AS & A Level · Practice questions · Inverse trig and hyperbolic integrals

Match the root, then substitute.

Six original Cambridge-style questions on standard integrals and substitutions: hyperbolic functions, \(\sin^{-1}\), \(\tan^{-1}\), \(\sinh^{-1}\) and \(\cosh^{-1}\) forms, completing the square, and trigonometric and hyperbolic substitutions with exact answers.

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

Tidy the quadratic, match the form, keep it exact.

Coefficient 1: take the coefficient of \(x^{2}\) out (its square root, if under a root).
Complete the square if there is an \(x\) term, and read off \(a\) (not \(a^{2}\)).
Match the form: \(a^{2} - x^{2} \to \sin^{-1}\); \(x^{2} + a^{2}\) (no root) \(\to \frac{1}{a}\tan^{-1}\); \(\sqrt{x^{2} + a^{2}} \to \sinh^{-1}\); \(\sqrt{x^{2} - a^{2}} \to \cosh^{-1}\).
Root on top: substitute \(x = a\sin u\), \(a\sinh u\) or \(a\cosh u\); include \(dx\) and change the limits.
Exact answers: convert \(\sinh^{-1}\) and \(\cosh^{-1}\) with \(\ln\!\big(k + \sqrt{k^{2} \pm 1}\big)\) and simplify the surd.

Recall

[3 marks]

(a) Find \(\displaystyle\int \operatorname{sech}^{2} 3x\,dx\).

(b) Find the exact value of \(\displaystyle\int_{0}^{\ln 3} \cosh x\,dx\).

(a) \(\dfrac{1}{3}\tanh 3x + c\). ✓
(b) \(\big[\sinh x\big]_{0}^{\ln 3} = \sinh(\ln 3) = \dfrac{3 - \frac{1}{3}}{2} = \dfrac{4}{3}\). ✓
(c) \(\tanh 2x = \dfrac{\sinh 2x}{\cosh 2x}\), and \(\dfrac{d}{dx}\cosh 2x = 2\sinh 2x\), so the integral is \(\dfrac{1}{2}\ln\cosh 2x + c\). ✓

Calculation

[4 marks]

Find the exact value of

(a) \(\displaystyle\int_{0}^{2} \frac{dx}{\sqrt{16 - x^{2}}}\)

(b) \(\displaystyle\int_{5}^{17/2} \frac{dx}{\sqrt{x^{2} - 16}}\), giving your answer as a single logarithm.

(a) \(\Big[\sin^{-1}\dfrac{x}{4}\Big]_{0}^{2} = \sin^{-1}\dfrac{1}{2} = \dfrac{\pi}{6}\). ✓
(b) \(\Big[\cosh^{-1}\dfrac{x}{4}\Big]_{5}^{17/2} = \cosh^{-1}\dfrac{17}{8} - \cosh^{-1}\dfrac{5}{4}\). ✓
\(\cosh^{-1}\dfrac{17}{8} = \ln\!\left(\dfrac{17}{8} + \sqrt{\dfrac{289}{64} - 1}\right) = \ln\!\left(\dfrac{17}{8} + \dfrac{15}{8}\right) = \ln 4\). ✓
\(\cosh^{-1}\dfrac{5}{4} = \ln\!\left(\dfrac{5}{4} + \dfrac{3}{4}\right) = \ln 2\). ✓
So the integral is \(\ln 4 - \ln 2 = \ln 2\). ✓

Calculation

[5 marks]

(a) Show that \(\displaystyle\int_{0}^{1} \frac{dx}{\sqrt{9x^{2} + 16}} = \frac{1}{3}\ln 2\).

(b) Find the exact value of \(\displaystyle\int_{0}^{3/2} \frac{dx}{4x^{2} + 9}\).

(a) \(\sqrt{9x^{2} + 16} = 3\sqrt{x^{2} + \tfrac{16}{9}}\), so \(a = \tfrac{4}{3}\): ✓
\[\int_{0}^{1} \frac{dx}{\sqrt{9x^{2} + 16}} = \frac{1}{3}\Big[\sinh^{-1}\frac{3x}{4}\Big]_{0}^{1} = \frac{1}{3}\sinh^{-1}\frac{3}{4}.\] ✓
\(\sinh^{-1}\dfrac{3}{4} = \ln\!\left(\dfrac{3}{4} + \sqrt{\dfrac{9}{16} + 1}\right) = \ln\!\left(\dfrac{3}{4} + \dfrac{5}{4}\right) = \ln 2\), giving \(\dfrac{1}{3}\ln 2\). ✓
(b) \(4x^{2} + 9 = 4\left(x^{2} + \tfrac{9}{4}\right)\), so \(a = \tfrac{3}{2}\): ✓
\[\frac{1}{4}\cdot\frac{2}{3}\Big[\tan^{-1}\frac{2x}{3}\Big]_{0}^{3/2} = \frac{1}{6}\tan^{-1} 1 = \frac{\pi}{24}.\] ✓

Analysis

[5 marks]

(a) Find the exact value of \(\displaystyle\int_{3}^{5} \frac{dx}{\sqrt{x^{2} - 6x + 13}}\), giving your answer in logarithmic form.

(b) Find the exact value of \(\displaystyle\int_{1}^{5/2} \frac{dx}{\sqrt{8 + 2x - x^{2}}}\).

(a) \(x^{2} - 6x + 13 = (x - 3)^{2} + 4\), so \(a = 2\): ✓
\[\Big[\sinh^{-1}\frac{x - 3}{2}\Big]_{3}^{5} = \sinh^{-1} 1 - \sinh^{-1} 0 = \ln\!\big(1 + \sqrt{2}\big).\] ✓
(b) \(8 + 2x - x^{2} = 9 - (x^{2} - 2x + 1) = 9 - (x - 1)^{2}\), so \(a = 3\): ✓
\[\Big[\sin^{-1}\frac{x - 1}{3}\Big]_{1}^{5/2} = \sin^{-1}\frac{1}{2} - \sin^{-1} 0 = \frac{\pi}{6}.\] ✓

Analysis

[5 marks]

Use the substitution \(x = 2\sinh u\) to show that \(\displaystyle\int_{0}^{2} \sqrt{x^{2} + 4}\,dx = 2\sqrt{2} + 2\ln\!\big(1 + \sqrt{2}\big)\).

\(dx = 2\cosh u\,du\) and \(\sqrt{4\sinh^{2} u + 4} = 2\cosh u\). Limits: \(u = 0\) to \(u = \sinh^{-1} 1 = \ln\!\big(1 + \sqrt{2}\big)\). ✓
\[\int_{0}^{2} \sqrt{x^{2} + 4}\,dx = \int 4\cosh^{2} u\,du = 2\int (\cosh 2u + 1)\,du = \Big[\sinh 2u + 2u\Big].\] ✓
\(\sinh 2u = 2\sinh u\cosh u\). At the upper limit \(\sinh u = 1\) and \(\cosh u = \sqrt{2}\), so \(\sinh 2u = 2\sqrt{2}\). ✓
\[\Big[2\sinh u\cosh u + 2u\Big]_{0}^{\ln(1 + \sqrt{2})} = 2\sqrt{2} + 2\ln\!\big(1 + \sqrt{2}\big) - 0.\] ✓

Analysis

[6 marks]

(a) Use the substitution \(x = 3\cosh u\) to show that \(\displaystyle\int_{3}^{5} \sqrt{x^{2} - 9}\,dx = 10 - \frac{9}{2}\ln 3\).

(b) The region \(R\) is enclosed by the right-hand branch of the hyperbola \(x^{2} - y^{2} = 9\) and the line \(x = 5\). Deduce the exact area of \(R\).

(a) \(dx = 3\sinh u\,du\) and \(\sqrt{9\cosh^{2} u - 9} = 3\sinh u\) for \(u \ge 0\). Limits: \(x = 3 \Rightarrow u = 0\); \(x = 5 \Rightarrow \cosh u = \tfrac{5}{3}\), \(u = \ln\!\left(\tfrac{5}{3} + \tfrac{4}{3}\right) = \ln 3\). ✓
\[\int_{0}^{\ln 3} 9\sinh^{2} u\,du = \frac{9}{2}\int_{0}^{\ln 3} (\cosh 2u - 1)\,du = \frac{9}{2}\Big[\sinh u\cosh u - u\Big]_{0}^{\ln 3}.\] ✓
At \(u = \ln 3\): \(\cosh u = \tfrac{5}{3}\), \(\sinh u = \tfrac{4}{3}\), so \(\sinh u\cosh u = \tfrac{20}{9}\). ✓
\[\frac{9}{2}\left(\frac{20}{9} - \ln 3\right) = 10 - \frac{9}{2}\ln 3.\] ✓
(b) The upper half of the branch is \(y = \sqrt{x^{2} - 9}\), and \(R\) is symmetric about the \(x\)-axis, so its area is twice the integral. ✓
Area of \(R = 20 - 9\ln 3\) (about 10.11 square units). ✓

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.