left\lvert a \right\rvert \left\lvert \sec(\theta) \right\rvert \\
&= a \sec(\theta)
\end{align*}
since \( a \gt 0 \), and \( \sec(\theta) \gt 0 \) in \( \left( \frac{\pi}{2}, \frac{\pi}{2} \right) \).
 Use the appropriate trigonometric substitution(s) to evaluate the following integrals.
 \( \displaystyle \int \dfrac{1}{\sqrt{1 + x^2}} ~ dx \)
 \( \displaystyle \int \dfrac{x + 1}{\sqrt{4  x^2}} ~ dx \)
 \( \displaystyle \int_{\tfrac{1}{2}}^{1} \dfrac{\sqrt{1  x^2}}{x} ~ dx \)
 \( \displaystyle \int_{0}^{3} \frac{x^3}{\left( x^2 + 9 \right)^{\frac{3}{2}}} ~ dx \)
 The formula for the area \( A \) of a circle with radius \( r \) is \( A = \pi r^2 \). This can be shown in the following manner:
 Imagine a circle of radius \( r \) centred at the origin as a region enclosed by two curves. Explain why the area of the circle is given by \[ A = 2 \int_{r}^{r} \sqrt{r^2  x^2} ~ dx \]
 Evaluate the integral from part i) to confirm that \( A = \pi r^2 \). (Hint: The identity \( \cos^2(\theta) = \frac{1}{2} \big( 1 + 2 \cos(\theta) \big) \) may be useful.)
Definitely Improper
The definition of \( \displaystyle \int_{a}^{b} f(x) ~ dx \) assumes that \( [a, b] \) is a finite interval. However, in many applications (both theoretical and physical), it is important to consider what happens to certain phenomena over intervals of (effectively) infinite size, such as great distances between heavenly bodies or very long stretches of time.
To help mathematically analyze these situations, the improper integral
\[ \int_{a}^{\infty} f(x) ~ dx := \lim_{b \to \infty} \int_{a}^{b} f(x) ~ dx \]
was defined as a natural way of extending the definite integral to the infinite interval \( [a, \infty) \).
Note that the symbol “\( := \)” does not mean “equal to”; instead, it means “is defined to be.”
If the limit on the right exists, we say that the integral on the left exists; otherwise, it does not exist.
 Determine whether each of the following improper integrals exist.
 \( \displaystyle \int_{0}^{\infty} x^2 e^{x} ~ dx \)
 \( \displaystyle \int_{1}^{\infty} \dfrac{\ln(t)}{t} ~ dt \)
 \( \displaystyle \int_{1}^{\infty} \dfrac{1}{x^p} ~ dx \) for \( p \gt 0 \)
 There is exactly one value of the constant \( k \) for which the improper integral \[ \int_{0}^{\infty} \left( \dfrac{2x}{x^2 + 1}  \dfrac{k}{x + 1} \right) ~ dx \] exists. Find the required antiderivative, determine the value of \( k \), and evaluate the integral for that value of \( k \).
 Let \( A(x) \) be the area between \( y = 0 \) and \( y = t^{2}e^{\frac{1}{t}} \) for \( 1 \leq t \leq x \).
 Find an expression for \( A(x) \) by evaluating a suitable definite integral having \( x \) as the upper limit.
 Determine whether \( \displaystyle \lim_{x \to \infty} A(x) \) exists, and interpret this result.

To propel a rocket of mass \( m \) radially outward from the Earth's surface, work must be done against the force of gravity.
If \( r \) measures the distance from the centre of the Earth to the rocket, then the work required to move it from the surface (located at a distance \( r = R_e \) from the centre) outward to \( r = b \) is given by
\[ W(b) = k \int_{R_e}^{b} \dfrac{1}{r^2} ~ dr \]
where \( k \) is a positive constant with \( k = GmM_e \). (Note: \( G \) is the universal gravitational constant, \( m \) is the mass of the rocket, and \( M_e \) is the mass of the Earth.)
 Find a closed form expression for \( W(b) \). (I.e., express it without using an integral.)

The work required to enable the rocket to escape Earth's gravity is given by \[ \displaystyle \lim_{b \to \infty} W(b) \] i.e., the amount needed to send the rocket infinitely far away from the centre of the Earth.
If all of this energy must come from (i.e., be equal to) the initial kinetic energy \( \frac{1}{2}mv_0^2 \), determine the value of the escape velocity, \( v_0 \).
Extension: Introducing Hyperbolic Trigonometric Functions
Two linear combinations of exponential functions occur so often in physical applications that it is convenient to define them as functions.
Specifically, we define
\( \cosh(x) := \dfrac{e^x + e^{x}}{2} \)
and
\( \sinh(x) := \dfrac{e^x  e^{x}}{2} \)
which are known as the hyperbolic cosine and hyperbolic sine functions, respectively.
 Verify the following properties of the hyperbolic trigonometric functions:
 \( \cosh^{2}(x)  \sinh^{2}(x) = 1 \)
 \( \cosh(x) \) is an even function and \( \sinh(x) \) is an odd function
 \( \dfrac{d}{dx} \Big( \cosh(x) \Big) = \sinh(x) \) and \( \dfrac{d}{dx} \Big( \sinh(x) \Big) = \cosh(x) \)
 Sketch the graphs of \( y = \cosh(x) \), \( y = \sinh(x) \), and \( y = \frac{1}{2}e^x \) on the same axes. What is the role of the graph of \( \frac{1}{2}e^x \) in relation to the graph of the two hyperbolic trignometric functions?
Cables suspended between supports play many important roles, such as in suspension bridges, telecommunication lines, and so on.
Basic physical principles reveal, in particular, that a cable of density \( \rho \), hanging under its own weight between two supports, assumes the shape \( y(x) \) satisfying the relation \[ \dfrac{d^2y}{dx^2} = a \sqrt{1 + \left( \dfrac{dy}{dx} \right)^2} \] with the positive constant \( a = \dfrac{\rho g}{T_h} \), where \( g \) is the gravitational acceleration and \( T_h \) is the tension in the cable at its lowest point.
 Show that \( y(x) = \frac{1}{a}\cosh{(ax)} + C \) satisfies this equation for any constant \( C \). Such a curve is called a caternary.
A cable has a length of \( 80 \) metres and is suspended between two supports of equal height, \( 60 \) metres apart. If the cable assumes a form whose shape can be described by the curve \( y = \frac{1}{a} \cosh{(ax)} \), show that \( 40a = \sinh{(30a)} \).
Find an approximate value for \( a \) using Newton's method by using a starting value of \( a_0 = 0.1 \). Hence, find an approximate height for the supports.
Hint: The length of the curve \( y = f(x) \) for \( a \leq x \leq b \) is given by \( L = \displaystyle \int_{a}^{b} \sqrt{1 + \big( f'(x) \big)^2} ~ dx \).
An architectural arch is to be constructed of the form \( y = h  a \cosh{(bx)} \), where \( y = 0 \) represents ground level. The arch is to be \( 100 \) m high at the centre (\( x = 0 \)), and \( 200 \) m wide at ground level. At the points where the arch meets the ground, the slope of the tangent line to the arch has a magnitude of \( 3 \).
Show that \( a \) and \( b \) must satisfy \( \left( 1 + \dfrac{100}{a} \right)^2  \dfrac{9}{a^2b^2} = 1 \).
Suppose some other construction needs dictate that \( b = 0.01 \). Find \( a \) and \( h \) in this case and obtain an equation for the arch.

The familiar trigonometric functions \( \cos(\theta) \) and \( \sin(\theta) \) are defined in terms of the relationship between the points on the unit circle \( x^2 + y^2 = 1 \) and the angle \( \theta \), measured counterclockwise from the positive \( x \)axis to the segment \( OP \), where \( O \) is the origin and \( P \) is the point \( ( \cos(\theta), \sin(\theta) ) \).
Recall that the formula for the area of a sector of measure \( \theta \) (measured in radians) in a circle of radius \( r \) is given by \( A = \frac{1}{2} r^2 \theta \). Noting that \( r = 1 \) in a unit circle gives \( \theta = 2A \), after rearranging.
Thus, we may think of the coordinates of \( P(x, y) \) as \( x = \cos(\theta) = \cos{(2A)} \) and \( y = \sin(\theta) = \sin{(2A)} \).
There is an analogous set of functions for hyperbolas. Such functions also describe the coordinates of a point on the hyperbola in terms of a suitably defined area \( A \).
Specifically, consider a “unit” hyperbola with equation \( x^2  y^2 = 1 \). Let \( O \) be the origin, \( P\left( x_0, y_0 \right) \) be a point on the hyperbola, and \( Q \left( x_0, 0 \right) \) a point on the \( x \)axis directly below \( P \). Define \( A \) to be the area between the \(x\)axis, the hyperbola, and line segment \( OP \), as shown in the diagram. Then,
\begin{align*}
A &= \left( \text{area}~\triangle{OPQ} \right)  \left( \text{area under}~y = \sqrt{x^2  1} \right) \\
&= \frac{1}{2} x_0 y_0  \int_{x = 1}^{x = x_0} \sqrt{x^2  1} ~ dx
\end{align*}
For the analogy with the circle to hold, we claim that \( x_0 = \cosh{(2A)} \) and \( y_0 = \sinh{(2A)} \). To verify this, proceed as follows:
 Show that \( \dfrac{d}{dx} \left( \dfrac{x}{2} \sqrt{x^2  1}  \dfrac{1}{2} \ln{\left( x + \sqrt{x^2  1} \right)} \right) = \sqrt{x^2  1} \), for \( x \gt 1 \).
 Use the result from i) to show that \( 2A = \ln(x_0 + y_0) \), given that \( y_0 = \sqrt{x_0^2  1} \).
 Assuming that \( x_0 \geq 1 \) and \( y_0 \geq 0 \), verify that
\[ \cosh{(2A)} = \frac{1}{2} \left( e^{2A} + e^{2A} \right) = x_0 \]
and
\[ \sinh{(2A)} = \frac{1}{2} \left( e^{2A}  e^{2A} \right) = y_0 \]
to complete the analogy with the trigonometric functions used to describe the circle.