\( \DeclareMathOperator{cosec}{cosec} \)

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International Baccalaureate Mathematics

Calculus

Syllabus Content

Derivatives of tanx, secx, cosecx, cotx, ax, logax, arcsinx, arccosx, arctanx.
Indefinite integrals of the derivatives of any of the above functions.
The composites of any of these with a linear function.
Use of partial fractions to rearrange the integrand.

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Furthermore

Official Guidance, clarification and syllabus links:

Indefinite integral interpreted as a family of curves.

$$ \text{Examples } \int \dfrac{1}{x^2+2x+5}dx=\dfrac{1}{2}\arctan{\frac{(x+1)}{2}+C} $$ $$ \int \sec^2{(2x+5)}dx = \dfrac{1}{2} \tan{(2x+5)}+C$$
$$ \int \dfrac{1}{x^2+3x+2}dx = \ln|\frac{x+1}{x+2}| + C $$

Link to: partial fractions (AHL1.11)


Formula Booklet:

Standard derivatives

\( \frac{d}{dx} \tan(x) = \sec^2(x) \)

\( \frac{d}{dx} \sec(x) = \sec(x) \tan(x) \)

\( \frac{d}{dx} \cosec(x) = -\cosec(x) \cot(x) \)

\( \frac{d}{dx} \cot(x) = -\cosec^2(x) \)

\( \frac{d}{dx} a^x = a^x \ln(a) \)

\( \frac{d}{dx} \log_a(x) = \frac{1}{x \ln(a)} \)

\( \frac{d}{dx} \arcsin(x) = \frac{1}{\sqrt{1 - x^2}} \)

\( \frac{d}{dx} \arccos(x) = -\frac{1}{\sqrt{1 - x^2}} \)

\( \frac{d}{dx} \arctan(x) = \frac{1}{1 + x^2} \)

Standard Integrals

\( \int a^x dx = \dfrac{1}{\ln{a}}a^x + C\)

\( \int \dfrac{1}{a^2+x^2} dx = \dfrac{1}{a} \arctan{\left(\frac{x}{a}\right)}+C \)

\( \int \dfrac{1}{\sqrt{a^2-x^2}} dx = \arcsin{\left(\frac{x}{a}\right)}+C, \quad |x|\lt a \)


See the main Partial Fractions section of the syllabus in Number and Algebra.


Here is an example of how to use partial fractions to decompose the rational function \( \dfrac{2x-5}{(x-2)(x+1)^2} \):

$$\begin{aligned} \frac{2x-5}{(x-2)(x+1)^2} &= \frac{A}{x-2} + \frac{B}{x+1} + \frac{C}{(x+1)^2} \\ &= \frac{A(x+1)^2 + B(x-2)(x+1) + C(x-2)}{(x-2)(x+1)^2} \end{aligned} $$

Multiplying both sides by the common denominator gives:

$$2x-5 = A(x+1)^2 + B(x-2)(x+1) + C(x-2)$$

Substituting \(x = 2\) gives:

$$-1 = 9A$$ $$A = -\dfrac{1}{9}$$

Substituting \(x = -1\) gives:

$$-7 = -3C$$ $$C = \dfrac{7}{3}$$

Finally, substituting \(x = 0\) gives:

$$-5 = A - 2B -2C$$

Substituting A and C gives:

$$-5 = \left(-\frac{1}{9}\right) - 2B + -2\left( \frac{7}{3} \right)$$ $$B = \dfrac{1}{9}$$

Therefore, we have:

$$\frac{2x-5}{(x-2)(x+1)^2} = -\frac{1}{9(x-2)} + \frac{1}{9(x+1)} + \frac{7}{3(x+1)^2}$$

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