AS & A Level · Practice questions · Integrating exponentials, 1/(ax + b) and trigonometric functions

Backwards from the derivative.

Six original Cambridge-style questions on the standard integrals of Pure 2 and Pure 3, with exact answers in terms of \(e\), \(\ln\), \(\pi\) and surds.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Integrate the outside, then divide by a.

Recognise the standard form: \(e^{ax+b}\), \(\dfrac{1}{ax+b}\), \(\sin\), \(\cos\), \(\sec^2\) or \(\dfrac{1}{x^2+a^2}\).
Divide by the coefficient of \(x\) inside the bracket.
Signs: \(\int \sin = -\cos\), \(\int \cos = +\sin\).
Radians for every trigonometric limit.
Exact answers: tidy logs with the log laws and keep \(\pi\), \(e\) and surds.

Recall

[2 marks]

Find \(\displaystyle\int \left(e^{-2x} + \dfrac{5}{2x-1}\right) dx\).

\(\displaystyle\int e^{-2x}\,dx = -\dfrac{1}{2}e^{-2x}\). ✓
\(\displaystyle\int \dfrac{5}{2x-1}\,dx = \dfrac{5}{2}\ln|2x-1|\). ✓
Answer: \(-\dfrac{1}{2}e^{-2x} + \dfrac{5}{2}\ln|2x-1| + c\). ✓

Recall

[2 marks]

Find \(\displaystyle\int \left(4\cos\dfrac{x}{2} + \sec^2 4x\right) dx\).

\(\displaystyle\int 4\cos\dfrac{x}{2}\,dx = 4 \times 2\sin\dfrac{x}{2} = 8\sin\dfrac{x}{2}\). ✓
\(\displaystyle\int \sec^2 4x\,dx = \dfrac{1}{4}\tan 4x\). ✓
Answer: \(8\sin\dfrac{x}{2} + \dfrac{1}{4}\tan 4x + c\). ✓

Calculation

[4 marks]

Show that \(\displaystyle\int_0^{\ln 3} \left(e^{x} + e^{-x}\right)^{2} dx = \dfrac{40}{9} + 2\ln 3\).

Expand: \((e^{x} + e^{-x})^2 = e^{2x} + 2 + e^{-2x}\). ✓
Integrate: \(\left[\dfrac{1}{2}e^{2x} + 2x - \dfrac{1}{2}e^{-2x}\right]_0^{\ln 3}\). ✓
Upper: \(e^{2\ln 3} = 9\) and \(e^{-2\ln 3} = \dfrac{1}{9}\), so \(\dfrac{9}{2} + 2\ln 3 - \dfrac{1}{18}\). ✓
Lower: \(\dfrac{1}{2} + 0 - \dfrac{1}{2} = 0\). ✓
\(\dfrac{9}{2} - \dfrac{1}{18} = \dfrac{81 - 1}{18} = \dfrac{40}{9}\), so the value is \(\dfrac{40}{9} + 2\ln 3\). ✓

Analysis

[4 marks]

Given that \(k > 1\) and \(\displaystyle\int_1^{k} \dfrac{4}{2x-1}\,dx = \ln 25\), find the value of \(k\).

\(\displaystyle\int \dfrac{4}{2x-1}\,dx = 2\ln(2x-1)\). ✓
\(\left[2\ln(2x-1)\right]_1^{k} = 2\ln(2k-1) - 2\ln 1 = 2\ln(2k-1)\). ✓
\(2\ln(2k-1) = \ln 25 \Rightarrow \ln(2k-1)^2 = \ln 25 \Rightarrow (2k-1)^2 = 25\). ✓
\(2k - 1 > 0\), so \(2k - 1 = 5\) and \(k = 3\). ✓

Calculation

[4 marks]

(Paper 3) Find the exact value of \(\displaystyle\int_0^{2/3} \dfrac{1}{4+9x^2}\,dx\).

\(\dfrac{1}{4+9x^2} = \dfrac{1}{9}\times\dfrac{1}{x^2+\frac{4}{9}}\), so \(a = \dfrac{2}{3}\). ✓
\(\displaystyle\int \dfrac{1}{4+9x^2}\,dx = \dfrac{1}{9}\times\dfrac{3}{2}\tan^{-1}\dfrac{3x}{2} = \dfrac{1}{6}\tan^{-1}\dfrac{3x}{2}\). ✓
At \(x = \dfrac{2}{3}\): \(\dfrac{1}{6}\tan^{-1}1 = \dfrac{1}{6}\times\dfrac{\pi}{4}\). At 0 the value is 0. ✓
Answer: \(\dfrac{\pi}{24}\). ✓

Analysis

[6 marks]

The curve \(y = e^{2x} - \dfrac{6}{2x+1}\) lies above the \(x\)-axis for \(1 \le x \le 2\).

(a) Verify that \(y > 0\) when \(x = 1\).

(b) Find the exact area of the region bounded by the curve, the \(x\)-axis and the lines \(x = 1\) and \(x = 2\), and give its value to 3 significant figures.

(a) At \(x = 1\): \(y = e^{2} - \dfrac{6}{3} = e^2 - 2 = 5.39 > 0\). ✓
(b) \(\displaystyle\int \left(e^{2x} - \dfrac{6}{2x+1}\right) dx = \dfrac{1}{2}e^{2x} - 3\ln(2x+1)\). ✓
Upper: \(\dfrac{1}{2}e^{4} - 3\ln 5\). Lower: \(\dfrac{1}{2}e^{2} - 3\ln 3\). ✓
Area \(= \dfrac{1}{2}(e^{4} - e^{2}) - 3\ln 5 + 3\ln 3 = \dfrac{1}{2}(e^{4} - e^{2}) - 3\ln\dfrac{5}{3}\). ✓
\(= 23.6045 - 1.5325 = 22.1\) square units (3 s.f.). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.