MATH102 Integration Comprehensive Study Notes

Course Overview and Guidelines

• Course Details:

  • Course Name: MATHEMATICS 2 (MATH102)
  • Intake: January 2026
  • Subject: Integration Comprehensive Lecture Notes
  • Institution: Manica University, School of Basic Sciences, Lusaka, Zambia
  • Author / Dean: Mr Robert Dozva (MSc Mathematics), Dean of the School of Basic Sciences

• How to Use These Notes:

  • Coverage: Complete integration component of MATH102.
  • Structural Pattern: Each section presents a rule in a shaded box, followed by worked examples demonstrating the rule, and closes with a set of practice problems.
  • Solutions: Answers to all practice problems and full worked solutions to selected problems are provided in Appendix B.
  • Study Strategy: Work through all examples actively with pen and paper. Reading a solution is insufficient; following and executing a solution prepares for assessments. Every practice problem should be attempted prior to consulting the answers.

Introduction: Reversing Differentiation

• The Central Question of Integration:

  • Given a function $f(x)$, can we find a function $F(x)$ whose derivative is $f(x)$? That is, can we find $F(x)$ such that:     $\frac{\mathrm{d}F}{\mathrm{d}x} = f(x)$
  • Definition: Such a function $F(x)$ is called an antiderivative (or primitive) of $f(x)$.

• The Constant of Integration ($C$):

  • Antiderivatives are never unique because the derivative of any constant is zero.
  • Example: If $F(x) = x^2$, then $F'(x) = 2x$. Similarly:     $\frac{\mathrm{d}}{\mathrm{d}x}\left(x^2\right) = 2x$$\frac{\mathrm{d}}{\mathrm{d}x}\left(x^2 + 5\right) = 2x$$\frac{\mathrm{d}}{\mathrm{d}x}\left(x^2 - \frac{3}{4}\right) = 2x$
  • Geometrical Interpretation: Antiderivatives form families of curves that differ only by a vertical shift. Every member of the family $y = x^2 + C$ possesses the exact same gradient $2x$ at any given $x$.

• Notation of the Indefinite Integral:

  • If $F'(x) = f(x)$, the indefinite integral is written as:     $\int f(x)\,\mathrm{d}x = F(x) + C$
  • Verbal Reading: "The integral of $f$ of $x$, with respect to $x$".
  • Components:
  • $\int$ is the integral sign (a stretched letter S, standing for sum).
  • $f(x)$ is the integrand (the function being integrated).
  • $\mathrm{d}x$ indicates the variable of integration.
  • $C$ is the constant of integration.
  • Terminology: Called an indefinite integral because of the presence of the unknown constant $C$.

• Common Error Warning:

  • Forgetting $+C$ is the single most common mistake in this course and will result in lost marks in every test.
  • Always write $+C$ down immediately upon completing the integration step, prior to simplifying.
  • The only exception where $+C$ is omitted is during the evaluation of a definite integral.

• Linearity Properties of the Integral:

  • For any constant $k$ and any integrable functions $f(x)$ and $g(x)$:
  • Constant Multiple Rule:       $\int k f(x)\,\mathrm{d}x = k \int f(x)\,\mathrm{d}x$
  • Sum and Difference Rule:       $\int [f(x) \pm g(x)]\,\mathrm{d}x = \int f(x)\,\mathrm{d}x \pm \int g(x)\,\mathrm{d}x$
  • Meaning: Integration can be performed term by term, and constant multipliers slide outside the integral sign.
  • Product/Quotient Warning: There is no direct product rule for integration:     $\int f(x)g(x)\,\mathrm{d}x \neq \left(\int f(x)\,\mathrm{d}x\right)\left(\int g(x)\,\mathrm{d}x\right)$

• Checking Answers by Differentiation:

  • Differentiating the calculated antiderivative must reproduce the original integrand.
  • Example 1.1: Verify that:     $\int (3x^2 + 4x)\,\mathrm{d}x = x^3 + 2x^2 + C$
  • Solution: Differentiate the right-hand side:       $\frac{\mathrm{d}}{\mathrm{d}x}\left(x^3 + 2x^2 + C\right) = 3x^2 + 4x + 0 = 3x^2 + 4x$
  • Since this matches the integrand, the result is verified.

• Practice Problems (Section 1):

  1. Write down three different antiderivatives of $f(x) = 6x^5$.
  2. By differentiating, verify that:      $\int (2x + 7)\,\mathrm{d}x = x^2 + 7x + C$
  3. By differentiating, decide whether the following is correct:      $\int x \cos(x)\,\mathrm{d}x = \frac{1}{2}x^2 \sin(x) + C$
  4. Explain in one sentence why:      $\int 5\,\mathrm{d}x = 5x + C$
  5. If $F'(x) = g(x)$ and $F(2) = 7$, is $F$ the only antiderivative of $g$ with this property? Explain.

• Solutions to Section 1 Practice Problems:

  1. $x^6$, $x^6 + 1$, $x^6 - 3$ (or any three functions differing by constants).
  2. Verification:      $\frac{\mathrm{d}}{\mathrm{d}x}\left(x^2 + 7x + C\right) = 2x + 7$
  3. Incorrect. Differentiating $\frac{1}{2}x^2 \sin(x) + C$ using the product rule yields $x \sin(x) + \frac{1}{2}x^2 \cos(x)$, which does not equal $x \cos(x)$.
  4. Correct because:      $\frac{\mathrm{d}}{\mathrm{d}x}(5x) = 5$
  5. Yes. Two antiderivatives of the same function differ only by a constant, and the initial condition $F(2) = 7$ uniquely fixes that constant.

The Power Rule

• The Power Rule Statement:

  • For any rational number $n \neq -1$:     $\int x^n\,\mathrm{d}x = \frac{x^{n+1}}{n+1} + C$
  • Verbal Rule: Add one to the power, then divide by the new power.

• Explanation of Why $n \neq -1$:

  • If $n = -1$, the power rule formula yields division by $n + 1 = 0$, which is undefined.
  • Special Rule for $n = -1$:     $\int x^{-1}\,\mathrm{d}x = \int \frac{1}{x}\,\mathrm{d}x = \ln|x| + C$
  • Absolute Value Necessity: $\ln(x)$ is undefined for $x < 0$, whereas $\frac{1}{x}$ is well-defined for all $x \neq 0$. Differentiating gives:     $\frac{\mathrm{d}}{\mathrm{d}x}\ln|x| = \frac{1}{x}\quad \text{for all } x \neq 0$

• Worked Examples:

  • Example 2.1: Find $\int x^4\,\mathrm{d}x$.
  • Solution: Here $n = 4$, so $n + 1 = 5$:       $\int x^4\,\mathrm{d}x = \frac{x^5}{5} + C$
  • Example 2.2: Find $\int \frac{1}{x^3}\,\mathrm{d}x$.
  • Solution: Rewrite integrand as a power first:       $\int \frac{1}{x^3}\,\mathrm{d}x = \int x^{-3}\,\mathrm{d}x = \frac{x^{-2}}{-2} + C = -\frac{1}{2x^2} + C$
  • Example 2.3: Find $\int \sqrt{x}\,\mathrm{d}x$.
  • Solution: Rewrite root as fractional power:       $\int \sqrt{x}\,\mathrm{d}x = \int x^{1/2}\,\mathrm{d}x = \frac{x^{3/2}}{3/2} + C = \frac{2}{3}x^{3/2} + C = \frac{2}{3}\sqrt{x^3} + C$
  • Note: Dividing by $\frac{3}{2}$ is equivalent to multiplying by $\frac{2}{3}$.
  • Example 2.4: Find $\int \left(4x^3 - 6x^2 + 2\sqrt{x} - 9\right)\,\mathrm{d}x$.
  • Solution: Rewrite roots as powers and integrate term by term:       $\int \left(4x^3 - 6x^2 + 2x^{-1/2} - 9\right)\,\mathrm{d}x = 4 \times \frac{x^4}{4} - 6 \times \frac{x^3}{3} + 2 \times \frac{x^{1/2}}{1/2} - 9x + C$$= x^4 - 2x^3 + 4\sqrt{x} - 9x + C$
  • Note: $\int 9\,\mathrm{d}x = 9x$, because $9 = 9x^0$ and the power rule gives $9 \times \frac{x^1}{1}$.

• Expanding Before Integrating:

  • Integrands involving products or quotients cannot be integrated directly factor-by-factor; expand or divide first.
  • Example 2.5: Find $\int x^2 \left(3x - \frac{5}{x}\right)\,\mathrm{d}x$.
  • Solution: Expand the expression:       $\int (3x^3 - 5x)\,\mathrm{d}x = \frac{3x^4}{4} - \frac{5x^2}{2} + C$
  • Example 2.6: Find $\int \frac{x^3 + 2x}{x}\,\mathrm{d}x$.
  • Solution: Divide each term in the numerator by $x$:       $\int (x^2 + 2)\,\mathrm{d}x = \frac{x^3}{3} + 2x + C$

• Powers of a Linear Expression (Linear Substitution Shortcut):

  • For constants $a \neq 0$, $b$, and rational $n \neq -1$:     $\int (ax + b)^n\,\mathrm{d}x = \frac{(ax + b)^{n+1}}{a(n + 1)} + C$$\int \frac{1}{ax + b}\,\mathrm{d}x = \frac{1}{a}\ln|ax + b| + C$
  • Justification: The extra factor $\frac{1}{a}$ compensates for the chain rule multiplier $a$ when differentiating back.
  • Example 2.7: Find $\int (5x - 2)^7\,\mathrm{d}x$.
  • Solution: Here $a = 5$, $n = 7$:       $\int (5x - 2)^7\,\mathrm{d}x = \frac{(5x - 2)^8}{5 \times 8} + C = \frac{(5x - 2)^8}{40} + C$
  • Verification: Differentiating yields:       $\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{(5x - 2)^8}{40}\right) = \frac{8(5x - 2)^7 \times 5}{40} = (5x - 2)^7$
  • Example 2.8: Find $\int \frac{3}{2x + 9}\,\mathrm{d}x$.
  • Solution:       $3 \int \frac{1}{2x + 9}\,\mathrm{d}x = \frac{3}{2}\ln|2x + 9| + C$

• Practice Problems (Section 2):

  1. $\int x^7\,\mathrm{d}x$
  2. $\int 12x^5\,\mathrm{d}x$
  3. $\int \frac{1}{x^5}\,\mathrm{d}x$
  4. $\int \sqrt[3]{x}\,\mathrm{d}x$
  5. $\int (x^3 - 4x + 6)\,\mathrm{d}x$
  6. $\int \left(2x^4 + \frac{3}{x}\right)\,\mathrm{d}x$
  7. $\int \frac{5}{x^2}\,\mathrm{d}x$
  8. $\int \left(\sqrt{x} + \frac{1}{\sqrt{x}}\right)\,\mathrm{d}x$
  9. $\int x(x^2 - 3)\,\mathrm{d}x$
  10. $\int \frac{2x^4 - x^2}{x^2}\,\mathrm{d}x$
  11. $\int (3x + 1)^6\,\mathrm{d}x$
  12. $\int \frac{1}{(4x - 7)^3}\,\mathrm{d}x$
  13. $\int \sqrt{2x + 5}\,\mathrm{d}x$
  14. $\int \frac{8}{5x - 1}\,\mathrm{d}x$
  15. $\int (x + 2)(x - 3)\,\mathrm{d}x$
  16. A curve passes through $(1, 4)$ and has gradient $\frac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 - 2$. Find its equation.

• Solutions to Section 2 Practice Problems:

  1. $\frac{x^8}{8} + C$
  2. $2x^6 + C$
  3. $-\frac{1}{4x^4} + C$
  4. $\frac{3}{4}x^{4/3} + C$
  5. $\frac{x^4}{4} - 2x^2 + 6x + C$
  6. $\frac{2x^5}{5} + 3\ln|x| + C$
  7. $-\frac{5}{x} + C$
  8. $\frac{2}{3}x^{3/2} + 2\sqrt{x} + C$
  9. $\frac{x^4}{4} - \frac{3x^2}{2} + C$
  10. $\frac{2x^3}{3} - x + C$
  11. $\frac{(3x + 1)^7}{21} + C$
  12. $-\frac{1}{8(4x - 7)^2} + C$
  13. $\frac{1}{3}(2x + 5)^{3/2} + C$
  14. $\frac{8}{5}\ln|5x - 1| + C$
  15. $\frac{x^3}{3} - \frac{x^2}{2} - 6x + C$
  16. Solution: Integrate gradient function $y = \int (3x^2 - 2)\,\mathrm{d}x = x^3 - 2x + C$. Substitute $(1, 4)$: $4 = 1^3 - 2(1) + C \implies 4 = -1 + C \implies C = 5$. Curve equation: $y = x^3 - 2x + 5$.

Integration of Exponential Functions

• Exponential Formulas:   $\int e^x\,\mathrm{d}x = e^x + C$$\int e^{ax+b}\,\mathrm{d}x = \frac{1}{a}e^{ax+b} + C \quad (a \neq 0)$$\int a^x\,\mathrm{d}x = \frac{a^x}{\ln(a)} + C \quad (a > 0, a \neq 1)$

• Worked Examples:

  • Example 3.1: Find $\int e^{4x}\,\mathrm{d}x$.
  • Solution: With $a = 4$:       $\int e^{4x}\,\mathrm{d}x = \frac{1}{4}e^{4x} + C$
  • Verification: $\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{1}{4}e^{4x}\right) = \frac{1}{4} \times 4e^{4x} = e^{4x}$.
  • Example 3.2: Find $\int \left(e^{-3x} + 5e^x\right)\,\mathrm{d}x$.
  • Solution: Here $a = -3$ in the first term:       $\int e^{-3x}\,\mathrm{d}x + 5 \int e^x\,\mathrm{d}x = -\frac{1}{3}e^{-3x} + 5e^x + C$
  • Example 3.3: Find $\int 2^x\,\mathrm{d}x$.
  • Solution: Apply base-a rule with $a = 2$:       $\int 2^x\,\mathrm{d}x = \frac{2^x}{\ln(2)} + C$
  • Example 3.4: Find $\int \frac{e^{2x} + e^x}{e^x}\,\mathrm{d}x$.
  • Solution: Simplify first using index laws:       $\int (e^x + 1)\,\mathrm{d}x = e^x + x + C$

• Sign Precaution Box:

  • A very common error is writing $\int e^{-x}\,\mathrm{d}x = e^{-x} + C$. This is incorrect.
  • Here $a = -1$, so:     $\int e^{-x}\,\mathrm{d}x = -e^{-x} + C$

• Practice Problems (Section 3):

  1. $\int e^{7x}\,\mathrm{d}x$
  2. $\int e^{-5x}\,\mathrm{d}x$
  3. $\int 3e^{2x}\,\mathrm{d}x$
  4. $\int e^{3x+1}\,\mathrm{d}x$
  5. $\int \left(e^x + x^2\right)\,\mathrm{d}x$
  6. $\int \left(4e^{-2x} - e^x\right)\,\mathrm{d}x$
  7. $\int 5^x\,\mathrm{d}x$
  8. $\int e^x\left(e^x + 3\right)\,\mathrm{d}x$
  9. $\int \frac{e^{3x}}{e^x}\,\mathrm{d}x$
  10. $\int \left(e^{x/2} + \frac{1}{x}\right)\,\mathrm{d}x$

• Solutions to Section 3 Practice Problems:

  1. $\frac{e^{7x}}{7} + C$
  2. $-\frac{e^{-5x}}{5} + C$
  3. $\frac{3}{2}e^{2x} + C$
  4. $\frac{1}{3}e^{3x+1} + C$
  5. $e^x + \frac{x^3}{3} + C$
  6. $-2e^{-2x} - e^x + C$
  7. $\frac{5^x}{\ln(5)} + C$
  8. $\frac{e^{2x}}{2} + 3e^x + C$
  9. $\frac{e^{2x}}{2} + C$
  10. $2e^{x/2} + \ln|x| + C$

Integration of Trigonometric Functions

• Standard Trigonometric Integrals:   $\int \sin(x)\,\mathrm{d}x = -\cos(x) + C$$\int \cos(x)\,\mathrm{d}x = \sin(x) + C$$\int \sec^2(x)\,\mathrm{d}x = \tan(x) + C$$\int \csc^2(x)\,\mathrm{d}x = -\cot(x) + C$$\int \sec(x)\tan(x)\,\mathrm{d}x = \sec(x) + C$$\int \csc(x)\cot(x)\,\mathrm{d}x = -\csc(x) + C$

• Linear Arguments ($a \neq 0$):   $\int \sin(ax + b)\,\mathrm{d}x = -\frac{1}{a}\cos(ax + b) + C$$\int \cos(ax + b)\,\mathrm{d}x = \frac{1}{a}\sin(ax + b) + C$

• Rule for Remembering Signs:

  • Going from $\sin(x)$ and $\cos(x)$ by differentiation, the minus sign attaches to $\cos(x)$: $(\cos(x))' = -\sin(x)$.
  • Going backwards by integration, the minus sign attaches to $\sin(x)$: $\int \sin(x)\,\mathrm{d}x = -\cos(x) + C$

• Worked Examples:

  • Example 4.1: Find $\int (3\sin(x) - 2\cos(x))\,\mathrm{d}x$.
  • Solution:       $3 \int \sin(x)\,\mathrm{d}x - 2 \int \cos(x)\,\mathrm{d}x = -3\cos(x) - 2\sin(x) + C$
  • Example 4.2: Find $\int \cos(5x)\,\mathrm{d}x$.
  • Solution: With $a = 5$:       $\int \cos(5x)\,\mathrm{d}x = \frac{1}{5}\sin(5x) + C$
  • Example 4.3: Find $\int \sin\left(2x + \frac{\pi}{3}\right)\,\mathrm{d}x$.
  • Solution: With $a = 2$, $b = \frac{\pi}{3}$:       $-\frac{1}{2}\cos\left(2x + \frac{\pi}{3}\right) + C$

• Using Identities First:

  • Three Identities Worth Memorising:     $\sin^2(x) = \frac{1 - \cos(2x)}{2}$$\cos^2(x) = \frac{1 + \cos(2x)}{2}$$\sec^2(x) = 1 + \tan^2(x)$
  • Example 4.4: Find $\int \sin^2(x)\,\mathrm{d}x$.
  • Solution: Convert using double-angle identity:       $\int \sin^2(x)\,\mathrm{d}x = \int \frac{1 - \cos(2x)}{2}\,\mathrm{d}x = \frac{1}{2} \int (1 - \cos(2x))\,\mathrm{d}x$$= \frac{1}{2}\left(x - \frac{1}{2}\sin(2x)\right) + C = \frac{x}{2} - \frac{\sin(2x)}{4} + C$
  • Example 4.5: Find $\int \tan^2(x)\,\mathrm{d}x$.
  • Solution: Rearrange Pythagorean identity $\tan^2(x) = \sec^2(x) - 1$:       $\int (\sec^2(x) - 1)\,\mathrm{d}x = \tan(x) - x + C$

• Practice Problems (Section 4):

  1. $\int 5\cos(x)\,\mathrm{d}x$
  2. $\int (\sin(x) + \cos(x))\,\mathrm{d}x$
  3. $\int \sin(4x)\,\mathrm{d}x$
  4. $\int \cos(3x - 1)\,\mathrm{d}x$
  5. $\int 2\sec^2(x)\,\mathrm{d}x$
  6. $\int (\csc^2(x) - \sin(x))\,\mathrm{d}x$
  7. $\int \cos^2(x)\,\mathrm{d}x$
  8. $\int \sec(x)(\sec(x) + \tan(x))\,\mathrm{d}x$
  9. $\int (1 + \tan^2(x))\,\mathrm{d}x$
  10. $\int \sin\left(\frac{x}{2}\right)\,\mathrm{d}x$

• Solutions to Section 4 Practice Problems:

  1. $5\sin(x) + C$
  2. $\sin(x) - \cos(x) + C$
  3. $-\frac{\cos(4x)}{4} + C$
  4. $\frac{\sin(3x - 1)}{3} + C$
  5. $2\tan(x) + C$
  6. $-\cot(x) + \cos(x) + C$
  7. $\frac{x}{2} + \frac{\sin(2x)}{4} + C$
  8. $\tan(x) + \sec(x) + C$
  9. $\tan(x) + C$
  10. $-2\cos\left(\frac{x}{2}\right) + C$

Integration of Products: Integration by Parts

• Derivation of Formula:

  • Start from product rule of differentiation for two functions $u(x)$ and $v(x)$:     $\frac{\mathrm{d}}{\mathrm{d}x}(uv) = u\frac{\mathrm{d}v}{\mathrm{d}x} + v\frac{\mathrm{d}u}{\mathrm{d}x}$
  • Integrate both sides with respect to $x$:     $uv = \int u \frac{\mathrm{d}v}{\mathrm{d}x}\,\mathrm{d}x + \int v \frac{\mathrm{d}u}{\mathrm{d}x}\,\mathrm{d}x$
  • Rearrange for the integration by parts formula:     $\int u \frac{\mathrm{d}v}{\mathrm{d}x}\,\mathrm{d}x = uv - \int v \frac{\mathrm{d}u}{\mathrm{d}x}\,\mathrm{d}x \quad \text{or compactly} \quad \int u\,\mathrm{d}v = uv - \int v\,\mathrm{d}u$

• Choosing $u$ and $\mathrm{d}v$: The LIATE Rule:

  • Select $u$ as the function type appearing first in the list:
  • L: Logarithmic ($\ln(x)$)
  • I: Inverse trigonometric ($\arctan(x)$, $\arcsin(x)$)
  • A: Algebraic ($x$, $x^2$, polynomials)
  • T: Trigonometric ($\sin(x)$, $\cos(x)$)
  • E: Exponential ($e^x$)
  • Whatever factor remains becomes $\mathrm{d}v$.
  • Rationale: $u$ gets differentiated (simplifies), $\mathrm{d}v$ gets integrated.

• Worked Examples:

  • Example 5.1: Find $\int x e^x\,\mathrm{d}x$.
  • Solution: Algebraic (A) precedes Exponential (E), so $u = x$, $\frac{\mathrm{d}v}{\mathrm{d}x} = e^x$.       $u = x \implies \frac{\mathrm{d}u}{\mathrm{d}x} = 1$$\frac{\mathrm{d}v}{\mathrm{d}x} = e^x \implies v = e^x$$\int x e^x\,\mathrm{d}x = x e^x - \int e^x \times 1\,\mathrm{d}x = x e^x - e^x + C = e^x(x - 1) + C$
  • Example 5.2: Find $\int x \ln(x)\,\mathrm{d}x$.
  • Solution: Logarithmic (L) precedes Algebraic (A), so $u = \ln(x)$, $\frac{\mathrm{d}v}{\mathrm{d}x} = x$.       $u = \ln(x) \implies \frac{\mathrm{d}u}{\mathrm{d}x} = \frac{1}{x}$$\frac{\mathrm{d}v}{\mathrm{d}x} = x \implies v = \frac{x^2}{2}$$\int x \ln(x)\,\mathrm{d}x = \frac{x^2}{2}\ln(x) - \int \frac{x^2}{2} \times \frac{1}{x}\,\mathrm{d}x = \frac{x^2}{2}\ln(x) - \frac{1}{2}\int x\,\mathrm{d}x$$= \frac{x^2}{2}\ln(x) - \frac{x^2}{4} + C$
  • Example 5.3 (The $\ln(x)$ trick): Find $\int \ln(x)\,\mathrm{d}x$.
  • Solution: Express as $\ln(x) \times 1$. Set $u = \ln(x)$, \frac{\mathrm{d}v}{\mathrm{d}x} = 1$.\n      u = \ln(x) \implies \frac{\mathrm{d}u}{\mathrm{d}x} = \frac{1}{x}\n      \frac{\mathrm{d}v}{\mathrm{d}x} = 1 \implies v = x\n      \int \ln(x)\,\mathrm{d}x = x\ln(x) - \int x \times \frac{1}{x}\,\mathrm{d}x = x\ln(x) - \int 1\,\mathrm{d}x = x\ln(x) - x + C\n\n- Repeated Application and Cyclic Cases:\n - Example 5.4: Find \int x^2 \sin(x)\,\mathrm{d}x.\n - Solution: First application: u = x^2$,$\frac{\mathrm{d}v}{\mathrm{d}x} = \sin(x) \implies \frac{\mathrm{d}u}{\mathrm{d}x} = 2x$,$v = -\cos(x).\n      \int x^2 \sin(x)\,\mathrm{d}x = -x^2 \cos(x) + 2 \int x \cos(x)\,\mathrm{d}x\n - Second application on \int x \cos(x)\,\mathrm{d}x$:$u = x$,$\frac{\mathrm{d}v}{\mathrm{d}x} = \cos(x) \implies \frac{\mathrm{d}u}{\mathrm{d}x} = 1$,$v = \sin(x).\n      \int x \cos(x)\,\mathrm{d}x = x \sin(x) - \int \sin(x)\,\mathrm{d}x = x \sin(x) + \cos(x)\n - Substituting back:\n      \int x^2 \sin(x)\,\mathrm{d}x = -x^2 \cos(x) + 2x \sin(x) + 2\cos(x) + C\n - The Cyclic Case (\int e^x \sin(x)\,\mathrm{d}x):\n - Applying integration by parts twice yields an equation returning to the original integral I:\n      I = \int e^x \sin(x)\,\mathrm{d}x \implies I = e^x \sin(x) - e^x \cos(x) - I\n      2I = e^x(\sin(x) - \cos(x)) \implies I = \frac{e^x(\sin(x) - \cos(x))}{2} + C\n\n- Practice Problems (Section 5):\n 1. \int x e^{2x}\,\mathrm{d}x\n 2. \int x \cos(x)\,\mathrm{d}x\n 3. \int x \sin(3x)\,\mathrm{d}x\n 4. \int x^2 \ln(x)\,\mathrm{d}x\n 5. \int x^2 e^x\,\mathrm{d}x\n 6. \int \ln(3x)\,\mathrm{d}x\n 7. \int x \sec^2(x)\,\mathrm{d}x\n 8. \int e^x \cos(x)\,\mathrm{d}x\n 9. \int (x + 1)e^{-x}\,\mathrm{d}x\n 10. \int \frac{\ln(x)}{x^2}\,\mathrm{d}x\n\n- Solutions to Section 5 Practice Problems:\n 1. \frac{x e^{2x}}{2} - \frac{e^{2x}}{4} + C\n 2. x \sin(x) + \cos(x) + C\n 3. -\frac{x \cos(3x)}{3} + \frac{\sin(3x)}{9} + C\n 4. \frac{x^3}{3}\ln(x) - \frac{x^3}{9} + C\n 5. e^x\left(x^2 - 2x + 2\right) + C\n 6. x\ln(3x) - x + C\n 7. x\tan(x) + \ln|\cos(x)| + C\n 8. \frac{e^x(\sin(x) + \cos(x))}{2} + C\n 9. -e^{-x}(x + 2) + C\n 10. -\frac{\ln(x)}{x} - \frac{1}{x} + C\n\n# Integration by Substitution\n\n- Principle of Substitution:\n - Reverses the chain rule. Applied when integrand contains an inner function g(x)$alongside its derivative$g'(x).\n - Formula: If u = g(x)$, then$\frac{\mathrm{d}u}{\mathrm{d}x} = g'(x) \implies \mathrm{d}u = g'(x)\,\mathrm{d}x, so:\n    \int f(g(x)) g'(x)\,\mathrm{d}x = \int f(u)\,\mathrm{d}u\n\n- 5-Step Method:\n 1. Choose u (typically the inner function, under a root, inside a bracket, or in a denominator).\n 2. Compute \frac{\mathrm{d}u}{\mathrm{d}x}$and rearrange for$\mathrm{d}x\n 3. Substitute into integral ensuring no variable x remains.\n 4. Integrate the resulting expression in terms of u\n 5. Substitute back u = g(x)$to restore variable$x\n\n- Worked Examples:\n - Example 6.1: Find \int 2x\left(x^2 + 1\right)^5\,\mathrm{d}x.\n - Solution: Set u = x^2 + 1 \implies \mathrm{d}u = 2x\,\mathrm{d}x.\n      \int u^5\,\mathrm{d}u = \frac{u^6}{6} + C = \frac{\left(x^2 + 1\right)^6}{6} + C\n - Example 6.2: Find \int x \sqrt{x^2 - 4}\,\mathrm{d}x.\n - Solution: Set u = x^2 - 4 \implies \mathrm{d}u = 2x\,\mathrm{d}x \implies x\,\mathrm{d}x = \frac{1}{2}\,\mathrm{d}u$.       $\frac{1}{2} \int u^{1/2}\,\mathrm{d}u = \frac{1}{2} \times \frac{2}{3}u^{3/2} + C = \frac{1}{3}\left(x^2 - 4\right)^{3/2} + C$
  • Example 6.3: Find $\int \frac{3x^2}{x^3 + 7}\,\mathrm{d}x$.
  • Solution: Set u = x^3 + 7 \implies \mathrm{d}u = 3x^2\,\mathrm{d}x$.\n      \int \frac{1}{u}\,\mathrm{d}u = \ln|u| + C = \ln\left|x^3 + 7\right| + C\n - Logarithmic Pattern Recognition:\n    \int \frac{f'(x)}{f(x)}\,\mathrm{d}x = \ln|f(x)| + C\n - Example: \int \tan(x)\,\mathrm{d}x = \int \frac{\sin(x)}{\cos(x)}\,\mathrm{d}x = -\ln|\cos(x)| + C = \ln|\sec(x)| + C\n - Example 6.4: Find \int \sin^4(x)\cos(x)\,\mathrm{d}x.\n - Solution: Set u = \sin(x) \implies \mathrm{d}u = \cos(x)\,\mathrm{d}x$.       $\int u^4\,\mathrm{d}u = \frac{u^5}{5} + C = \frac{\sin^5(x)}{5} + C$
  • Example 6.5: Find $\int x e^{x^2}\,\mathrm{d}x$.
  • Solution: Set u = x^2 \implies \mathrm{d}u = 2x\,\mathrm{d}x \implies x\,\mathrm{d}x = \frac{1}{2}\,\mathrm{d}u$.\n      \frac{1}{2} \int e^u\,\mathrm{d}u = \frac{1}{2}e^u + C = \frac{1}{2}e^{x^2} + C\n\n- When Substitution Fails Warning:\n - Substitution works only if remaining factor is a constant multiple of \mathrm{d}u.\n - In \int e^{x^2}\,\mathrm{d}x$, no factor of$x$exists to absorb into$\mathrm{d}u; it cannot be expressed in terms of elementary functions.\n\n- Substitution in Definite Integrals:\n - Method: Change limits of integration to corresponding u$-values. Never revert back to$x$.
  • Example 6.6: Evaluate $\int_{0}^{2} 2x\left(x^2 + 1\right)^3\,\mathrm{d}x$
  • Solution: Let u = x^2 + 1 \implies \mathrm{d}u = 2x\,\mathrm{d}x$.\n - Update limits:\n      x = 0 \implies u = 0^2 + 1 = 1\n      x = 2 \implies u = 2^2 + 1 = 5\n - Evaluate integral:\n      \int_{1}^{5} u^3\,\mathrm{d}u = \left[\frac{u^4}{4}\right]{1}^{5} = \frac{625}{4} - \frac{1}{4} = 156\n\n- Practice Problems (Section 6):\n 1. \int 3x^2\left(x^3 + 2\right)^4\,\mathrm{d}x\n 2. \int 2x e^{x^2}\,\mathrm{d}x\n 3. \int \frac{2x}{x^2 + 5}\,\mathrm{d}x\n 4. \int x\left(4 - x^2\right)^3\,\mathrm{d}x\n 5. \int \cos^3(x)\sin(x)\,\mathrm{d}x\n 6. \int \frac{\ln(x)}{x}\,\mathrm{d}x\n 7. \int \frac{1}{\sqrt{3x + 1}}\,\mathrm{d}x\n 8. \int \sec^2(x)e^{\tan(x)}\,\mathrm{d}x\n 9. \int \frac{x}{\sqrt{x^2 + 9}}\,\mathrm{d}x\n 10. \int \cot(x)\,\mathrm{d}x\n 11. \int{0}^{1} x\left(x^2 + 3\right)^2\,\mathrm{d}x\n 12. \int_{1}^{e} \frac{(\ln(x))^2}{x}\,\mathrm{d}x\n\n- Solutions to Section 6 Practice Problems:\n 1. \frac{\left(x^3 + 2\right)^5}{5} + C\n 2. e^{x^2} + C\n 3. \ln\left(x^2 + 5\right) + C\n 4. -\frac{\left(4 - x^2\right)^4}{8} + C\n 5. -\frac{\cos^4(x)}{4} + C\n 6. \frac{(\ln(x))^2}{2} + C\n 7. \frac{2}{3}\sqrt{3x + 1} + C\n 8. e^{\tan(x)} + C\n 9. \sqrt{x^2 + 9} + C\n 10. \ln|\sin(x)| + C\n 11. Solution: u = x^2 + 3 \implies \mathrm{d}u = 2x\,\mathrm{d}x$. Limits: $x=0 \implies u=3$, x=1 \implies u=4$. \frac{1}{2}\int_3^4 u^2\,\mathrm{d}u = \frac{1}{6}\left[u^3\right]3^4 = \frac{1}{6}(64 - 27) = \frac{37}{6}.\n 12. Solution: u = \ln(x) \implies \mathrm{d}u = \frac{1}{x}\,\mathrm{d}x$. Limits: $x=1 \implies u=0$, x=e \implies u=1$. \int_0^1 u^2\,\mathrm{d}u = \left[\frac{u^3}{3}\right]_0^1 = \frac{1}{3}.\n\n# Integration Giving Inverse Trigonometric Functions\n\n- Standard Forms (a > 0):\n  \int \frac{1}{\sqrt{a^2 - x^2}}\,\mathrm{d}x = \arcsin\left(\frac{x}{a}\right) + C\n  \int \frac{1}{a^2 + x^2}\,\mathrm{d}x = \frac{1}{a}\arctan\left(\frac{x}{a}\right) + C\n  \int \frac{1}{x\sqrt{x^2 - a^2}}\,\mathrm{d}x = \frac{1}{a}\mathrm{arcsec}\left(\frac{x}{a}\right) + C\n\n- Form Identification Rules:\n - Square root of difference a^2 - x^2$in denominator$\implies \arcsin\n - No root, sum a^2 + x^2$in denominator$\implies \arctan\n - Root of difference x^2 - a^2$with extra factor$x$outside$\implies \mathrm{arcsec}\n - Note Asymmetry: \arctan$and$\mathrm{arcsec}$carry multiplier$\frac{1}{a}$outside;$\arcsin does not.\n\n- Worked Examples:\n - Example 7.1: Find \int \frac{1}{\sqrt{9 - x^2}}\,\mathrm{d}x.\n - Solution: Here a^2 = 9 \implies a = 3.\n      \int \frac{1}{\sqrt{9 - x^2}}\,\mathrm{d}x = \arcsin\left(\frac{x}{3}\right) + C\n - Example 7.2: Find \int \frac{1}{16 + x^2}\,\mathrm{d}x.\n - Solution: Here a^2 = 16 \implies a = 4.\n      \int \frac{1}{16 + x^2}\,\mathrm{d}x = \frac{1}{4}\arctan\left(\frac{x}{4}\right) + C\n - Example 7.3: Find \int \frac{5}{4 + 9x^2}\,\mathrm{d}x.\n - Solution: Factor out 9 to ensure coefficient of x^2 is 1:\n      \int \frac{5}{4 + 9x^2}\,\mathrm{d}x = \frac{5}{9} \int \frac{1}{\frac{4}{9} + x^2}\,\mathrm{d}x\n - Here a^2 = \frac{4}{9} \implies a = \frac{2}{3}:\n      = \frac{5}{9} \times \frac{1}{2/3}\arctan\left(\frac{x}{2/3}\right) + C = \frac{5}{6}\arctan\left(\frac{3x}{2}\right) + C\n\n- Completing the Square Method:\n - General quadratics in denominators are integrated by completing the square.\n - Example 7.4: Find \int \frac{1}{x^2 + 6x + 13}\,\mathrm{d}x.\n - Solution: Complete square in denominator:\n      x^2 + 6x + 13 = (x^2 + 6x + 9) + 4 = (x + 3)^2 + 2^2\n - Let u = x + 3$,$a = 2:\n      \int \frac{1}{(x + 3)^2 + 4}\,\mathrm{d}x = \frac{1}{2}\arctan\left(\frac{x + 3}{2}\right) + C\n - Example 7.5: Find \int \frac{1}{\sqrt{5 - 4x - x^2}}\,\mathrm{d}x.\n - Solution: Complete square under root:\n      5 - 4x - x^2 = 5 - (x^2 + 4x) = 5 - ((x + 2)^2 - 4) = 9 - (x + 2)^2\n - Here a = 3:\n      \int \frac{1}{\sqrt{9 - (x + 2)^2}}\,\mathrm{d}x = \arcsin\left(\frac{x + 2}{3}\right) + C\n\n- Practice Problems (Section 7):\n 1. \int \frac{1}{\sqrt{25 - x^2}}\,\mathrm{d}x\n 2. \int \frac{1}{4 + x^2}\,\mathrm{d}x\n 3. \int \frac{3}{1 + x^2}\,\mathrm{d}x\n 4. \int \frac{1}{\sqrt{1 - 4x^2}}\,\mathrm{d}x\n 5. \int \frac{2}{9 + 4x^2}\,\mathrm{d}x\n 6. \int \frac{1}{x^2 + 2x + 5}\,\mathrm{d}x\n 7. \int \frac{1}{\sqrt{8 - 2x - x^2}}\,\mathrm{d}x\n 8. \int{0}^{1} \frac{1}{1 + x^2}\,\mathrm{d}x\n\n- Solutions to Section 7 Practice Problems:\n 1. \arcsin\left(\frac{x}{5}\right) + C\n 2. \frac{1}{2}\arctan\left(\frac{x}{2}\right) + C\n 3. 3\arctan(x) + C\n 4. \frac{1}{2}\arcsin(2x) + C\n 5. \frac{1}{3}\arctan\left(\frac{2x}{3}\right) + C\n 6. \frac{1}{2}\arctan\left(\frac{x + 1}{2}\right) + C\n 7. \arcsin\left(\frac{x + 1}{3}\right) + C\n 8. Solution: [\arctan(x)]0^1 = \arctan(1) - \arctan(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4}.\n\n# The Definite Integral\n\n- Riemann Sum Definition:\n - Divide interval [a, b]$into$n$subintervals of width$\Delta x = \frac{b - a}{n}.\n - Approximate total area using rectangle sums:\n    \text{Total Area} \approx \sum{i=1}^{n} f(x_i)\Delta x\n - Taking the limit as n \to \infty$($\Delta x \to 0) defines the definite integral:\n    \int_{a}^{b} f(x)\,\mathrm{d}x = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i)\Delta x\n\n- Fundamental Theorem of Calculus:\n - If f(x)$is continuous on$[a, b]$and$F(x)$is any antiderivative of$f(x)$(such that$F' = f):\n    \int_{a}^{b} f(x)\,\mathrm{d}x = [F(x)]{a}^{b} = F(b) - F(a)\n - Cancellation of C$: Using$F(x) + C gives:\n    (F(b) + C) - (F(a) + C) = F(b) - F(a)\n - Thus, C drops out completely and is omitted in definite integrals.\n\n- Worked Examples:\n - Example 8.1: Evaluate \int{1}^{3} x^2\,\mathrm{d}x.\n - Solution:\n      \int_{1}^{3} x^2\,\mathrm{d}x = \left[\frac{x^3}{3}\right]{1}^{3} = \frac{3^3}{3} - \frac{1^3}{3} = 9 - \frac{1}{3} = \frac{26}{3}\n - Example 8.2: Evaluate \int{0}^{\pi/2} \cos(x)\,\mathrm{d}x.\n - Solution (angles in radians):\n      \int_{0}^{\pi/2} \cos(x)\,\mathrm{d}x = [\sin(x)]{0}^{\pi/2} = \sin\left(\frac{\pi}{2}\right) - \sin(0) = 1 - 0 = 1\n - Example 8.3: Evaluate \int{0}^{1} e^{2x}\,\mathrm{d}x.\n - Solution:\n      \left[\frac{1}{2}e^{2x}\right]{0}^{1} = \frac{1}{2}e^2 - \frac{1}{2}e^0 = \frac{e^2 - 1}{2} \approx 3.195\n\n- Properties of Definite Integrals:\n 1. Zero Width:\n     \int{a}^{a} f(x)\,\mathrm{d}x = 0\n 2. Swapping Limits:\n     \int_{a}^{b} f(x)\,\mathrm{d}x = -\int_{b}^{a} f(x)\,\mathrm{d}x\n 3. Interval Additivity (a < c < b):\n     \int_{a}^{b} f(x)\,\mathrm{d}x = \int_{a}^{c} f(x)\,\mathrm{d}x + \int_{c}^{b} f(x)\,\mathrm{d}x\n 4. Constant Factorization:\n     \int_{a}^{b} k f(x)\,\mathrm{d}x = k \int_{a}^{b} f(x)\,\mathrm{d}x\n\n- Practice Problems (Section 8):\n 1. \int_{0}^{2} x^3\,\mathrm{d}x\n 2. \int_{1}^{4} (2x + 1)\,\mathrm{d}x\n 3. \int_{1}^{2} \frac{1}{x}\,\mathrm{d}x\n 4. \int_{0}^{\pi} \sin(x)\,\mathrm{d}x\n 5. \int_{0}^{1} (e^x + 1)\,\mathrm{d}x\n 6. \int_{1}^{9} \sqrt{x}\,\mathrm{d}x\n 7. \int_{-1}^{1} (x^2 - x)\,\mathrm{d}x\n 8. \int_{0}^{\pi/4} \sec^2(x)\,\mathrm{d}x\n 9. \int_{2}^{2} (x^5 + 3x)\,\mathrm{d}x\n 10. \int_{0}^{2} x e^x\,\mathrm{d}x\n\n- Solutions to Section 8 Practice Problems:\n 1. 4\n 2. 18\n 3. \ln(2)\n 4. 2\n 5. e\n 6. \frac{52}{3}\n 7. \frac{2}{3}\n 8. 1\n 9. 0\n 10. e^2 + 1\n\n# Areas Under and Between Curves\n\n- Area Under a Curve:\n - If f(x) \ge 0$on$[a, b]$, area$A$bounded by$y = f(x)$, the x-axis,$x = a$, and$x = b is:\n    A = \int_{a}^{b} f(x)\,\mathrm{d}x\n - Example 9.1: Find area enclosed by y = x^2$, x-axis,$x = 0$, and$x = 3.\n - Solution:\n      A = \int_{0}^{3} x^2\,\mathrm{d}x = \left[\frac{x^3}{3}\right]{0}^{3} = 9 - 0 = 9\,\text{square units}\n\n- Regions Below the Axis (Signed Area):\n - Definite integrals calculate signed area. Regions where f(x) < 0 yield negative values.\n - Total Physical Area Procedure:\n 1. Find x-intercepts by solving f(x) = 0\n 2. Split the integral into subintervals bounded by these roots\n 3. Evaluate each integral separately, take absolute values, and sum them\n - Example 9.2: Total area bounded by y = x^2 - 4$and x-axis from$x = 0$to$x = 3\n - Solution: Intercept at x^2 - 4 = 0 \implies x = 2$(inside$[0, 3]).\n      \int{0}^{2} (x^2 - 4)\,\mathrm{d}x = \left[\frac{x^3}{3} - 4x\right]{0}^{2} = \frac{8}{3} - 8 = -\frac{16}{3}\n      \int{2}^{3} (x^2 - 4)\,\mathrm{d}x = \left[\frac{x^3}{3} - 4x\right]{2}^{3} = (9 - 12) - \left(\frac{8}{3} - 8\right) = -3 + \frac{16}{3} = \frac{7}{3}\n      \text{Total Area} = \left|-\frac{16}{3}\right| + \frac{7}{3} = \frac{16}{3} + \frac{7}{3} = \frac{23}{3}\,\text{square units}\n - Note: Net area integrated directly gives -\frac{16}{3} + \frac{7}{3} = -3, which is physically incorrect for area.\n\n- Area Between Two Curves:\n - If f(x) \ge g(x)$on$[a, b]:\n    A = \int{a}^{b} [f(x) - g(x)]\,\mathrm{d}x = \int_{a}^{b} (y_{\text{upper}} - y_{\text{lower}})\,\mathrm{d}x\n - Intersection boundaries found by setting f(x) = g(x).\n - Example 9.3: Area enclosed between y = x^2$and$y = x + 2\n - Solution:\n 1. Find intersections: x^2 = x + 2 \implies x^2 - x - 2 = 0 \implies (x - 2)(x + 1) = 0 \implies x = -1, 2\n 2. Test point x = 0$: Line gives$y = 2$, parabola gives$y = 0 (line is upper curve).\n 3. Integrate:\n         A = \int_{-1}^{2} ((x + 2) - x^2)\,\mathrm{d}x = \left[\frac{x^2}{2} + 2x - \frac{x^3}{3}\right]{-1}^{2}\n         = \left(2 + 4 - \frac{8}{3}\right) - \left(\frac{1}{2} - 2 + \frac{1}{3}\right) = \frac{10}{3} - \left(-\frac{7}{6}\right) = \frac{27}{6} = \frac{9}{2} = 4.5\,\text{square units}\n\n- Exam Technique Notes:\n - Always sketch the region prior to integrating to determine upper/lower boundaries and roots.\n - Always include units in answers as square units.\n\n- Area With Respect to the y-Axis:\n - For horizontal strips, express curve as x = h(y) and integrate over y-limits:\n    A = \int{c}^{d} h(y)\,\mathrm{d}y\n - Example 9.4: Area bounded by y = \sqrt{x}$, y-axis, and$y = 2\n - Solution: Rewrite curve as x = y^2$. Integrate from$y = 0$to$y = 2\n      A = \int_{0}^{2} y^2\,\mathrm{d}y = \left[\frac{y^3}{3}\right]{0}^{2} = \frac{8}{3}\,\text{square units}\n\n- Practice Problems (Section 9):\n 1. Find area under y = 3x^2$from$x = 1$to$x = 2\n 2. Find area under y = \sin(x)$from$x = 0$to$x = \pi\n 3. Find area under y = e^x$from$x = 0$to$x = 2\n 4. Find area enclosed by y = 4 - x^2 and the x-axis\n 5. Find total area between y = x^3$and x-axis from$x = -1$to$x = 1. Compare with net area.\n 6. Find area enclosed between y = x^2$and$y = 4\n 7. Find area enclosed between y = x^2$and$y = x\n 8. Find area between y = x^2$and$y = \sqrt{x}$from$x = 0$to$x = 1\n 9. Find area bounded by y = x^2 + 1$, x-axis,$x = 0$, and$x = 2\n 10. Find area bounded by x = y^2$, y-axis, and$y = 3\n\n- Solutions to Section 9 Practice Problems:\n 1. 7\n 2. 2\n 3. e^2 - 1\n 4. \frac{32}{3}\n 5. Total area = \frac{1}{2}; Net area = 0. Net area cancels due to odd symmetry around origin.\n 6. \frac{32}{3}\n 7. \frac{1}{6}\n 8. \frac{1}{3}\n 9. \frac{14}{3}\n 10. 9\n\n# Mixed Revision Exercise\n\n- Problems:\n 1. \int \left(x^3 - \frac{2}{x}\right)\,\mathrm{d}x\n 2. \int x \cos\left(x^2\right)\,\mathrm{d}x\n 3. \int x e^{-x}\,\mathrm{d}x\n 4. \int \frac{4}{9 + x^2}\,\mathrm{d}x\n 5. \int (2x + 1)^5\,\mathrm{d}x\n 6. \int \sin^2(3x)\,\mathrm{d}x\n 7. \int \frac{x^2}{x^3 + 1}\,\mathrm{d}x\n 8. \int \frac{1}{\sqrt{16 - x^2}}\,\mathrm{d}x\n 9. \int{0}^{1} (3x^2 - 2x)\,\mathrm{d}x\n 10. \int_{0}^{\pi/2} \sin(2x)\,\mathrm{d}x\n 11. \int x^2 \cos(x)\,\mathrm{d}x\n 12. \int \frac{e^x}{e^x + 1}\,\mathrm{d}x\n 13. Find area enclosed between curve y = x^2 - 2x$and line$y = x\n 14. Particle velocity v(t) = 3t^2 - 6t\,\text{m\,s}^{-1}. Find total distance in first 3 seconds.\n 15. Show that \int_{0}^{1} \frac{1}{1 + x^2}\,\mathrm{d}x = \frac{\pi}{4}\n\n- Solutions to Mixed Revision Problems:\n 1. \frac{x^4}{4} - 2\ln|x| + C\n 2. \frac{1}{2}\sin\left(x^2\right) + C\n 3. -e^{-x}(x + 1) + C\n 4. \frac{4}{3}\arctan\left(\frac{x}{3}\right) + C\n 5. \frac{(2x + 1)^6}{12} + C\n 6. \frac{x}{2} - \frac{\sin(6x)}{12} + C\n 7. \frac{1}{3}\ln\left|x^3 + 1\right| + C\n 8. \arcsin\left(\frac{x}{4}\right) + C\n 9. 0\n 10. 1\n 11. x^2 \sin(x) + 2x \cos(x) - 2\sin(x) + C\n 12. \ln\left(e^x + 1\right) + C\n 13. \frac{9}{2}\text{ square units}\n 14. 8\,\text{m}\n 15. Proof: [\arctan(x)]0^1 = \arctan(1) - \arctan(0) = \frac{\pi}{4}\n\n# Table of Standard Integrals\n\n- Function and Corresponding Integral Table:\n - x^n \quad (n \neq -1) \implies \frac{x^{n+1}}{n+1} + C\n - \frac{1}{x} \implies \ln|x| + C\n - e^x \implies e^x + C\n - e^{ax+b} \implies \frac{1}{a}e^{ax+b} + C\n - a^x \quad (a > 0, a \neq 1) \implies \frac{a^x}{\ln(a)} + C\n - \sin(x) \implies -\cos(x) + C\n - \cos(x) \implies \sin(x) + C\n - \sec^2(x) \implies \tan(x) + C\n - \csc^2(x) \implies -\cot(x) + C\n - \sec(x)\tan(x) \implies \sec(x) + C\n - \csc(x)\cot(x) \implies -\csc(x) + C\n - \tan(x) \implies \ln|\sec(x)| + C\n - \cot(x) \implies \ln|\sin(x)| + C\n - \ln(x) \implies x\ln(x) - x + C\n - \frac{1}{\sqrt{a^2 - x^2}} \implies \arcsin\left(\frac{x}{a}\right) + C\n - \frac{1}{a^2 + x^2} \implies \frac{1}{a}\arctan\left(\frac{x}{a}\right) + C\n - \frac{f'(x)}{f(x)} \implies \ln|f(x)| + C\n\n# Appendix: Fully Worked Solutions to Selected Problems\n\n- S1. Cyclic Case Integration by Parts (I = \int e^x \sin(x)\,\mathrm{d}x):\n - Step 1: Let u = \sin(x)$,$\frac{\mathrm{d}v}{\mathrm{d}x} = e^x \implies \frac{\mathrm{d}u}{\mathrm{d}x} = \cos(x)$,$v = e^x\n    I = e^x \sin(x) - \int e^x \cos(x)\,\mathrm{d}x\n - Step 2: Apply parts again on \int e^x \cos(x)\,\mathrm{d}x$with$u = \cos(x)$,$\frac{\mathrm{d}v}{\mathrm{d}x} = e^x\n    \int e^x \cos(x)\,\mathrm{d}x = e^x \cos(x) + \int e^x \sin(x)\,\mathrm{d}x = e^x \cos(x) + I\n - Step 3: Substitute back:\n    I = e^x \sin(x) - (e^x \cos(x) + I) \implies 2I = e^x(\sin(x) - \cos(x))\n    \int e^x \sin(x)\,\mathrm{d}x = \frac{e^x(\sin(x) - \cos(x))}{2} + C\n - Warnings: Use consistent assignment for u$in both steps.$C is added after dividing by 2.\n\n- S2. Section 5 Question 8 (\int e^x \cos(x)\,\mathrm{d}x):\n - From S1 derivation, \int e^x \cos(x)\,\mathrm{d}x = e^x \cos(x) + I\n    = e^x \cos(x) + \frac{e^x(\sin(x) - \cos(x))}{2} = \frac{e^x(\sin(x) + \cos(x))}{2} + C\n\n- S3. Section 2 Question 16 (Boundary Value Problem):\n - Given \frac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 - 2$and point$(1, 4).\n - Integrate: y = \int (3x^2 - 2)\,\mathrm{d}x = x^3 - 2x + C\n - Substitute x = 1, y = 4:\n    4 = 1 - 2 + C \implies C = 5 \implies y = x^3 - 2x + 5\n\n- S4. Section 6 Question 11 (Definite Substitution):\n - Evaluate \int{0}^{1} x\left(x^2 + 3\right)^2\,\mathrm{d}x\n - Let u = x^2 + 3 \implies \mathrm{d}u = 2x\,\mathrm{d}x \implies x\,\mathrm{d}x = \frac{1}{2}\,\mathrm{d}u$.
  • Change limits: $x = 0 \implies u = 3$; x = 1 \implies u = 4$.\n - Evaluate:\n    \frac{1}{2}\int_{3}^{4} u^2\,\mathrm{d}u = \frac{1}{6}[u^3]{3}^{4} = \frac{1}{6}(64 - 27) = \frac{37}{6}\n\n- S5. Section 7 Question 7 (Completing Square Under Root):\n - Evaluate \int \frac{1}{\sqrt{8 - 2x - x^2}}\,\mathrm{d}x\n - Rearrange quadratic: 8 - 2x - x^2 = 8 - (x^2 + 2x) = 8 - ((x + 1)^2 - 1) = 9 - (x + 1)^2\n - Match arcsin form with a = 3:\n    \int \frac{1}{\sqrt{9 - (x + 1)^2}}\,\mathrm{d}x = \arcsin\left(\frac{x + 1}{3}\right) + C\n\n- S6. Section 9 Question 5 (Total vs Net Area for Odd Function):\n - Curve y = x^3$on$[-1, 1]$. Root at$x = 0.\n    \int{-1}^{0} x^3\,\mathrm{d}x = -\frac{1}{4}, \quad \int_{0}^{1} x^3\,\mathrm{d}x = \frac{1}{4}\n - Total physical area = \left|-\frac{1}{4}\right| + \frac{1}{4} = \frac{1}{2}\text{ square units}.\n - Net area = -\frac{1}{4} + \frac{1}{4} = 0 (regions below and above axis cancel).\n\n- S7. Section 10 Question 14 (Distance vs Displacement from Velocity):\n - v(t) = 3t^2 - 6t = 3t(t - 2)$. Reverses direction at$t = 2.\n - Distance requires \int |v|\,\mathrm{d}t:\n    \int_{0}^{2} (3t^2 - 6t)\,\mathrm{d}t = [t^3 - 3t^2]{0}^{2} = 8 - 12 = -4\n    \int{2}^{3} (3t^2 - 6t)\,\mathrm{d}t = [t^3 - 3t^2]_{2}^{3} = (27 - 27) - (8 - 12) = 4\n - Total Distance = \left|-4\right| + 4 = 8\,\text{metres}. (Displacement is -4 + 4 = 0\,\text{metres}$$).