Category: Exercises

Exponential Fourier Series

The function $f$ is defined by $f(t) = e^t$ for $-1 < t \leq 1$ and extended periodically by $f(t) = f(t+2)$ .

Problem 1. For any nonnegative integer $n$ , evaluate the integrals

$\displaystyle \int f(t) \cos (n \pi t)\, \mathrm dt,\quad \int f(t) \sin(n \pi t)\, \mathrm dt.$

(Click for Solution)

Solution. Integrating by parts twice,

$\begin{aligned} \int e^{t} \cos (n \pi t)\, \mathrm dt &= \underbrace{e^t}_{\mathrm I} \cdot \underbrace{\cos (n \pi t)}_{\mathrm S} - \int \underbrace{e^t}_{\mathrm I} \cdot \underbrace{(-n\pi \cdot \sin(n \pi t))}_{\mathrm D}\, \mathrm dt \\ &= e^t \cos(n\pi t) + n \pi \int e^t \sin(n \pi t)\, \mathrm dt \\ &= e^t \cos(n \pi t) + n\pi \left( \underbrace{e^t}_{\mathrm I} \cdot \underbrace{\sin (n \pi t)}_{\mathrm S} - \int \underbrace{e^t}_{\mathrm I} \cdot \underbrace{(n\pi \cdot \cos(n \pi t))}_{\mathrm D}\, \mathrm dt \right) \\ &= e^t ( \cos(n \pi t) + n\pi \sin(n \pi t)) - n^2 \pi^2 \int e^t \cos(n\pi t)\, \mathrm dt. \end{aligned}$

By careful algebruh,

$\displaystyle \int e^{t} \cos (n \pi t)\, \mathrm dt = \frac{e^t ( \cos(n \pi t) + n\pi \sin(n \pi t))}{1 + n^2 \pi^2} + C_1.$

In a similar though tedious manner,

$\displaystyle \int e^{t} \sin (n \pi t)\, \mathrm dt = \frac{e^t ( \sin(n \pi t) - n\pi \cos(n \pi t))}{1 + n^2 \pi^2} + C_2.$

Remark 1. For an alternate solution, recall Euler’s formula which states that

$e^{i n \pi t} = \cos(n \pi t) + i \sin(n \pi t).$

Multiplying by $f(t)$ then integrating on the left-hand side yields

$\begin{aligned} \int f(t) e^{i n \pi t} \, \mathrm dt &= \int e^t \cdot e^{i n \pi t}\, \mathrm dt \\ &= \int e^{(1 + i n \pi )t}\, \mathrm dt \\ &= \frac {e^{(1 + i n \pi )t}}{1 + i n \pi} + C \\ &= e^t \cdot \frac{1 - i n \pi}{1 + n^2 \pi^2} (\cos (n \pi t) + i \sin(n \pi t)) + C \\ &= e^t \cdot \frac{(\cos (n \pi t) + n \pi \sin(n \pi t)) + i (\sin(n \pi t) - n \pi \cos(n \pi t))}{1 + n^2 \pi^2} + C \\ &= e^t \cdot \frac{\cos (n \pi t) + n \pi \sin(n \pi t)}{1 + n^2 \pi^2} + i \cdot e^t \cdot \frac{\sin(n \pi t) - n \pi \cos(n \pi t)}{1 + n^2 \pi^2} + C . \end{aligned}$

The real and imaginary parts then agree with our solutions in Problem 1.

Problem 2. Denoting $T = 2$ , the Fourier series of $f$ is defined by

$\displaystyle f(t) = \frac{a_0}{2} + \sum_{n=1}^\infty \left( a_n \cos\left( \frac{2n\pi t}{T} \right) + b_n \sin\left( \frac{2n\pi t}{T} \right) \right),$

where the Fourier coefficients $a_n, b_n$ are defined by

$\displaystyle a_n = \frac 2{T} \int_{-T/2}^{T/2} f(t) \cos\left( \frac{2n\pi t}{T} \right)\, \mathrm dt,\quad b_n = \frac 2{T} \int_{-T/2}^{T/2} f(t) \sin\left( \frac{2n\pi t}{T} \right)\, \mathrm dt.$

Use Problem 1 to evaluate the Fourier series of $f$ .

(Click for Solution)

Solution. By Problem 1,

$\begin{aligned} a_n &= \frac 2{2} \int_{-2/2}^{2/2} f(t) \cos \left( \frac{2n\pi t}{2} \right)\, \mathrm dt \\ &= \int_{-1}^1 e^t \cos (n \pi t)\, \mathrm dt \\ &= \left[ \frac{e^t ( \cos(n \pi t) + n\pi \sin(n \pi t))}{1 + n^2 \pi^2} \right]_{-1}^1 \\ &= \frac{(-1)^n ( e - e^{-1} )}{1 + n^2 \pi^2} . \end{aligned}$

Similarly,

$\begin{aligned} b_n &= \frac 2{2} \int_{-2/2}^{2/2} f(t) \sin \left( \frac{2n\pi t}{2} \right)\, \mathrm dt \\ &= \int_{-1}^1 e^t \sin (n \pi t)\, \mathrm dt \\ &= \left[ \frac{e^t ( \sin(n \pi t) - n\pi \cos(n \pi t))}{1 + n^2 \pi^2} \right]_{-1}^1 \\ &= -n \pi \cdot \frac {(-1)^n (e - e^{-1} )}{1 + n^2 \pi^2} = -n\pi a_n. \end{aligned}$

By the definition of the Fourier series,

$\displaystyle \begin{aligned} f(t) &= \frac{a_0}{2} + \sum_{n=1}^\infty \left( a_n \cos\left( \frac{2n\pi t}{T} \right) + b_n \sin\left( \frac{2n\pi t}{T} \right) \right) \\ &= \frac{a_0}{2} + \sum_{n=1}^\infty a_n (( \cos (n \pi t) - \pi n \sin (n \pi t) ) \\ &= \frac{e - e^{-1}}{2} + \sum_{n=1}^\infty \frac{(-1)^n ( e - e^{-1} )}{1 + n^2 \pi^2} (( \cos (n \pi t) - \pi n \sin (n \pi t) ). \end{aligned}$

Remark 2. If instead we used the complex-exponential form of the Fourier series

$\displaystyle f(t) = \sum_{n = -\infty}^\infty c_n e^{i \cdot 2\pi n t/T},\quad c_n := \frac 1{T} \int_{-T/2}^{T/2} f(t) \cdot e^{i \cdot 2\pi n t/T}\, \mathrm dt,$

then complex-valued integration yields

$\displaystyle c_n = \frac 12 \left[ \frac{e^{(1 + i \cdot \pi n) t}}{1 + i \cdot \pi n} \right]_{-1}^1 = \frac {(-1)^n (e - e^{-1})(1 - (\pi n) \cdot i)}{2(1 + n^2 \pi^2)},$

recovering the results in Problem 2 via the relationships

$a_n = c_n + \bar{c}_n,\quad b_n = -i \cdot (c_n - \bar{c}_n) =-n\pi a_n.$

—Joel Kindiak, 14 May 25, 1911H

May 14, 2025
The Wallis Product

Problem 1. By considering the integral $\displaystyle I(n) := \int_0^{\pi} \sin^n(x)\, \mathrm dx$ , prove the Wallis product

$\displaystyle \frac 21 \cdot \frac 23 \cdot \frac 43 \cdot \frac 45 \cdot \cdots := \lim_{n \to \infty} \prod_{k=1}^n \frac{2n \cdot 2n}{(2n-1)(2n+1)} = \frac{\pi}{2}.$

(Click for Solution)

Solution. We leave it as an exercise to check that $I(0) = \pi$ and $I(1) = 2$ . These observations motivate an expression of $I(n+2)$ in terms of $I(n)$ . To that end, we integrate by parts to obtain

$\begin{aligned} & \int \sin^{n+2}(x)\, \mathrm dx = \int \sin(x) \sin^{n+1}(x)\, \mathrm dx \\ &= \underbrace{-{\cos(x)}}_{\mathrm I} \underbrace{\sin^{n+1}(x)}_{\mathrm S} - \int \underbrace{-{\cos(x)}}_{\mathrm I} \underbrace{(n+1) \sin^n(x) \cos(x) }_{\mathrm D}\, \mathrm dx \\ &= -{\cos(x)} \sin^{n+1}(x) + (n+1) \int \cos^2(x) \sin^n(x)\, \mathrm dx \\ &= -{\cos(x)} \sin^{n+1}(x) + (n+1) \int (1-\sin^2(x)) \sin^n(x)\, \mathrm dx \\ &= -{\cos(x)} \sin^{n+1}(x) + (n+1) \left( \int \sin^n(x)\, \mathrm dx - \int \sin^{n+2}(x)\, \mathrm dx\right). \end{aligned}$

By algebruh,

$\displaystyle (n+2) \int \sin^{n+2}(x)\, \mathrm dx = -{\cos(x)} \sin^{n+1}(x) + (n+1) \int \sin^n(x)\, \mathrm dx.$

Plugging in the limits of $[0, \pi]$ ,

$\displaystyle (n+2) \int_0^{\pi} \sin^{n+2}(x)\, \mathrm dx = (n+1) \int_0^{\pi} \sin^n(x)\, \mathrm dx.$

Hence,

$\displaystyle \frac{I(n+2)}{I(n)} = \frac{n+1}{n+2}.$

In particular,

$\displaystyle I(2k+2) = \frac{2k+1}{2k+2} \cdot I(2k) \quad\Rightarrow \quad I(2n+2) = \pi \cdot \prod_{k=1}^n \frac{2k-1}{2k}$

Similarly,

$\displaystyle I(2n+1) = 2 \cdot \prod_{k=1}^n \frac{2k}{2k+1}.$

Dividing the terms appropriately,

$\displaystyle \prod_{k=1}^n \frac{2k \cdot 2k}{(2k-1) (2k+1)} = \frac{\pi}{2} \cdot \frac{I(2n+1)}{I(2n+2)}.$

It suffices to prove that $I(2n+1)/I(2n+2) \to 1$ . Since $0 \leq \sin(x) \leq 1$ for $x \in [0, \pi]$ ,

$0 \leq \sin^{n+1}(x) \leq \sin^n(x) \leq \sin^{n-1}(x) \leq 1.$

By the monotonicity of integration, $I(n+1) \leq I(n)$ so that

$\displaystyle 1 \leq \frac{I(2n+1)}{I(2n+2)} \leq \frac{I(2n)}{I(2n+2)} = \frac{2n+2}{2n+1} \to 1.$

By the squeeze theorem, $I(2n+1)/I(2n+2) \to 1$ , as required.

—Joel Kindiak, 21 Apr 25, 2129H

April 21, 2025
The Intermediate Value Derivative

Problem 1. Let $f : [a, b] \to \mathbb R$ differentiable. Assume $f'(a) < f'(b)$ . Prove that for any $m \in (f'(a), f'(b))$ , there exists $c \in [a, b]$ such that $f'(c) = m$ .

(Click for Solution)

Solution. We observe that $f'(c) = m \iff (f(x)-mx)'(c) = 0$ . Define the differentiable function $g : [a, b] \to \mathbb R$ by $g(x) = f(x) - mx$ . Then it suffices to find $c \in [a, b]$ such that $g'(c) = 0$ .

Since $g$ is differentiable on $[a, b]$ , it is continuous on $[a, b]$ . By the extreme value theorem, there exists $c \in [a, b]$ such that $g(x) \geq g(c)$ for any $x \in [a, b]$ . We claim that $g'(c) = 0$ .

Firstly, we note that $g'(a) = f'(a) - m < 0$ , so that there exists $\delta_1 > 0$ such that $g(c) \leq g(a+\delta_1) < g(a)$ . Similarly, $g'(b) = f'(b) - m > 0$ , so that there exists $\delta_2 > 0$ such that $g(c) \leq g(b-\delta_2) < g(b)$ . Hence, $g(c) \neq g(a), g(b)$ . Therefore, $c \in (a, b)$ , which implies $g'(c) = 0$ , as required.

Remark 1. Any function that satisfies the conclusion of the intermediate value theorem is called a Darboux function. The intermediate value theorem demonstrates that continuous functions are Darboux functions. This problem demonstrates that derivatives are automatically Darboux functions, even if they are not continuous.

—Joel Kindiak, 22 Jan 25, 1234H

April 3, 2025
Several Laborious Integrals

Problem 1. Evaluate the integral $\displaystyle \int \frac{(2x+1) \ln(9x+5)}{(3x+2)^4} \, \mathrm dx$ .

(Click for Solution)

Solution. We first evaluate $\displaystyle \int \frac{(2x+1) }{(3x+2)^4} \, \mathrm dx$ for simplicity. Integrating by parts,

$\displaystyle \begin{aligned} \int \frac{(2x+1) }{(3x+2)^4} \, \mathrm dx &= \underbrace{ \frac{(3x+2)^{-3}}{-3 \cdot 3} }_{\mathrm I} \cdot \underbrace{(2x+1)}_{\mathrm S} - \int \underbrace{ \frac{(3x+2)^{-3}}{-3 \cdot 3} }_{\mathrm I} \cdot \underbrace{2}_{\mathrm D}\, \mathrm dx \\ &= -\frac{2x+1}{9(3x+2)^3} + \frac 29 \int (3x+2)^{-3}\, \mathrm dx \\ &= -\frac{2x+1}{9(3x+2)^3} + \frac 29 \cdot \frac{(3x+2)^{-2}}{-2 \cdot 3} +C \\ &= -\frac{2x+1}{9(3x+2)^3} - \frac 1{27(3x+2)^2} +C \\ &= -\frac{3(2x+1)}{27(3x+2)^3} - \frac {3x+2}{27(3x+2)^3} +C \\ &= -\frac{3(2x+1) + 3x+2}{27(3x+2)^3} +C \\ &= -\frac{9x+5}{27(3x+2)^3} + C.\end{aligned}$

Therefore, integrating by parts again,

$\displaystyle \begin{aligned} & \int \frac{(2x+1) \ln(9x+5)}{(3x+2)^4} \, \mathrm dx \\ &= \underbrace{-\frac{9x+5}{27(3x+2)^3}}_{\mathrm I} \cdot \underbrace{\ln(9x+5)}_{\mathrm S} - \int \underbrace{-\frac{9x+5}{27(3x+2)^3}}_{\mathrm I}\cdot \underbrace{\frac 1{9x+5} \cdot 9}_{\mathrm D}\, \mathrm dx \\ &= -\frac{(9x+5)\ln(9x+5)}{27(3x+2)^3} + \frac 13 \int \frac{1}{(3x+2)^3}\, \mathrm dx \\ &= -\frac{(9x+5)\ln(9x+5)}{27(3x+2)^3} + \frac 13 \int (3x+2)^{-3}\, \mathrm dx \\ &= -\frac{(9x+5)\ln(9x+5)}{27(3x+2)^3} + \frac 13 \cdot \frac 13 \cdot \frac{(3x+2)^{-2}}{-2} + C \\ &= -\frac{(9x+5)\ln(9x+5)}{27(3x+2)^3} - \frac 1{18 (3x+2)^2} + C.\end{aligned}$

Problem 2. Evaluate the integral $\displaystyle \int_4^5 \frac{9x^3 - 14x^2 + 8x - 11}{(x-1)(x-2)(x^2+3)} \, \mathrm dx$ .

(Click for Solution)

Solution. We decompose into a sum of partial fractions as follows:

$\displaystyle \frac{9x^3 - 14x^2 + 8x - 11}{(x-1)(x-2)(x^2+3)} = \frac{A}{x-1} + \frac{B}{x-2} + \frac{Cx+D}{x^2+3}.$

Assuming we have evaluated $A,B,C,D$ , integrating yields

$\displaystyle \begin{aligned} &\int_4^5 \frac{9x^3 - 14x^2 + 8x - 11}{(x-1)(x-2)(x^2+3)} \, \mathrm dx \\ &= \int_4^5 \left(A \cdot \frac{1}{x-1} + B \cdot \frac{1}{x-2} + \frac{C}{2} \cdot \frac{2x}{x^2+3} + D \cdot \frac{1}{(\sqrt 3)^2 + x^2}\right)\, \mathrm dx \\ &= \left[A \ln|x-1| + B \ln |x-2| + \frac C2 \ln(x^2+3) + \frac{D}{\sqrt 3} \tan^{-1}\left( \frac{x}{\sqrt 3}\right)\right]_4^5 \\ &= A \ln \frac 43 + B \ln \frac 32 + \frac C2 \ln \frac{28}{19} + \frac{D}{\sqrt 3} \left( \tan^{-1}\frac{5}{\sqrt 3} - \tan^{-1}\frac{4}{\sqrt 3}\right). \end{aligned}$

It remains to evaluate $A,B,C,D$ . Cross-multiplying yields

$\displaystyle \begin{aligned} 9x^3 - 14x^2 + 8x - 11 &= A(x-2)(x^2+3) \\ &\phantom{=} + B(x-1)(x^2+3)\\ &\phantom{=} + (Cx+D)(x-1)(x-2).\end{aligned}$

Setting $x = 1$ yields $A = 2$ .
Setting $x = 2$ yields $B = 3$ .

Comparing the terms involving $x^3$ ,

$9 = A + B + C \quad \Rightarrow \quad C = 4.$

Comparing the constant terms,

$-11 = -6A + (-3B) + 2D \quad \Rightarrow \quad D = 5.$

Therefore, the definite integral evaluates to

$\begin{aligned} &\int_4^5 \frac{9x^3 - 14x^2 + 8x - 11}{(x-1)(x-2)(x^2+3)} \, \mathrm dx \\ &=2 \ln \frac 43 + 3 \ln \frac 32 + \frac 42 \ln \frac{28}{19} + \frac{5}{\sqrt 3} \left( \tan^{-1}\frac{5}{\sqrt 3} - \tan^{-1}\frac{4}{\sqrt 3}\right)\\ &= 2 \ln \frac {112}{57} + 3 \ln \frac 32 + \frac{5}{\sqrt 3} \left( \tan^{-1}\frac{5}{\sqrt 3} - \tan^{-1}\frac{4}{\sqrt 3}\right) \\ &= 2 \ln \frac {112}{57} + 3 \ln \frac 32 + \frac{5}{\sqrt 3}\tan^{-1}\frac{\sqrt 3}{23},\end{aligned}$

where the last equality follows from the identity

$\displaystyle \tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1} \left(\frac{a-b}{1+ab}\right).$

Problem 3. Given $a > 0$ and $b^2 - 4ac < 0$ , evaluate $\displaystyle \int \frac{1}{\sqrt{ax^2 + bx + c}}\, \mathrm dx$ . Deduce the value of

$\displaystyle \lim_{n \to \infty} \left( \frac{1}{\sqrt{n^2+1^2}} + \frac{1}{\sqrt{n^2+2^2}} + \cdots + \frac{1}{\sqrt{n^2+n^2}} \right).$

(Click for Solution)

Solution. For the integral, we first complete the square

$ax^2 + bx + c = a(x+h)^2 + k,$

where $h$ and $k$ are constants to be determined in terms of $a,b,c$ . To determine $h, k$ , we expand the right-hand side to obtain

$ax^2 + bx + c = ax^2 + (2ah)x + ah^2 + k.$

Comparing coefficients $h = b/(2a)$ and

$\begin{aligned} k = c - ah^2 &= c - a \left(\frac b{2a}\right)^2 \\ &= c - a \cdot \frac{b^2}{4a^2} \\ &= \frac{4ac- b^2}{4a} \\ &= \frac{-(b^2 - 4ac)}{4a} > 0. \end{aligned}$

Therefore,

$\begin{aligned} &\int \frac{1}{\sqrt{ax^2 + bx + c}}\, \mathrm dx \\ &= \int \frac{1}{\sqrt{a(x+h)^2 + k}}\, \mathrm dx \\ &= \frac{1}{\sqrt a} \int \frac{1}{\sqrt{(x+h)^2 + (\sqrt{k/a})^2 }}\, \mathrm dx \\ &= \frac 1{\sqrt a} \ln \left|(x+h) + \sqrt{(x+h)^2 + (\sqrt{k/a})^2} \right| + C \\ &= \frac 1{\sqrt a} \ln \left| x+\frac b{2a} + \sqrt{ \left(x+\frac b{2a} \right)^2 + \frac{4ac-b^2}{4a^2} } \right| + C. \end{aligned}$

For the limit, we factorise $n^2$ to obtain

$\displaystyle \begin{aligned} &\lim_{n \to \infty} \left( \frac{1}{\sqrt{n^2+1^2}} + \frac{1}{\sqrt{n^2+2^2}} + \cdots + \frac{1}{\sqrt{n^2+n^2}} \right) \\ &=\lim_{n \to \infty} \frac 1n \left( \frac{1}{\sqrt{1+\left( \frac 1n\right)^2}} + \frac{1}{\sqrt{1+\left( \frac 2n\right)^2}} + \cdots + \frac{1}{\sqrt{1+\left( \frac nn\right)^2}} \right) \\ &= \int_0^1 \frac{1}{\sqrt{1+x^2}}\, \mathrm dx, \end{aligned}$

where the last equality follows from the definition of integration. Setting $a = 1, b = 0, c = 1$ in the indefinite integral,

$\begin{aligned} \int_0^1 \frac{1}{\sqrt{1+x^2}}\, \mathrm dx &= \left[ \frac 1{\sqrt 1} \ln \left| x+ 0 + \sqrt{ (x+ 0 )^2 + \frac{4 \cdot 1 \cdot 1 - 0^2}{4 \cdot 1^2} } \right| \right]_0^1 \\ &= \left[ \ln \left| x + \sqrt{ x^2 + 1 } \right| \right]_0^1 \\ &= \ln ( 1 + \sqrt{ 2 } ) - 0 \\ &= \ln ( 1 + \sqrt{ 2 } ).\end{aligned}$

—Joel Kindiak, 24 Feb 25, 1800H

March 26, 2025

Calculus, math, mathematics, physics, science
Calculus Recap (Polytechnic)

Limits

Problem 1. Evaluate the limit $\displaystyle \lim_{x \to 1} (x+1)$ .

(Click for Solution)

Solution. Substituting $x = 1$ ,

$\displaystyle \begin{aligned} \lim_{x \to 1} (x+1) &= 1 + 1 = 2. \end{aligned}$

Problem 2. Evaluate the limit $\displaystyle \lim_{x \to 1} \frac{x^2 - 1}{x-1}$ .

(Click for Solution)

Solution. Factorising for $x \neq 1$ ,

$\displaystyle \begin{aligned} \lim_{x \to 1} \frac{x^2 - 1}{x-1} &= \lim_{x \to 1} \frac{(x-1)(x+1)}{x-1} \\ &= \lim_{x \to 1} (x+1) = 2, \end{aligned}$

where the last $=$ follows from Problem 1.

Problem 3. Evaluate the limit $\displaystyle \lim_{x \to 1} \frac{\sqrt x - 1}{x-1}$ .

(Click for Solution)

Solution. Rationalising the numerator then substituting $x = 1$ ,

$\displaystyle \begin{aligned} \lim_{x \to 1} \frac{\sqrt x - 1}{x-1} &= \lim_{x \to 1} \frac{\sqrt x - 1}{x-1} \cdot \frac{\sqrt x + 1}{\sqrt x + 1} \\ &= \lim_{x \to 1} \frac{(\sqrt x)^2 - 1^2}{(x-1)(\sqrt x + 1)} \\ &= \lim_{x \to 1} \frac{x - 1}{(x-1)(\sqrt x + 1)} \\ &= \lim_{x \to 1} \frac 1{\sqrt x + 1} = \frac 1{\sqrt 1 + 1} = \frac 12.\end{aligned}$

Problem 4. Without using differentiation, compute the gradient of the tangent to $y =x^2$ at $(1, 1)$ .

(Click for Solution)

Solution. Denote $(x_1, y_1) = (1, 1)$ and $(x_2, y_2) = (x, x^2)$ . By definition of the gradient of the tangent,

$\displaystyle \begin{aligned} \lim_{x \to 1} \frac{\text{(rise)}}{\text{(run)}} &= \lim_{x \to 1} \frac{y_2 - y_1}{x_2 - x_1} = \lim_{x \to 1} \frac{x^2 - 1}{x - 1} = 2,\end{aligned}$

where the last $=$ follows from Problem 2.

Differentiation

Problem 5. Using differentiation, compute the gradient of the tangent to $y = x^2$ at $(1, 1)$ .

(Click for Solution)

Solution. Differentiating,

$\displaystyle \frac{\mathrm dy}{\mathrm dx} = \frac{\mathrm d}{\mathrm dx} (x^2) = 2x.$

Setting $x = 1$ , $\displaystyle \frac{\mathrm dy}{\mathrm dx} = 2 \cdot 1 = 2$ .

Problem 6. Defining $f(x) = x^2$ , evaluate $f''(1)$ .

(Click for Solution)

Solution. Differentiating twice,

$\begin{aligned} f(x) &= x^2, \\ f'(x) &= 2x, \\ f''(x) &= 2. \end{aligned}$

Setting $x = 1$ , $f''(1) = 2$ .

Problem 7. Evaluate $\displaystyle \frac{\mathrm d}{\mathrm dx}\left(\ln(x^2+9) \right)$ .

(Click for Solution)

Solution. Employing the chain rule,

$\displaystyle \begin{aligned} \frac{ \mathrm d }{ \mathrm dx }\left( \ln(x^2+1) \right) &= \frac 1{x^2 + 9} \cdot \frac{\mathrm d}{\mathrm dx}(x^2 + 1) \\ &= \frac 1{x^2 + 9} \cdot 2x \\ &= \frac { 2x }{x^2 + 9}. \end{aligned}$

Problem 8. Evaluate $\displaystyle \frac{\mathrm d}{\mathrm dx}\left(\sin \left(\ln(x^2+9) \right) \right)$ .

(Click for Solution)

Solution. Employing the chain rule and Problem 7,

$\displaystyle \begin{aligned} \frac{\mathrm d}{\mathrm dx}\left(\sin \left(\ln(x^2+1) \right) \right) &= \cos\left(\ln(x^2+9) \right) \cdot \frac{\mathrm d}{\mathrm dx}\ln(x^2+9) \\ &= \cos\left(\ln(x^2+9) \right) \cdot \frac {2x}{x^2 + 9} \\ &= \frac {2x \cos\left(\ln(x^2+9) \right) }{x^2 + 9}. \end{aligned}$

Problem 9. Evaluate $\displaystyle \frac{\mathrm d}{\mathrm dx} \left( (\ln x) \cdot (x^2 + 9) \right)$ .

(Click for Solution)

Solution. Employing the product rule,

$\displaystyle \begin{aligned} \frac{\mathrm d}{\mathrm dx} \left( (\ln x) \cdot (x^2 + 9) \right) &= (x^2 + 9) \cdot \frac{\mathrm d}{\mathrm dx} ( \ln x ) + \ln x \cdot \frac{\mathrm d}{\mathrm dx} (x^2 + 9) \\ &= (x^2 + 9) \cdot \frac 1x + \ln x \cdot (2x) \\ &= x + \frac 9x + 2x \ln x. \end{aligned}$

Problem 10. Evaluate $\displaystyle \frac{\mathrm d}{\mathrm dx} \left( \frac{\ln x}{x^2 + 9}\right)$ .

(Click for Solution)

Solution. Employing the quotient rule,

$\displaystyle \begin{aligned} \frac{\mathrm d}{\mathrm dx} \left( \frac{\ln x}{x^2 + 9} \right) &= \frac{(x^2 +9) \cdot \frac{\mathrm d}{\mathrm dx} ( \ln x ) - \ln x \cdot \frac{\mathrm d}{\mathrm dx} (x^2 + 9) }{(x^2 + 9)^2} \\ &= \frac{ (x^2 + 9) \cdot \frac 1x - \ln x \cdot 2x}{(x^2 + 9)^2 } \\ &= \frac{x^2 + 9 -2x^2 \ln x}{x(x^2 + 9)^2}. \end{aligned}$

Integration

Problem 11. Evaluate $\displaystyle \int \left(2\ln x + \frac 5{x^2+9}\right)\, \mathrm dx$ .

(Click for Solution)

Solution. Applying the linearity of integration and using common formulas (here and here),

$\displaystyle \begin{aligned} \int \left(2\ln x + \frac 5{x^2+9}\right)\, \mathrm dx &= \int \left(2 \cdot\ln x + 5 \cdot \frac 1{x^2 + 9}\right)\, \mathrm dx \\ &= 2 \cdot \int \ln x \, \mathrm dx + 5 \cdot \int \frac 1{3^2 + x^2} \, \mathrm dx \\ &= 2 \cdot (x \ln x - x) + 5 \cdot \frac 1{3} \tan^{-1} \left(\frac x3 \right) + C \\ &= 2x \ln x - 2x + \frac 53 \tan^{-1} \left(\frac x3 \right) + C. \end{aligned}$

Problem 12. Evaluate $\displaystyle \int_1^2 x\ln (x^2+9)\, \mathrm dx$ .

(Click for Solution)

Solution. Making the substitution $\displaystyle u = x^2 + 9 \Rightarrow \mathrm du = 2x\, \mathrm dx$ ,

$\displaystyle \begin{aligned} \int x\ln (x^2+9 )\, \mathrm dx &= \frac 12 \int \ln (x^2+9 ) \cdot 2x\, \mathrm dx \\ &= \frac 12 \int \ln u \, \mathrm du \\ &= \frac 12 (u \ln u - u) + C \\ &= \frac 12 \left((x^2 + 9) \ln (x^2 + 9) - (x^2 + 9) \right) + C. \end{aligned}$

Hence, substituting the limits,

$\displaystyle \begin{aligned} \int_1^2 x\ln (x^2+1)\, \mathrm dx &= \left[ \frac 12 \left( (x^2 + 9) \ln (x^2 + 9) - (x^2 + 9) \right) \right]_1^2 \\ &= \frac 12 \left( 13 \ln 13 - 13 \right) - \frac 12 \left( 10 \ln 10 - 10 \right) \\ &= \frac {13}2 \ln 13 - 5 \ln 10 - \frac 32.\end{aligned}$

Problem 13. Evaluate $\displaystyle \int \tan^3 x\, \mathrm dx$ .

(Click for Solution)

Solution. We solved this before, and solve it again for revision. Using the Pythagorean identity $\sec^2 x = \tan^2 x +1$ ,

$\displaystyle \begin{aligned} \int \tan^3 x\, \mathrm dx &= \int \tan x \tan^2 x\, \mathrm dx \\ &= \int \tan x (\sec^2 x - 1)\, \mathrm dx \\ &= \int \tan x \sec^2 x \, \mathrm dx - \int \tan x\, \mathrm dx \\ &= \int \tan x \sec^2 x \, \mathrm dx -(- \ln|{\cos x}|) + C.\end{aligned}$

For the first integral, make the substitution $\displaystyle u = \tan x \Rightarrow \mathrm dx = \frac{1}{\sec^2 x}\, \mathrm du$ . Then

$\displaystyle \begin{aligned}\int \tan x \sec^2 x \, \mathrm dx &= \int u \sec^2 x \cdot \frac{1}{\sec^2 x}\, \mathrm du \\ &= \int u\, \mathrm du \\ &= \frac 12 u^2 + C\\ &= \frac 12 \tan^2 x + C.\end{aligned}$

Therefore, $\displaystyle \int \tan^3 x\, \mathrm dx = \frac 12 \tan^2 x + \ln|{\cos x}| + C.$

Problem 14. Evaluate $\displaystyle \int \ln (x^2+9)\, \mathrm dx$ .

(Click for Solution)

Solution. Writing $\ln (x^2+9) = 1 \cdot \ln (x^2+9)$ and integrating by parts,

$\displaystyle \begin{aligned} \int \ln (x^2+9)\, \mathrm dx &= \underbrace{x}_{\mathrm I} \cdot \underbrace{\ln (x^2+9)}_{\mathrm S} - \int \underbrace{x}_{\mathrm I} \cdot \underbrace{\frac{2x}{x^2+9}}_{\mathrm D}\, \mathrm dx \\ &= x \ln(x^2 + 9) - \int \frac{2x^2}{x^2 + 9}\, \mathrm dx \\ &= x \ln(x^2 + 9) - 2\int \frac{(x^2 + 9) - 9}{x^2 + 1}\, \mathrm dx \\ &= x \ln(x^2 + 9) - 2\int \left( 1 - 9 \cdot \frac{1}{3^2 + x^2}\right) \, \mathrm dx \\ &= x \ln(x^2 + 9) - 2 \left( x - 9 \cdot \frac 1{3} \tan^{-1} \left( \frac x3 \right) \right) + C\\ &= x \ln(x^2 + 9) - 2x + 6\tan^{-1} \left( \frac x3 \right) + C.\end{aligned}$

Problem 15. Evaluate $\displaystyle \int \frac {2x+5}{x(x^2+9)}\, \mathrm dx$ .

(Click for Solution)

Solution. If we can decompose the integrand into

$\displaystyle \frac {2x+5}{x(x^2+9)} = \frac Ax + \frac{Bx+C}{x^2+9},$

Then a bit of basic integration yields

$\displaystyle \begin{aligned} \int \frac {2x+5}{x(x^2+9)}\, \mathrm dx &= \int \left(\frac Ax + \frac{Bx+C}{x^2+9}\right)\, \mathrm dx \\ &= \int \left(A \cdot \frac 1x + \frac B2 \cdot \frac{2x}{x^2+9} + C \cdot \frac 1{x^2 + 9} \right)\, \mathrm dx \\ &= A \int \frac 1x\, \mathrm dx + \frac B2 \int \frac {2x}{x^2 + 1}\, \mathrm dx + C \int \frac 1{3^2 + x^2}\, \mathrm dx \\ &= A \ln|x| + \frac B2 \ln (x^2 + 1) + C \cdot \frac 13 \tan^{-1} \left(\frac x3 \right) + D. \end{aligned}$

It remains to complete the decomposition. Clearing denominators,

$\begin{aligned}\frac {2x+5}{x(x^2+9)} &= \frac Ax + \frac{Bx+C}{x^2+9} \\ 2x+5 &= \frac {Ax(x^2+9)}x + \frac{(Bx+C)x(x^2+9)}{x^2+9} \\ 2x+5 &= A(x^2 + 9) + (Bx + C)x\\ 2x+5 &= (A+B)x^2 + Cx + 9A.\end{aligned}$

Setting $x = 0$ yields $A = 5/9$ .

Comparing coefficients yields $B = -5/9$ , and $C = 2$ .

Hence, by back-substituting,

$\displaystyle \begin{aligned} \int \frac 1{x(x^2+9)}\, \mathrm dx &= \frac 59 \ln|x| - \frac 5{18} \ln (x^2 + 9) + \frac 23 \tan^{-1} \left(\frac x3 \right) + C. \end{aligned}$

Problem 16. Evaluate $\displaystyle \int \frac {(2 \sin \theta +5)\cos \theta}{\sin \theta (\sin^2 \theta+9)}\, \mathrm d\theta$ .

(Click for Solution)

Solution. Making the substitution $\displaystyle u = \sin \theta \Rightarrow \mathrm d\theta = \frac{1}{\cos \theta}\, \mathrm du$ and applying the result from Problem 15,

$\displaystyle \begin{aligned} & \int \frac {(2 \sin \theta +5)\cos \theta}{\sin \theta (\sin^2 \theta+9)}\, \mathrm dx \\ &= \int \frac {(2u+5)\cos \theta}{u (u^2+9)} \cdot \frac{1}{\cos \theta}\, \mathrm du \\ &= \int \frac {2u+5}{u (u^2+9)} \, \mathrm du \\ &= \frac 59 \ln|u| - \frac 5{18} \ln (u^2 + 9) + \frac 23 \tan^{-1} \left(\frac u3 \right) + C \\ &= \frac 59 \ln| {\sin \theta} | - \frac 5{18} \ln ( \sin^2 \theta + 9 ) + \frac 23 \tan^{-1} \left(\frac {\sin \theta}3 \right) + C. \end{aligned}$

—Joel Kindiak, 13 Feb 25, 1757H

February 13, 2025

Calculus, math, mathematics, physics, Probability
Several Differentiation Drills

For any integer $n$ , you may freely use the results

$\begin{aligned} (x^n)' &= nx^{n-1},\\ \sin'(x) &= \cos(x),\\ \cos'(x) &= -{\sin(x)},\\ \tan'(x) &= \sec^2(x), \\ (e^x)' &= e^x, \\ \ln'(x) &= \frac 1x, \end{aligned}$

from either previous posts or from real analysis. You may also apply the chain rule, product rule, and quotient rule however you wish.

Problem 1. For any $a > 0$ , prove that

$\displaystyle \begin{aligned} (a^x)' = a^x \ln(a), \quad \log_a'(x) = \frac{1}{x \ln(a)}. \end{aligned}$

(Click for Solution)

Solution. By exponential properties,

$a^x = (e^{\ln(a)})^x = e^{x \ln(a)}.$

Hence, by the chain rule,

$\displaystyle \begin{aligned} (a^x)' &= (e^{x \ln(a)})'\\ &= e^{x \ln(a)} \cdot \frac{\mathrm d}{\mathrm dx}(x \ln a) \\ &= e^{x \ln(a)} \cdot \ln(a) \\ &= a^x \ln(a). \end{aligned}$

By logarithm properties and linearity,

$\displaystyle \begin{aligned} \log_a'(x) &= \left( \frac{\ln(x)}{\ln(a)} \right)'\\ &= \frac{1}{\ln(a)} \ln'(x) \\ &= \frac{1}{\ln(a)} \cdot \frac 1x = \frac 1{x \ln (a)}. \end{aligned}$

Up till now, we have proven that $(x^n)' = nx^{n-1}$ only for integer values of $n$ . We extend this result to all real values of $n$ .

Problem 2. For any real $r$ , use Problem 1 to prove that $(x^r)' = rx^{r-1}$ .

(Click for Solution)

Solution. By exponential properties,

$x^r = (e^{\ln x})^r = e^{r \ln x}.$

Hence, by the chain rule,

$\displaystyle \begin{aligned} (x^r)' &= (e^{r \ln x})'\\ &= e^{r \ln x} \cdot \frac{\mathrm d}{\mathrm dx}(r \ln x) \\ &= x^r \cdot r \cdot \frac 1x = rx^{r-1}. \end{aligned}$

Example 1. By Problem 2,

$\displaystyle \frac{\mathrm d}{\mathrm dx}(x^{\sqrt 2}) = \sqrt 2 \cdot x^{\sqrt 2 - 1}, \quad \frac{\mathrm d}{\mathrm dx}(x^e) = e \cdot x^{e - 1}, \quad \frac{\mathrm d}{\mathrm dx}(x^\pi) = \pi \cdot x^{\pi - 1}.$

In particular,

$\displaystyle \frac{\mathrm d}{\mathrm dx}(x^e) = e \cdot x^{e - 1}\neq e^x = \frac{\mathrm d}{\mathrm dx}(e^x).$

Problem 3. Prove that

$\displaystyle \sec'(x) = \sec(x)\tan(x),\quad \csc'(x) = -{\csc(x)\cot(x)}.$

(Click for Solution)

Solution. Writing $\displaystyle \sec(x) = \frac 1{\cos(x)}$ , apply the quotient rule to obtain

$\displaystyle \begin{aligned} \sec'(x) &= \frac{ (1)' \cdot \cos(x) - 1 \cdot \cos'(x) }{ \cos^2(x) } \\ &= \frac{ 0 \cdot \cos(x) - 1 \cdot (-\sin(x)) }{ \cos^2(x) } \\ &= \frac{\sin(x)}{\cos^2(x)} = \frac{1}{\cos(x)} \cdot \frac{\sin(x)}{\cos(x)} = \sec(x) \tan(x). \end{aligned}$

We recall that $\csc(x) = \sec(\pi/2-x)$ so that

$\displaystyle \begin{aligned}\csc'(x) &= (\sec(\pi/2-x))' \\ &= \sec'(\pi/2-x)) \cdot (-1) \\ &= \sec(\pi/2-x) \tan(\pi/2-x) \cdot (-1) \\ &= \csc(x) \cot(x) \cdot (-1) \\ &= - \csc(x) \cot(x). \end{aligned}$

Problem 4. Given that $t > 0$ , $x \equiv x(t) > 0$ , and $y \equiv y(t) > 0$ , evaluate $\displaystyle \frac{\mathrm d}{\mathrm dt}(x^y)$ .

(Click for Solution)

Solution. Writing $x^y = e^{y \ln(x)}$ and using the chain rule,

$\begin{aligned} \frac{\mathrm d}{\mathrm dt}(x^y) &= \frac{\mathrm d}{\mathrm dt} (e^{y \ln(x)}) \\ &= e^{y \ln(x)} \left( \frac{\mathrm d}{\mathrm dt}(y \ln(x)) \right) \\ &= x^y \left( \frac{\mathrm dy}{\mathrm dt} \cdot \ln(x) + y \cdot \frac{\mathrm d}{\mathrm dt} ( \ln(x)) \right) \\ &= x^y \left( \frac{\mathrm dy}{\mathrm dt} \cdot \ln(x) + y \cdot \frac 1x \cdot \frac{\mathrm dx}{\mathrm dt} \right) \\ &= x^y \ln(x) \cdot \frac{\mathrm dy}{\mathrm dt} + yx^{y-1} \cdot \frac{\mathrm dx}{\mathrm dt}. \end{aligned}$

Remark 1. If $x = t$ and $y$ is constant, we recover the power rule. If $x$ is constant and $y = t$ , we recover Problem 1. Furthermore, if $y = x = t$ , we obtain

$\displaystyle \frac{\mathrm d}{\mathrm dx}(x^x) = x^x (\ln(x) + 1).$

In the language of partial derivatives, denoting $z = x^y \equiv { x(t) }^{ y(t) }$ ,

$\displaystyle \frac{\mathrm d z}{\mathrm dt} = \frac{\partial z}{\partial y} \cdot \frac{\mathrm dy}{\mathrm dt} + \frac{\partial z}{\partial x} \cdot \frac{\mathrm dx}{\mathrm dt}.$

Problem 5. Suppose $f, f^{-1}$ are differentiable on $\mathbb R$ . Evaluate $(f^{-1})'(x)$ .

(Click for Solution)

Solution. By definition of the inverse function,

$\begin{aligned} \frac{\mathrm d}{\mathrm dx} (x) &= \frac{\mathrm d}{\mathrm dx} (f(f^{-1}(x))) \\ 1 &= f'(f^{-1}(x)) \cdot \frac{\mathrm d}{\mathrm dx}(f^{-1}(x)) \\ &= f'(f^{-1}(x)) \cdot f'(f^{-1}(x)) \\ (f^{-1})'(x) &= \frac 1{f'(f^{-1}(x))}. \end{aligned}$

—Joel Kindiak, 9 Feb 25, 2126H

February 9, 2025

ai, artificial-intelligence, math, mathematics, physics
Several Limit Drills

Problem 1. For any real number $a$ , evaluate the limit

$\displaystyle \lim_{h \to 0} \frac{(a+h)^2 - a^2}{h}.$

(Click for Solution)

Solution. Expanding the numerator yields

$\displaystyle \begin{aligned} \frac{(a+h)^2 - a^2}{h} &= \frac{a^2 + 2ah + h^2 - a^2}{h} \\ &= \frac{2ah + h^2}{h}\\ &= \frac{(2a+h)h}{h}\\ &= 2a + h. \end{aligned}$

Taking $h \to 0$ yields

$\displaystyle \begin{aligned} \lim_{h \to 0} \frac{(a+h)^2 - a^2}{h} &=\lim_{h \to 0} (2a+h) \\ &= 2a+0 \\ &= 2a. \end{aligned}$

Problem 2. For any real number $a > 0$ , evaluate the limit

$\displaystyle \lim_{h \to 0} \frac{\sqrt{a+h} - \sqrt{a}}{h}.$

(Click for Solution)

Solution. Rationalising the numerator,

$\displaystyle \begin{aligned} \frac{\sqrt{a+h} - \sqrt{a}}{h} &= \frac{\sqrt{a+h} - \sqrt{a}}{h} \cdot \frac{\sqrt{a+h} + \sqrt{a}}{\sqrt{a+h} + \sqrt{a}} \\ &= \frac{(\sqrt{a+h})^2 - (\sqrt{a})^2}{h(\sqrt{a+h} + \sqrt{a})} \\ &= \frac{(a+h) - a}{h(\sqrt{a+h} + \sqrt{a})} \\ &= \frac{h}{h(\sqrt{a+h} + \sqrt{a})} \\ &= \frac{1}{\sqrt{a+h} + \sqrt{a}}.\end{aligned}$

Taking $h \to 0$ yields

$\displaystyle \begin{aligned} \lim_{h \to 0} \frac{(a+h)^2 - a^2}{h} &=\lim_{h \to 0} \frac{1}{\sqrt{a+h} + \sqrt{a}} \\ &= \frac{1}{\sqrt{a + 0} + \sqrt a} \\ &= \frac{1}{2\sqrt a}. \end{aligned}$

Problem 3. For any real number $a \neq 0$ , evaluate the limit

$\displaystyle \lim_{h \to 0} \frac{\frac 1{a+h} - \frac 1a}{h}.$

(Click for Solution)

Solution. Combining denominators,

$\displaystyle \begin{aligned} \frac{\frac 1{a+h} - \frac 1a}{h} &= \frac{1}{h} \left(\frac 1{a+h} - \frac 1a\right) \\ &= \frac{1}{h} \cdot \frac{a - (a+h)}{a(a+h)} \\ &= \frac{1}{h} \cdot \frac{-h}{a(a+h)} \\ &= - \frac{1}{a(a+h)}.\end{aligned}$

Taking $h \to 0$ yields

$\displaystyle \begin{aligned} \lim_{h \to 0} \frac{\frac 1{a+h} - \frac 1a}{h} &=\lim_{h \to 0} - \frac{1}{a(a+h)} \\ &= - \frac{1}{a(a+0)} \\ &= -\frac{1}{a^2}. \end{aligned}$

Remark 1. The derivative $f'$ of a function $f$ is defined by

$f'(a) := \displaystyle \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$

whenever the right-hand side exists. In this exercise, we applied this definition to the functions $f(x) = x^2$ , $f(x) = \sqrt x$ , $f(x) = 1/x$ , and evaluated their derivatives respectively:

$\displaystyle (x^2)' = 2x,\quad (\sqrt x)' = \frac{1}{2\sqrt x},\quad \left(\frac 1x\right)' = -\frac 1{x^2}.$

—Joel Kindiak, 9 Feb 25, 1116H

February 9, 2025

Calculus, machine-learning, math, mathematics, physics
Some Integration Drills

Problem 1. Evaluate the integral $\displaystyle \int(x^2 + \tan x - 3)\, \mathrm dx$ .

(Click for Solution)

Solution. By the linearity of integration,

$\displaystyle \begin{aligned}\int(x^2 + \tan x - 3)\, \mathrm dx &= \int x^2\, \mathrm dx + \int \tan x \, \mathrm dx - \int 3\, \mathrm dx \\ &= \frac 13 x^3 - \ln|{\cos x}| - 3x + C.\end{aligned}$

Problem 2. Evaluate the integral $\displaystyle \int \left(\frac{1}{9 + x^2} + \frac{1}{\sqrt{4-x^2}} \right)\, \mathrm dx$ .

(Click for Solution)

Solution. By the linearity of integration,

$\displaystyle \begin{aligned} \int \left(\frac{1}{9 + x^2} + \frac{1}{\sqrt{4-x^2}} \right)\, \mathrm dx &= \int \frac{1}{9 + x^2} \, \mathrm dx + \int \frac{1}{ \sqrt{4-x^2} } \, \mathrm dx \\ &= \int \frac{1}{3^2 + x^2} \, \mathrm dx + \int \frac{1}{ \sqrt{2^2-x^2} } \, \mathrm dx \\ &= \frac 13 \tan^{-1} \left( \frac x3 \right) + \sin^{-1} \left( \frac x2 \right) + C. \end{aligned}$

Problem 3. Evaluate the integral $\displaystyle \int_0^1 x \sec(x^2) \tan(x^2)\, \mathrm dx$ .

(Click for Solution)

Solution. By the substitution $\displaystyle u = x^2 \Rightarrow \mathrm dx = \frac 1{2x}\, \mathrm du$ ,

$\displaystyle \begin{aligned} \int x \sec(x^2) \tan(x^2)\, \mathrm dx &= \int x \sec(u) \tan(u) \cdot \frac{1}{2x}\, \mathrm du \\ &= \frac 12 \int \sec(u) \tan(u)\, \mathrm du \\ &= \frac 12 \sec(u) + C \\ &= \frac 12 \sec(x^2) + C. \end{aligned}$

Hence, substituting the limits,

$\displaystyle \begin{aligned} \int_0^1 x \sec(x^2) \tan(x^2)\, \mathrm dx &= \frac 12 [\sec(x^2)]_0^1\\ &= \frac 12 (\sec(1) - \sec(0)) \\ &= \frac 1{2 \cos(1)} - \frac 12 \\ &\approx 0.425, \end{aligned}$

when approximated to 3 decimal places.

Problem 4. Evaluate the integral $\displaystyle \int \tan^3 x\, \mathrm dx$ .

(Click for Solution)

Solution. Using the Pythagorean identity $\sec^2 x = \tan^2 x +1$ ,

$\displaystyle \begin{aligned} \int \tan^3 x\, \mathrm dx &= \int \tan x \tan^2 x\, \mathrm dx \\ &= \int \tan x (\sec^2 x - 1)\, \mathrm dx \\ &= \int \tan x \sec^2 x \, \mathrm dx - \int \tan x\, \mathrm dx \\ &= \int \tan x \sec^2 x \, \mathrm dx -(- \ln|{\cos x}|) + C.\end{aligned}$

For the first integral, make the substitution $\displaystyle u = \tan x \Rightarrow \mathrm dx = \frac{1}{\sec^2 x}\, \mathrm du$ . Then

$\displaystyle \begin{aligned}\int \tan x \sec^2 x \, \mathrm dx &= \int u \sec^2 x \cdot \frac{1}{\sec^2 x}\, \mathrm du \\ &= \int u\, \mathrm du \\ &= \frac 12 u^2 + C\\ &= \frac 12 \tan^2 x + C.\end{aligned}$

Therefore, $\displaystyle \int \tan^3 x\, \mathrm dx = \frac 12 \tan^2 x + \ln|{\cos x}| + C.$

Problem 5. Evaluate the integral $\displaystyle \int \sin(\ln \theta)\, \mathrm d\theta$ .

(Click for Solution)

Solution. Make the substitution $u = \ln \theta \Rightarrow \mathrm d\theta = e^u\, \mathrm du$ . Then

$\displaystyle \int \sin(\ln \theta)\, \mathrm d\theta = \int e^u \sin(u)\, \mathrm du.$

Integrating by parts twice,

$\displaystyle \begin{aligned}\int e^u \sin(u)\, \mathrm du &= \underbrace{e^u}_{\mathrm I}\, \underbrace{\sin(u)}_{\mathrm S} - \int \underbrace{e^u}_{\mathrm I}\, \underbrace{\cos(u)}_{\mathrm D}\, \mathrm du \\ &= e^u \sin(u) - \int e^u \cos(u)\, \mathrm du\\ &= e^u \sin(u) - \left(\underbrace{e^u}_{\mathrm I}\, \underbrace{\cos(u)}_{\mathrm S} - \int \underbrace{e^u}_{\mathrm I}\, \underbrace{-\sin(u)}_{\mathrm D}\, \mathrm du\right) \\ &= e^u \sin(u) - \left(e^u \cos(u) + \int e^u \sin(u) \, \mathrm du\right) \\ &= e^u \sin(u) - e^u \cos(u) - \int e^u \sin(u)\, \mathrm du \\&= e^u (\sin(u) - \cos(u)) - \int e^u \sin(u)\, \mathrm du.\end{aligned}$

By algebruh,

$\displaystyle \begin{aligned} \int e^u \sin(u)\, \mathrm du + \int e^u \sin(u)\, \mathrm du &= e^u (\sin(u) - \cos(u)) \\ 2 \int e^u \sin(u)\, \mathrm du &= e^u (\sin(u) - \cos(u)) \\ \int e^u \sin(u)\, \mathrm du &= \frac 12 e^u (\sin(u) - \cos(u)) + C\\ \int \sin(\ln \theta)\, \mathrm d\theta &= \frac 12 \theta (\sin(\ln \theta) - \cos(\ln \theta)) + C.\end{aligned}$

Problem 6. Evaluate the integral $\displaystyle \int \left(\frac{1}{36 - x^2} + \frac{1}{\sqrt{36-x^2}} \right)\, \mathrm dx$ .

(Click for Solution)

Solution. By the linearity of integration,

$\displaystyle \begin{aligned} \int \left(\frac{1}{36 - x^2} + \frac{1}{\sqrt{36 -x^2}} \right)\, \mathrm dx &= \int \frac{1}{36 - x^2} \, \mathrm dx + \int \frac{1}{ \sqrt{36-x^2} } \, \mathrm dx \\ &= \int \frac{1}{6^2 - x^2} \, \mathrm dx + \int \frac{1}{ \sqrt{6^2-x^2} } \, \mathrm dx \\ &= \frac 1{2 \cdot 6} \ln \left| \frac{6+x}{6-x} \right| + \sin^{-1} \left( \frac x6 \right) + C\\ &= \frac 1{12} \ln \left| \frac{6+x}{6-x} \right| + \sin^{-1} \left( \frac x6 \right) + C. \end{aligned}$

Problem 7. Evaluate the integral $\displaystyle \int_0^2 x \sin(x^2 + 5)\, \mathrm dx$ .

(Click for Solution)

Solution. Making the substitution $u = x^2 + 5$ yields

$\displaystyle \mathrm du = 2x\, \mathrm dx,\quad u(0) = 5, \quad u(2) = 9.$

Therefore,

$\displaystyle \begin{aligned} \int_0^2 x \sin(x^2 + 5)\, \mathrm dx &= \frac 12 \int_0^2 \sin(x^2 + 5) \cdot 2x\, \mathrm dx \\ &= \frac 1{2}\int_5^9 \sin(u)\, \mathrm du \\ &= \frac 12 [-\cos(u)]_5^9 \\ &= \frac 12 (-\cos(9) - (-\cos(5))) \\ &= \frac 12 (\cos(5) - \cos(9)) \approx 0.597, \end{aligned}$

when approximated to 3 decimal places.

Problem 8. Evaluate the integral $\displaystyle \int \cot^3 x\, \mathrm dx$ .

(Click for Solution)

Solution. Recall that $\cot(x) = \tan(\pi/2 - x)$ . Making the substitution $u = \pi/2-x \Rightarrow \mathrm du = -\mathrm dx$ then using the result in Problem 4,

$\displaystyle \begin{aligned} \int \cot^3 x\, \mathrm dx &= \int \tan^3 (\pi/2-x)\, \mathrm dx \\ &= \int \tan^3(u) \cdot -1\, \mathrm du \\ &= -\int \tan^3(u)\, \mathrm du \\ &= -\left( \frac 12 \tan^2 u + \ln|{\cos u}| \right) + C \\ &= -\left( \frac 12 \tan^2 (\pi/2-x) + \ln|{\cos (\pi/2-x)}| \right) + C \\ &= -\left( \frac 12 \cot^2 x + \ln|{\sin x}| \right) + C \\ &= - \frac 12 \cot^2 x - \ln|{\sin x}| + C \end{aligned}$

Problem 9. Evaluate the integral $\displaystyle \int e^{2\theta} \sin (e^\theta) \, \mathrm d\theta$ .

(Click for Solution)

Solution. Making the substitution $\displaystyle u = e^{\theta} \Rightarrow \mathrm d\theta = \frac 1{u}\, \mathrm du$ then integrating by parts,

$\displaystyle \begin{aligned} \int e^{2\theta} \sin(e^\theta) \, \mathrm d\theta &= \int u^2 \sin(u)\cdot \frac 1{u}\, \mathrm du \\ &= \int u \sin(u)\, \mathrm du\\ &= \underbrace{-\cos(u)}_{\mathrm I} \underbrace{u}_{\mathrm S} - \int \underbrace{-\cos(u)}_{\mathrm I}\underbrace{1}_{\mathrm D}\, \mathrm du \\ &= -u \cos u + \int \cos u\, \mathrm du\\ &= -u \cos u + \sin u + C\\ &= -e^\theta \cos(e^\theta) + \sin(e^\theta) + C. \end{aligned}$

Problem 10. Given that

$f(x) = \begin{cases} 1 - |x|, & |x| \leq 1, \\ 0, &\text{otherwise},\end{cases}$

evaluate $\displaystyle \int_0^{2} f(x^2-1)\, \mathrm dx$ .

(Click for Solution)

Solution. By the definition of $f$ ,

$f(x^2 - 1) = \begin{cases} 1 - |x^2 - 1|, & |x^2 - 1| \leq 1, \\ 0, &\text{otherwise}.\end{cases}$

Parsing the inequality $|x^2 - 1| \leq 1 \iff |x| \leq \sqrt 2$ ,

$f(x^2 - 1) = \begin{cases} 1 - |x^2 - 1|, & |x| \leq \sqrt 2, \\ 0, &\text{otherwise}.\end{cases}$

By definition of the absolute value,

$|x^2 - 1| = \begin{cases} x^2 - 1, & |x| \geq 1, \\ 1 - x^2, &|x| \leq 1.\end{cases}$

Parsing the inequalities again,

$f(x^2 - 1) = \begin{cases} x^2, & 0 \leq |x| \leq 1, \\ 2-x^2 , & 1 \leq |x| \leq \sqrt 2, \\ 0, &\text{otherwise}.\end{cases}$

Therefore, we can evaluate the integral as

$\begin{aligned} \displaystyle \int_0^{2} f(x^2-1)\, \mathrm dx &= \int_0^1 f(x^2 - 1)\, \mathrm dx + \int_1^{\sqrt 2} f(x^2 - 1)\, \mathrm dx + \int_{\sqrt 2}^{2} 0\, \mathrm dx \\ &= \int_0^1 x^2\, \mathrm dx + \int_1^{\sqrt 2} (2 - x^2)\, \mathrm dx+ 0 \\ &= \left[ \frac{x^3}{3}\right]_0^1 + \left[ 2x - \frac{x^3}{3} \right]_1^{\sqrt 2} \\ &= \left( \frac 13 - 0 \right) + \left( 2\sqrt 2 - \frac{2\sqrt 2}{3} \right) \\ &= \frac 13 + \frac{4}{3}\sqrt 2. \end{aligned}$

—Joel Kindiak, 20 Jan 25, 0925H

January 20, 2025

ai, Calculus, math, mathematics, physics
When Integrals Approximate One

In the following problems, let $f$ be a function that is continuous on $\mathbb R$ .

Problem 1. Given that $\displaystyle \lim_{x \to \infty} \frac{1}{x} \int_0^x f(t)\, \mathrm dt = 1$ , prove that $\displaystyle \lim_{x \to \infty} f(x) = 1$ if the limit exists.

(Click for Solution)

Solution. We first claim that $\displaystyle \int_0^x f(t)\, \mathrm dt \to \infty$ . Fix $M > 0$ . By hypothesis, for any $\epsilon > 0$ , there exists $K > 0$ such that $x > K$ implies

$\displaystyle \left|\frac{1}{x} \int_0^x f(t)\, \mathrm dt - 1\right| < \epsilon \quad \Rightarrow \quad \int_0^x f(t)\, \mathrm dt > (1 - \epsilon)x.$

Setting $\epsilon = 1/2$ , choose $N := \max\{K, 2M\}$ so that $x > N$ implies

$\displaystyle \int_0^x f(t)\, \mathrm dt > (1-1/2) \cdot 2M = M.$

Therefore $\displaystyle \int_0^x f(t)\, \mathrm dt \to \infty$ . By L’Hôpital’s rule and the fundamental theorem of calculus,

$\displaystyle \lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{\frac{\mathrm d}{\mathrm dx} \int_0^x f(t)\, \mathrm dt}{\frac{\mathrm d}{\mathrm dx} (x)} = \lim_{x \to \infty} \frac{\int_0^x f(t)\, \mathrm dt}{x} = 1.$

But what if we don’t know that the limit exists? The added feature of $f$ being increasing helps.

Problem 2. Given that $\displaystyle \lim_{x \to \infty} \frac{1}{x} \int_0^x f(t)\, \mathrm dt = 1$ , prove that $\displaystyle \lim_{x \to \infty} f(x) = 1$ if $f$ is increasing.

(Click for Solution)

Solution. Fix $\epsilon > 0$ . By the given condition, for any $k > 0$ to be chosen, there exists $K > 0$ such that for $x > K$ ,

$\displaystyle -k \epsilon < \frac{1}{x} \int_0^x f(t)\, \mathrm dt - 1 < k \epsilon.$

We claim that $f(x) \leq 1$ for any $x$ . Suppose $f(N) > 1$ for some $N$ . Define $\epsilon_0 := (f(N)-1)/2$ so that for $x > N$ , $f(x) \geq f(N) > 1 + \epsilon_0$ . Then

$\displaystyle \begin{aligned} \frac{1}{x} \int_0^x f(t)\, \mathrm dt &= \frac{1}{x} \int_0^N f(t)\, \mathrm dt + \frac{1}{x} \int_N^x f(t)\, \mathrm dt \\ &\geq \frac{1}{x} \int_0^N f(t)\, \mathrm dt + (1+\epsilon_0) (1-N/x) \\ &\geq \frac{N}{x} \cdot (f(0) - (1+\epsilon_0)) + (1+\epsilon_0) . \end{aligned}$

Taking $x \to \infty$ yields

$\displaystyle 1 = \lim_{x \to \infty} \frac{1}{x} \int_0^x f(t)\, \mathrm dt \geq 0 \cdot (f(0) - (1+\epsilon_0)) + (1+\epsilon_0) > 1,$

a contradiction. Therefore, $f(x) \leq 1$ for any $x$ . On the other hand, apply the mean value theorem and the fundamental theorem of calculus to find $c \in (0, x)$ such that

$\displaystyle f(c) = \frac{1}{x} \int_0^x f(t)\, \mathrm dt > 1 - k\epsilon.$

Setting $k = 1$ and using the fact that $f$ is increasing,

$1 - \epsilon < f(c) \leq f(x) \leq 1 < 1 +\epsilon\quad \Rightarrow \quad |f(x) - 1| < \epsilon.$

It turns out that we can we relax our hypothesis. However, this will require more rudimentary ideas.

Problem 3. Given that $\displaystyle \lim_{x \to \infty} \frac{1}{x} \int_0^x f(t)\, \mathrm dt = 1$ , prove that $\displaystyle \lim_{x \to \infty} f(x) = 1$ .

(Click for Solution)

Solution. Fix $\epsilon > 0$ . By the given condition, for any $k > 0$ to be chosen, there exists $K > 0$ such that for $x > K$ ,

$\displaystyle -k \epsilon < \frac{1}{x} \int_0^x f(t)\, \mathrm dt - 1 < k \epsilon.$

By algebra, since $x > K > 0$ ,

$\displaystyle -k \epsilon x < \int_0^x f(t)\, \mathrm dt - x < k \epsilon x.$

In particular, the inequality holds replacing $x$ with $x+h$ for $|h| < (x-K)/2$ so that

$\displaystyle -k \epsilon (x+h) < \int_0^{x+h} f(t)\, \mathrm dt - (x+h) < k \epsilon (x+h).$

Subtracting the inequalities then dividing by $h$ ,

$\displaystyle -k \epsilon < \frac 1h \int_x^{x+h} f(t)\, \mathrm dt - 1 < k \epsilon.$

Taking $h \to 0$ and applying the fundamental theorem of calculus,

$\displaystyle -k \epsilon \leq f(x)-1 \leq k \epsilon.$

Setting $k = 1/2$ ,

$\displaystyle -\epsilon < -k \epsilon \leq f(x)-1 \leq k \epsilon < \epsilon\quad \Rightarrow \quad |f(x) - 1| < \epsilon.$

By the $\epsilon$ – $K$ definition for limits, $\displaystyle \lim_{x \to \infty} f(x) = 1$ .

—Joel Kindiak, 26 Nov 24, 0110H

November 26, 2024

Calculus, machine-learning, math, mathematics, physics
The Classic Secant Integral

No course in calculus is complete without the famous problem of integrating $\sec^3$ .

Problem 1. Evaluate $\displaystyle \int \sec^3(x)\, \mathrm dx$ .

(Click for Solution)

Solution. Write $\sec^3(x) = \sec^2(x) \sec(x)$ . Since $\sec^2(x)$ integrates to $\tan(x)$ , we integrate by parts to obtain

$\displaystyle \begin{aligned} \int \sec^3(x)\, \mathrm dx &= \int \sec^2(x) \sec(x)\, \mathrm dx \\ &= \underbrace{\tan(x)}_{\text I} \underbrace{\sec(x)}_{\text S} - \int \underbrace{\tan(x)}_{\text I} \cdot \underbrace{\sec(x)\tan(x)}_{\text D}\,\mathrm dx. \end{aligned}$

Writing $\tan^2(x) = \sec^2(x) - 1$ ,

$\displaystyle \begin{aligned} \int \sec^3(x)\, \mathrm dx &= \tan(x) \sec(x) - \int \tan(x) \cdot \sec(x)\tan(x)\,\mathrm dx \\ &= \tan(x) \sec(x) - \int (\sec^2(x) - 1) \cdot \sec(x)\,\mathrm dx \\ &= \tan(x) \sec(x) - \int \sec^3(x)\, \mathrm dx + \int \sec(x)\,\mathrm dx. \end{aligned}$

By algebra and recalling our result for the integral of $\sec$ ,

$\displaystyle \begin{aligned} \int \sec^3(x)\, \mathrm dx &= \frac 12 \tan(x) \sec(x) + \frac 12 \int \sec(x)\,\mathrm dx \\ &= \frac 12 \tan(x) \sec(x) + \frac 12 \ln|{\sec(x) + \tan(x)}| + C. \end{aligned}$

—Joel Kindiak, 28 Oct 24, 1042H

October 28, 2024

ai, geometry, machine-learning, math, mathematics

Category: Exercises

Limits

Differentiation

Integration