IB Maths AA HL 5.10 Indefinite integration Question Bank

Practise IB Mathematics SL/HL 5.10 by applying indefinite integration methods to exam-style questions.

Exam points

identify the mathematical structure, variable or representation
select and apply the correct theorem, formula or algorithm
check the result using units, domain, graph or logical reasoning

[Maximum number: 6]

The functions f and g are defined by

\begin{aligned} & f(x)=\frac{\mathrm{e}^{x}+\mathrm{e}^{-x}}{2}, x \in \mathbb{R} \\ & g(x)=\frac{\mathrm{e}^{x}-\mathrm{e}^{-x}}{2}, x \in \mathbb{R} \end{aligned}

Use the substitution $u=\mathrm{e}^{x}$ to find $\int_{0}^{\ln 3} \frac{1}{4 f(x)-2 g(x)} \mathrm{d} x$ . Give your answer in the form $\frac{\pi \sqrt{a}}{b}$ where $a, b \in \mathbb{Z}^{+}$ .

$\quad u=\mathrm{e}^{x} \Rightarrow \mathrm{~d} u=\mathrm{e}^{x} \mathrm{~d} x$

\int \frac{\mathrm{e}^{x}}{\mathrm{e}^{2 x}+3} \mathrm{~d} x=\int \frac{1}{u^{2}+3} \mathrm{~d} u

(when x=0, u=1 and when $x=\ln 3, u=3$ )

\begin{aligned} & \int_{1}^{3} \frac{1}{u^{2}+3} \mathrm{~d} u=\left[\frac{1}{\sqrt{3}} \arctan \left(\frac{u}{\sqrt{3}}\right)\right]_{1}^{3} \\ & \left.=\left[\frac{1}{\sqrt{3}} \arctan \left(\frac{\mathrm{e}^{x}}{\sqrt{3}}\right)\right]_{0}^{\ln 3}\right) \\ & =\frac{\pi \sqrt{3}}{9}-\frac{\pi \sqrt{3}}{18} \\ & =\frac{\pi \sqrt{3}}{18} \end{aligned}

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SL 5.10—Indefinite integration question 1