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IB Mathematics AA AHL Derivatives of secx , cscx , cotx Study Notes

IB Mathematics AA AHL Derivatives of secx , cscx , cotx Study Notes - New Syllabus

IB Mathematics AA AHL Derivatives of secx , cscx , cotx Study Notes

LEARNING OBJECTIVE

  • Derivatives

Key Concepts: 

  • Derivatives of Functions
  • Partial fractions

MAA HL and SL Notes – All topics

Derivatives of Special Functions

Trigonometric Functions

FunctionDerivative
\( \tan x \)\( \sec^2 x \)
\( \sec x \)\( \sec x \tan x \)
\( \csc x \)\( -\csc x \cot x \)
\( \cot x \)\( -\csc^2 x \)

 Exponential and Logarithmic Functions

FunctionDerivative
\( a^x \)\( a^x \ln a \)
\( \log_a x \)\( \dfrac{1}{x \ln a} \)

Inverse Trigonometric Functions

FunctionDerivative
\( \arcsin x \)\( \dfrac{1}{\sqrt{1 – x^2}} \)
\( \arccos x \)\( \dfrac{-1}{\sqrt{1 – x^2}} \)
\( \arctan x \)\( \dfrac{1}{1 + x^2} \)

Indefinite Integrals of Derivatives

Concept:

When we integrate the derivative of a function, we obtain the original function up to a constant of integration. That is, the indefinite integral represents a family of curves that differ by a vertical shift.

General form:

If \( \dfrac{d}{dx}f(x) = f'(x) \), then:

$\boxed{\int f'(x) \, dx = f(x) + C \quad \text{(C is an arbitrary constant)}}$

Linear Composite Function Integration:

For a linear function inside another function, if \( u = ax + b \), then:

$\boxed{ \int f'(ax + b) \, dx = \frac{1}{a} f(ax + b) + C}$

 Integration of a composite function with a linear inner function requires dividing by the coefficient of \(x\) to reverse the chain rule effect.

Common Indefinite Integrals:

  • \(\boxed{ \int \sec^2 x \, dx = \tan x + C }\)
  • \(\boxed{ \int \sec x \tan x \, dx = \sec x + C }\)
  • \(\boxed{ \int -\csc x \cot x \, dx = \csc x + C }\)
  • \(\boxed{ \int -\csc^2 x \, dx = \cot x + C }\)
  • \(\boxed{ \int a^x \ln a \, dx = a^x + C }\)
  • \(\boxed{ \int \dfrac{1}{x \ln a} \, dx = \log_a x + C }\)
  • \(\boxed{ \int \dfrac{1}{\sqrt{1 – x^2}} \, dx = \arcsin x + C }\)
  • \(\boxed{ \int \dfrac{-1}{\sqrt{1 – x^2}} \, dx = \arccos x + C }\)
  • \(\boxed{ \int \dfrac{1}{1 + x^2} \, dx = \arctan x + C }\)

 Interpretation:

Each indefinite integral gives a family of antiderivatives, meaning all functions that differ from the original by a constant. These graphs are vertical translations of each other.

Example 

Evaluate: \( \int e^x \sin x \, dx \)

▶️ Answer/Explanation
Let \( I = \int e^x \sin x \, dx \)

Use integration by parts:

  • First, let \( u = \sin x \), \( dv = e^x dx \)
  • Then \( du = \cos x dx \), \( v = e^x \)

$ I = e^x \sin x – \int e^x \cos x \, dx $ Now let \( J = \int e^x \cos x \, dx \) — apply integration by parts again:

  • Let \( u = \cos x \), \( dv = e^x dx \)
  • Then \( du = -\sin x dx \), \( v = e^x \)

$ J = e^x \cos x + \int e^x \sin x \, dx $

So now:
$ I = e^x \sin x – \left( e^x \cos x + \int e^x \sin x \, dx \right) $
$ I = e^x \sin x – e^x \cos x – I $
Add \( I \) to both sides: $ 2I = e^x(\sin x – \cos x) $
$ \boxed{I = \dfrac{e^x(\sin x – \cos x)}{2} + C} $

Example 

Evaluate: \( \int \frac{1}{x^2 + 2x + 5} \, dx \)

▶️ Answer/Explanation
Solution:
Complete the square: \( x^2 + 2x + 5 = (x + 1)^2 + 2^2 \)
Use standard result: \( \int \frac{1}{x^2 + a^2} \, dx = \frac{1}{a} \arctan\left(\frac{x}{a}\right) + C \)
$ \int \frac{1}{x^2 + 2x + 5} \, dx = \frac{1}{2} \arctan\left( \frac{x + 1}{2} \right) + C $

Example 

Evaluate: \( \int \sec^2(2x + 5) \, dx \)

▶️ Answer/Explanation
Solution:
Let \( u = 2x + 5 \Rightarrow du = 2dx \Rightarrow dx = \frac{1}{2} du \)
$ \int \sec^2(2x + 5) \, dx = \frac{1}{2} \int \sec^2(u) \, du = \frac{1}{2} \tan(u) + C = \frac{1}{2} \tan(2x + 5) + C $

Partial Fractions for Integration

Partial Fractions for Integration

When the integrand is a rational function (a ratio of polynomials), and the degree of the numerator is less than the degree of the denominator, we can often simplify the expression using partial fractions. This is especially useful when the denominator can be factored into linear or irreducible quadratic terms.

General form of partial fractions decomposition:

Suppose we have a rational function

\(\displaystyle\boxed{ \frac{P(x)}{Q(x)}}\)

where \(Q(x)\) factors into linear factors like \((ax + b)\) and/or irreducible quadratic factors like \((cx^2 + dx + e)\). Then the partial fractions take the form:

For each distinct linear factor \((ax + b)^m\), include terms of the form:

\(\displaystyle\boxed{ \frac{A_1}{ax + b} + \frac{A_2}{(ax + b)^2} + \cdots + \frac{A_m}{(ax + b)^m}}\)

For each distinct irreducible quadratic factor \((cx^2 + dx + e)^n\), include terms of the form:

\(\displaystyle\boxed{ \frac{B_1 x + C_1}{cx^2 + dx + e} + \frac{B_2 x + C_2}{(cx^2 + dx + e)^2} + \cdots + \frac{B_n x + C_n}{(cx^2 + dx + e)^n}}\)

Key Steps:

  1. Factor the denominator completely.
  2. Set up the decomposition into partial fractions.
  3. Solve for unknown constants.
  4. Integrate each term separately.

Example

Evaluate: \( \int \frac{1}{x^2 + 3x + 2} \, dx \)

▶️ Answer/Explanation
Factor the denominator:
\( x^2 + 3x + 2 = (x + 1)(x + 2) \)
Decompose into partial fractions:
$ \frac{1}{(x + 1)(x + 2)} = \frac{A}{x + 1} + \frac{B}{x + 2} $
Multiply through by \( (x + 1)(x + 2) \):
$ 1 = A(x + 2) + B(x + 1) $
\( 1 = Ax + 2A + Bx + B = (A + B)x + (2A + B) \)
Match coefficients: $ A + B = 0,\quad 2A + B = 1 $
$ B = -A,\quad 2A – A = 1 \Rightarrow A = 1,\ B = -1 $
$ \int \left( \frac{1}{x + 1} – \frac{1}{x + 2} \right) dx = \ln|x + 1| – \ln|x + 2| + C $
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