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2025 Paper 3 Q1
D: 1500.0 B: 1500.0
integration beta function trigonometric substitution substitution integration by parts symmetry definite integral Wallis integral

You need not consider the convergence of the improper integrals in this question. For \(p, q > 0\), define $$b(p,q) = \int_0^1 x^{p-1}(1-x)^{q-1} \, dx$$

  1. Show that \(b(p,q) = b(q,p)\).
  2. Show that \(b(p+1,q) = b(p,q) - b(p,q+1)\) and hence that \(b(p+1,p) = \frac{1}{2}b(p,p)\).
  3. Show that $$b(p,q) = 2\int_0^{\pi/2} (\sin\theta)^{2p-1}(\cos\theta)^{2q-1} \, d\theta$$ Hence show that \(b(p,p) = \frac{1}{2^{2p-1}}b(p,\frac{1}{2})\).
  4. Show that $$b(p,q) = \int_0^\infty \frac{t^{p-1}}{(1+t)^{p+q}} \, dt$$
  5. Evaluate $$\int_0^\infty \frac{t^{3/2}}{(1+t)^6} \, dt$$

Solution:

  1. \begin{align*} && b(p,q) &= \int_0^1 x^{p-1}(1-x)^{q-1}\, \d x \\ u = 1-x, \d u = -\d x && &= \int_{u=1}^{u = 0} (1-u)^{p-1}u^{q-1} (-1) \, \d u \\ &&&= \int_0^1 (1-u)^{p-1}u^{q-1} \d u \\ &&&= \int_0^1 u^{q-1}(1-u)^{p-1} \d u \\ &&&= b(q,p) \end{align*}
  2. \begin{align*} b(p+1,q) + b(p,q+1) &= \int_0^1 x^p(1-x)^{q-1} \d x + \int_0^1 x^{p-1}(1-x)^{q} \d x \\ &= \int_0^1 \left (x^p(1-x)^{q-1} + x^{p-1}(1-x)^{q}\right) \d x \\ &= \int_0^1 x^{p-1}(1-x)^{q-1} \left (x + (1-x) \right) \d x \\ &= \int_0^1 x^{p-1}(1-x)^{q-1} \d x \\ &= b(p,q) \end{align*} Therefore \(b(p+1,q) = b(p,q) - b(p,q+1)\), in particular \(2b(p+1,p) = b(p+1,p)+b(p,p+1) = b(p,p) \Rightarrow b(p+1,p) = \frac12 b(p,p)\) as required.
  3. \begin{align*} && b(p,q) &= \int_0^1 x^{p-1} (1-x)^{q-1} \d x \\ x = \sin^2 \theta, \d x = 2 \sin \theta \cos \theta \d \theta && &= \int_{u=0}^{u = \pi/2} \sin^{2p-2} \theta (1-\sin^2 \theta)^{q-1} \cdot 2 \sin \theta \cos \theta \d \theta \\ &&&= 2 \int_0^{\pi/2} \sin^{2p-1} \theta \cos^{2q-2} \cos \theta \d \theta \\ &&&= 2 \int_0^{\pi/2} \sin^{2p-1} \theta \cos^{2q-1} \theta \d \theta \end{align*} \begin{align*} b(p,p) &= 2\int_0^{\pi/2} (\sin \theta)^{2p-1}(\cos \theta)^{2p-1} \d \theta \\ &= 2 \int_0^{\pi/2} \left (\frac12 \sin 2\theta \right)^{2p-1} \d \theta \\ &= \frac1{2^{2p-1}} 2 \int_0^{\pi/2} (\sin 2 \theta)^{2p-1} \d \theta \\ &= \frac1{2^{2p-1}} 2 \int_{x=0}^{x=\pi} (\sin x)^{2p-1} 2 \d x\\ &= \frac1{2^{2p-1}} 2 \int_{x=0}^{x=\pi/2} (\sin x)^{2p-1} \d x\\ &= \frac1{2^{2p-1}} 2 \int_{0}^{\pi/2} (\sin x)^{2p-1} (\cos x)^{0} \d x\\ &= \frac1{2^{2p-1}} b(p,\tfrac12) \end{align*}
  4. \begin{align*} &&b(p,q) &= \int_0^1 x^{p-1}(1-x)^{q-1} \d x \\ t = \frac{x}{1-x}, \d t = (1-x)^{-2} \d x &&&= \int_{t=0}^{t = \infty} \left ( \frac{t}{1+t} \right)^{p-1} \left ( 1-\frac{t}{1+t} \right)^{q+1} \d t\\ x = \frac{t}{1+t} && &=\int_0^\infty t^{p-1} (1+t)^{-(p-1)-(q+1)} \d t \\ &&&= \int_0^{\infty} \frac{t^{p-1}}{(1+t)^{p+q}} \d t \end{align*}
  5. \begin{align*} I &= \int_0^\infty \frac{t^{3/2}}{(1+t)^6} \, dt \\ &= b( \tfrac52, \tfrac72) \\ &= b( \tfrac52, \tfrac52+1) \\ &= \tfrac12 b( \tfrac52, \tfrac52) \\ &= \frac12 \cdot \frac1{2^{4}} b(\tfrac52, \tfrac12) \\ &= \frac{1}{2^5} \cdot 2 \int_0^{\pi/2} (\sin \theta)^{4} \d \theta \\ &= \frac1{2^4} \int_0^{\pi/2}\left (\frac{1-\cos 2 \theta}{2} \right)^2 \d \theta \\ &= \frac1{2^6} \int_0^{\pi/2}\left (1 - 2 \cos 2 \theta + \cos^{2} 2 \theta \right) \d \theta \\ &= \frac1{2^6} \int_0^{\pi/2}\left (1 - 2 \cos 2 \theta + \frac{\cos 4 \theta + 1}{2} \right) \d \theta \\ &= \frac1{2^6} \left [\frac32 \theta - \sin 2 \theta + \frac18 \sin 4 \theta \right]_0^{\pi/2} \\ &= \frac1{2^6} \frac{3 \pi}{4} \\ &= \frac{3 \pi}{2^8} \end{align*}

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2024 Paper 3 Q12
D: 1500.0 B: 1500.0
cumulative distribution function geometric probability continuous probability distributions integration expected value trigonometric substitution hyperbolic substitution

  1. A point is chosen at random in the square \(0 \leqslant x \leqslant 1\), \(0 \leqslant y \leqslant 1\), so that the probability that a point lies in any region is equal to the area of that region. \(R\) is the random variable giving the distance of the point from the origin. Show that the cumulative distribution function of \(R\) is given by \[\mathrm{P}(R \leqslant r) = \sqrt{r^2 - 1} + \tfrac{1}{4}\pi r^2 - r^2 \cos^{-1}(r^{-1}),\] when \(1 \leqslant r \leqslant \sqrt{2}\). What is the cumulative distribution function when \(0 \leqslant r \leqslant 1\)?
  2. Show that \(\displaystyle\mathrm{E}(R) = \frac{2}{3}\int_1^{\sqrt{2}} \frac{r^2}{\sqrt{r^2-1}}\,\mathrm{d}r\).
  3. Show further that \(\mathrm{E}(R) = \frac{1}{3}\Bigl(\sqrt{2} + \ln\bigl(\sqrt{2}+1\bigr)\Bigr)\).

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2023 Paper 2 Q1
D: 1500.0 B: 1500.0
integration substitution definite integral trigonometric substitution rational functions symmetry

  1. Show that making the substitution \(x = \frac{1}{t}\) in the integral \[\int_a^b \frac{1}{(1+x^2)^{\frac{3}{2}}}\,\mathrm{d}x\,,\] where \(b > a > 0\), gives the integral \[\int_{b^{-1}}^{a^{-1}} \frac{-t}{(1+t^2)^{\frac{3}{2}}}\,\mathrm{d}t\,.\]
  2. Evaluate:
    1. \(\displaystyle\int_{\frac{1}{2}}^{2} \frac{1}{(1+x^2)^{\frac{3}{2}}}\,\mathrm{d}x\,;\)
    2. \(\displaystyle\int_{-2}^{2} \frac{1}{(1+x^2)^{\frac{3}{2}}}\,\mathrm{d}x\,.\)
    1. Show that \[\int_{\frac{1}{2}}^{2} \frac{1}{(1+x^2)^2}\,\mathrm{d}x = \int_{\frac{1}{2}}^{2} \frac{x^2}{(1+x^2)^2}\,\mathrm{d}x = \frac{1}{2}\int_{\frac{1}{2}}^{2} \frac{1}{1+x^2}\,\mathrm{d}x\,,\] and hence evaluate \[\int_{\frac{1}{2}}^{2} \frac{1}{(1+x^2)^2}\,\mathrm{d}x\,.\]
    2. Evaluate \[\int_{\frac{1}{2}}^{2} \frac{1-x}{x(1+x^2)^{\frac{1}{2}}}\,\mathrm{d}x\,.\]

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2016 Paper 3 Q1
D: 1700.0 B: 1500.0
integration reduction formula proof by induction trigonometric substitution improper integral completing the square binomial coefficient integration by parts

Let \[ \displaystyle I_n= \int_{-\infty}^\infty \frac 1 {(x^2+2ax+b)^n} \, \d x \] where \(a\) and \(b\) are constants with \(b > a^2\), and \(n\) is a positive integer.

  1. By using the substitution \(x + a = \sqrt{b- a^2} \, \tan u\,\), or otherwise, show that \[ I_1 = \dfrac \pi {\sqrt{b-a^2}}\, . \]
  2. Show that \(2n(b - a^2)\, I_{n+1} =(2n - 1) \, I_n\,\).
  3. Hence prove by induction that \[ I_n =\frac{\pi}{2^{2n-2}( b - a^2)^{n-\frac12}} \, \binom {2n-2}{n-1} \]

Solution:

  1. \(\,\) \begin{align*} && I_1 &= \int_{-\infty}^{\infty} \frac{1}{x^2+2ax+b} \d x \\ &&&= \int_{-\infty}^{\infty} \frac{1}{b-a^2 +(x+a)^2} \d x \\ &&&= \left [ \frac{1}{\sqrt{b-a^2}} \tan^{-1} \frac{x+a}{\sqrt{b-a^2}} \right]_{-\infty}^{\infty} \\ &&&= \frac{\pi}{\sqrt{b-a^2}} \end{align*}
  2. \(\,\) Here is the corrected LaTeX code for the second part, maintaining your exact styling and notation.
  3. \(\,\) \begin{align*} && I_{n} &= \int_{-\infty}^{\infty} \frac{1}{(x^2+2ax+b)^{n}} \d x \\ &&&= \left[ \frac{x}{(x^2+2ax+b)^n} \right]_{-\infty}^{\infty} - \int_{-\infty}^\infty x \cdot \frac{-n(2x+2a)}{(x^2+2ax+b)^{n+1}} \d x \\ &&&= 0 + n \int_{-\infty}^\infty \frac{2x^2+2ax}{(x^2+2ax+b)^{n+1}} \d x \\ &&&= n \int_{-\infty}^\infty \frac{2(x^2+2ax+b) - (2ax+2b)}{(x^2+2ax+b)^{n+1}} \d x \\ &&&= 2n I_n - n \int_{-\infty}^\infty \frac{2ax+2b}{(x^2+2ax+b)^{n+1}} \d x \\ &&&= 2n I_n - n \int_{-\infty}^\infty \frac{a(2x+2a) + 2(b-a^2)}{(x^2+2ax+b)^{n+1}} \d x \\ &&&= 2n I_n - n \int_{-\infty}^\infty \frac{a(2x+2a)}{(x^2+2ax+b)^{n+1}} \d x - 2n(b-a^2) I_{n+1} \\ &&&= 2n I_n - n \left[ \frac{-a}{n(x^2+2ax+b)^n} \right]_{-\infty}^\infty - 2n(b-a^2) I{n+1} \\ &&&= 2n I_n - 0 - 2n(b-a^2) I_{n+1} \\ \Rightarrow && 2n(b-a^2)I_{n+1} &= (2n-1)I_n \end{align*}
  4. \(\,\) \begin{align*} && I_{n+1} &= \frac{2n-1}{2n(b-a^2)} I_n \\ &&&= \frac{2n-1}{2n(b-a^2)} \cdot \frac{2n-3}{2(n-1)(b-a^2)} I_{n-1} \\ &&&= \frac{2n-1}{2n(b-a^2)} \cdot \frac{2n-3}{2(n-1)(b-a^2)} \cdots I_{1} \\ &&&= \frac{(2n-1)(2n-3) \cdots 1}{2n \cdot 2(n-1) \cdots 2 (b-a^2)^n} \frac{\pi}{\sqrt{b-a^2}} \\ &&&= \frac{(2n-1)(2n-3) \cdots 1}{2^n n!} \frac{\pi}{(b-a^2)^{n+\frac12}} \\ &&&= \frac{(2n-1)(2n-3) \cdots 1 \cdot 2n \cdot 2(n-1) \cdots 2}{2^{2n} n!n!} \frac{\pi}{(b-a^2)^{n+\frac12}} \\ &&&= \frac{(2n)!}{2^{2n}n!n!}\frac{\pi}{(b-a^2)^{n+\frac12}} \\ &&&= \frac{\pi}{2^{2n}(b-a^2)^{n+\frac12}} \binom{2n}{n} \\ \Rightarrow && I_n &= \frac{\pi}{2^{2n-2}(b-a^2)^{n-\frac12}} \binom{2n-2}{n-1} \\ \end{align*}

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2013 Paper 2 Q2
D: 1600.0 B: 1500.0
integration reduction formula substitution proof by induction factorial Beta function trigonometric substitution

For \(n\ge 0\), let \[ I_n = \int_0^1 x^n(1-x)^n\d x\,. \]

  1. For \(n\ge 1\), show by means of a substitution that \[ \int_0^1 x^{n-1}(1-x)^n\d x = \int_0^1 x^n(1-x)^{n-1}\d x\, \] and deduce that \[ 2 \int_0^1 x^{n-1}(1-x)^n\d x = I_{n-1}\,. \] Show also, for \(n\ge1\), that \[ I_n = \frac n {n+1} \int_0^1 x^{n-1} (1-x)^{n+1} \d x \] and hence that \(I_n = \dfrac{n}{2(2n+1)} I_{n-1}\,.\)
  2. When \(n\) is a positive integer, show that \[ I_n = \frac{(n!)^2}{(2n+1)!}\,. \]
  3. Use the substitution \(x= \sin^2 \theta\) to show that \(I_{\frac12}= \frac \pi 8\), and evaluate \(I_{\frac32}\).

Solution:

  1. \(\,\) \begin{align*} u = 1-x, \d u = -\d x && \int_0^1 x^{n-1}(1-x)^n \d x &= \int_{u=1}^{u=0} (1-u)^{n-1}u^n (-1) \d u \\ &&&= \int_0^1 u^n (1-u)^{n-1} \d u \\ &&&= \int_0^1 x^n (1-x)^{n-1} \d x \\ \\ \Rightarrow && 2\int_0^1 x^{n-1}(1-x)^n \d x &= \int_0^1 \left ( x^{n-1}(1-x)^n + x^{n}(1-x)^{n-1} \right)\d x \\ &&&= \int_0^1x^{n-1}(1-x)^{n-1} \left ( (1-x) + x \right) \d x\\ &&&= I_{n-1} \\ \\ && I_n &= \left [x^n \cdot (-1)\frac{(1-x)^{n+1}}{n+1}\right]_0^1 + \int_0^1 n x^{n-1} \frac{(1-x)^{n+1}}{n+1} \d x\\ &&&= \frac{n}{n+1} \int_0^1 x^{n-1} (1-x)^{n+1} \d x \\ \\ && I_n &= \frac{n}{n+1} \int_0^1 x^{n-1} (1-x)^{n+1} \d x \\ &&&= \frac{n}{n+1} \int_0^1 \left ( x^{n-1} (1-x)^{n} - x^n(1-x)^n \right) \d x \\ &&&= \frac{n}{n+1} \left (\frac12 I_{n-1} - I_n \right) \\ \Rightarrow && I_n \cdot \left ( \frac{2n+1}{n+1} \right) &= \frac{n}{2(n+1)} I_{n-1}\\ \Rightarrow && I_n &= \frac{n}{2(2n+1)} I_{n-1} \end{align*}
  2. \(\,\) \begin{align*} && I_0 &= \int_0^1 1 \d x = 1 \\ \Rightarrow && I_1 &= \frac{1}{2 \cdot 3} \\ && I_n &= \frac{n}{2(2n+1)} \cdot \frac{n-1}{2(2n-1)}\cdot \frac{n-2}{2(2n-3)} \cdots \frac{1}{2 \cdot 3} \\ &&&= \frac{n!}{2^n (2n+1)(2n-1)(2n-3) \cdots 3} \\ &&&= \frac{n! (2n)(2n-2)\cdots 2}{2^n (2n+1)!} \\ &&&= \frac{(n!)^2 2^n}{2^n(2n+1)!} \\ &&&= \frac{(n!)^2}{(2n+1)^2} \end{align*}
  3. \(\,\) \begin{align*} && I_{\frac12} &= \int_0^1 \sqrt{x(1-x)} \d x\\ x = \sin^2 \theta, \d x = 2 \sin \theta \cos \theta \d \theta: &&&= \int_{\theta =0}^{\theta = \frac{\pi}{2}} \sin \theta \cos \theta 2 \sin \theta \cos \theta \d \theta \\ &&&= \frac12 \int_0^{\pi/2} \sin^2 2 \theta \d \theta \\ &&&= \frac12 \int_0^{\pi/2} \frac{1-\cos 2 \theta}{2} \d \theta \\ &&&= \frac14 \left [\theta - \frac12 \sin 2 \theta \right]_0^{\pi/2} \\ &&&= \frac{\pi}{8} \\ \\ && I_{\frac32} &= \frac{3/2}{2 \cdot ( 2 \cdot \frac32 + 1)} I_{\frac12} \\ &&&= \frac{3}{4 \cdot 4} \frac{\pi}{8} \\ &&&= \frac{3 \pi}{128} \end{align*}

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2012 Paper 2 Q3
D: 1600.0 B: 1516.0
integration substitution trigonometric substitution rational function definite integral algebraic manipulation t-substitution

Show that, for any function f (for which the integrals exist), \[ \int_0^\infty \f\big(x+\sqrt{1+x^2}\big) \,\d x = \frac12 \int_1^\infty \left(1+\frac 1 {t^2}\right) \f(t)\, \d t \,. \] Hence evaluate \[ \int_0^\infty \frac1 {2x^2 +1 + 2 x\sqrt{x^2+1} \ } \, \, \d x \,, \] and, using the substitution \(x=\tan\theta\), \[ \int_0^{\frac12\pi} \frac{1}{(1+\sin\theta)^3}\,\d \theta \,. \]

Solution: \begin{align*} && t &= x + \sqrt{1+x^2} \\ &&\frac1t &= \frac{1}{x+\sqrt{1+x^2}} \\ &&&= \frac{\sqrt{1+x^2}-x}{1+x^2-1} \\ &&&= \sqrt{1+x^2}-x \\ \Rightarrow && x &=\frac12 \left ( t - \frac1t\right) \\ \Rightarrow && \d x &=\frac12 \left (1 + \frac1{t^2} \right)\d t \\ \\ \Rightarrow && \int_0^\infty \f\big(x+\sqrt{1+x^2}\big) \,\d x &= \int_{t=1}^{t = \infty}f(t) \frac12\left (1 + \frac1{t^2} \right)\d t \\ &&&= \frac12 \int_1^\infty \left (1 + \frac1{t^2} \right)f(t) \d t \end{align*} \begin{align*} && I &= \int_0^\infty \frac1 {2x^2 +1 + 2 x\sqrt{x^2+1} \ } \d x \\ &&&= \int_0^\infty \frac1 {(x+\sqrt{x^2+1})^2} \d x \\ &&&= \frac12 \int_1^\infty \left (1 + \frac1{t^2} \right) \frac{1}{t^2} \d t \\ &&&= \frac12 \left [-\frac1t-\frac13\frac1{t^3} \right]_1^{\infty} \\ &&&= \frac12 \cdot \frac43 = \frac23 \end{align*} \begin{align*} && J &= \int_0^{\frac12 \pi} \frac{1}{(1+\sin \theta)^3} \d \theta \\ &&x &= \tan \theta\\ && \d x &= \sec^2 \theta = (1+x^2) \d \theta\\ && \tan\theta &= \frac{s}{\sqrt{1-s^2}}\\ \Rightarrow && \tan^2 \theta &= \frac{s^2}{1-s^2} \\ \Rightarrow && \sin \theta &= \frac{\tan\theta}{\sqrt{1+\tan^2\theta}} \\ && J &= \int_0^{\frac12 \pi} \frac{1}{(1+\sin \theta)^3} \d \theta \\ &&&= \int_0^{\frac12 \pi} \frac{1}{\left (1+ \frac{\tan\theta}{\sqrt{1+\tan^2\theta}} \right )^3} \d \theta \\ &&&= \int_{x=0}^{x=\infty} \frac{1}{\left(1 + \frac{x}{\sqrt{1+x^2}} \right)^3} \frac{1}{1+x^2} \d x \\ &&&= \int_0^{\infty} \frac{\sqrt{1+x^2}}{(\sqrt{1+x^2}+x)^3} \d x \\ &&J_a &= \int_0^{\infty} \frac{\sqrt{1+x^2}+x}{(\sqrt{1+x^2}+x)^3} \d x \\ &&&= \frac23 \\ &&J_b &= \int_0^{\infty} \frac{\sqrt{1+x^2}-x}{(\sqrt{1+x^2}+x)^3} \d x \\ &&&= \int_0^{\infty} \frac{1}{(\sqrt{1+x^2}+x)^4} \d x\\ &&&= \frac12\int_1^{\infty} \left (1 +\frac1{t^2} \right)\frac{1}{t^4} \d t \\ &&&= \frac12 \left [-\frac13 t^{-3}-\frac15t^{-5} \right]_1^{\infty} \\ &&&= \frac12 \cdot \frac8{15} = \frac4{15} \\ \Rightarrow && J &= \frac12(J_a+J_b) = \frac7{15} \end{align*}

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2007 Paper 2 Q3
D: 1600.0 B: 1575.2
integration trigonometric substitution t-substitution Weierstrass substitution arctan definite integral partial fractions

By writing \(x=a\tan\theta\), show that, for \(a\ne0\), $\displaystyle \int \frac 1 {a^2+x^2}\, \d x =\frac 1 a \arctan \frac x a + \text{constant}\,$.

  1. Let $\displaystyle I=\int_0^{\frac{1}{2}\pi} \frac {\cos x}{1+\sin^2 x} \, \d x\,$.
    1. Evaluate \(I\).
    2. Use the substitution \(t=\tan \frac12 x\) to show that \(\displaystyle \int_0^1 \frac {1-t^2}{1+6t^2+t^4} \, \d t = \tfrac12 I\,\).
  2. Evaluate \(\displaystyle \int_0^1 \frac {1-t^2}{1+14t^2+t^4} \, \d t \,\).

Solution: \begin{align*} && I &= \int \frac{1}{a^2+x^2} \d x\\ x = a \tan \theta, \d x =a \sec^2 \theta \d \theta &&&= \int \frac{1}{a^2+a^2\tan^2 x} a \sec^2 \theta \d \theta \\ &&&=\int \frac{\sec^2 \theta}{a \sec^2 \theta} \d \theta \\ &&&= \frac1a \theta + C \\ &&&= \frac1a \arctan \frac{x}{a} + C \end{align*}

    1. \(\,\) \begin{align*} && I &= \int_0^{\frac{1}{2}\pi} \frac {\cos x}{1+\sin^2 x} \d x \\ &&&= \left [ \arctan (\sin x) \right]_0^{\pi/2} \\ &&&= \arctan(1) - \arctan(0) = \frac{\pi}{4} \end{align*}
    2. \(\,\) \begin{align*} && t &= \tan \frac{x}{2} \\ \Rightarrow && \sin x &= \frac{2t}{1+t^2} \\ && \cos x &= \frac{1-t^2}{1+t^2} \\ && \d x &= \frac{2}{1+t^2} \d t \\ \Rightarrow && I &= \int_0^{\pi/2} \frac{\cos x}{1 + \sin^2 x } \d x \\ &&&= \int_{t=0}^{t = 1} \frac{\frac{1-t^2}{1+t^2}}{1 + \left (\frac{2t}{1+t^2} \right)^2} \frac{2}{1+t^2} \d t \\ &&&= 2 \int_0^1 \frac{1-t^2}{(1+t^2)^2+(2t)^2} \d t\\ &&&= 2 \int_0^1 \frac{1-t^2}{1+6t^2+t^4} \d t\\ \end{align*} From which the conclusion follows
  2. \(\,\) \begin{align*} && J &= \int_0^1 \frac {1-t^2}{1+14t^2+t^4} \, \d t \\ &&&= \int_0^1 \frac {\frac{1-t^2}{1+t^2}}{\frac{1+14t^2+t^4}{(1+t^2)^2}} \frac{1}{1+t^2} \, \d t \\ &&&= \int_0^1 \frac {\frac{1-t^2}{1+t^2}}{\frac{(t^2+1)^2+3(2t)^2}{(1+t^2)^2}} \frac{1}{1+t^2} \, \d t \\ &&&= \frac12\int_{x=0}^{x=\pi/2} \frac {\cos x}{1+3 \sin^2 x} \d x \\ &&&= \frac{1}{6}\left[ \sqrt{3} \arctan(\sin \sqrt{3}x)\right]_0^{\pi/2} \\ &&&= \frac16 \sqrt{3} \frac{\pi}{3} \\ &&&= \frac{\sqrt{3}\pi}{18} \end{align*}

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2004 Paper 1 Q4
D: 1516.0 B: 1484.0
integration trigonometric substitution sec substitution definite integral completing the square algebraic manipulation differentiation

Differentiate \(\sec {t}\) with respect to \(t\).

  1. Use the substitution \(x=\sec t\) to show that $\displaystyle \int^2_{\sqrt 2} \frac{1}{ x^3\sqrt {x^2-1} } \; \mathrm{d}x =\frac{\sqrt 3 - 2}{8} + \frac {\pi}{24} \;.$
  2. Determine $\displaystyle \int \frac{1} {( x+2) \sqrt {(x+1)(x+3)} } \; \mathrm{d}x \;$.
  3. Determine $\displaystyle \int \frac {1} {(x+2) \sqrt {x^2+4x-5} } \; \mathrm{d}x \;$.

Solution: \[\frac{\d}{\d t} \left ( \sec t \right) = \frac{\sin t }{\cos^2 t} = \sec t \tan t \]

  1. \(\,\) \begin{align*} && I_1 &= \int_{\sqrt{2}}^2 \frac{1}{x^3 \sqrt{x^2-1}} \\ x = \sec t, \d x = \sec t \tan t:&&&= \int_{t=\pi/4}^{t=\pi/3} \frac{1}{\sec^3 t \tan t} \sec t \tan t \d t \\ &&&= \int_{t=\pi/4}^{t=\pi/3} \cos^2 t \d t \\ &&&= \int_{t=\pi/4}^{t=\pi/3} \frac{1+\cos 2t}{2} \d t \\ &&&= \frac12 \frac{\pi}{12} + \frac12 \left (\sin \frac{\pi}{3} - \sin \frac{\pi}{4} \right) \\ &&&= \frac{\pi}{24} + \frac{\sqrt{3}-2}{8} \\ \end{align*}
  2. \(\,\) \begin{align*} && I_2 &= \int \frac{1}{(x+2)\sqrt{(x+1)(x+3)}} \d x \\ &&&= \int \frac{1}{(x+2)\sqrt{(x+2)^2-1}} \d x \\ &&&= \int \frac{1}{u\sqrt{u^2-1}} \d u \\ &&&= \sec^{-1} u + C \\ &&&= \sec^{1} (x+2) + C \end{align*}
  3. \(\,\) \begin{align*} && I_3 &= \int \frac{1}{(x+2)\sqrt{(x+2)^2 - 9}} \d x \\ &&&= \int\frac{1}{9(\frac{x+2}{3})\sqrt{(\frac{x+2}3)^2 - 1}} \d x \\ u = \frac{x+2}{3}, 3\d u =\d x &&&= \frac19 \int \frac{1}{u\sqrt{u^2-1}} 3 \d u \\ &&&= \frac13 \sec^{-1} u + C \\ &&&= \frac13 \sec^{-1} \frac{x+2}{3} + C \end{align*}

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2000 Paper 3 Q2
D: 1700.0 B: 1484.2
integration trigonometric substitution definite integral square root generalisation trigonometric identities

Use the substitution \(x = 2-\cos \theta \) to evaluate the integral $$ \int_{3/2}^2 \left(x - 1 \over 3 - x\right)^{\!\frac12}\! \d x. $$ Show that, for \(a < b\), $$ \int_p^q \left( x - a \over b - x\right)^{\!\frac12} \!\d x = \frac{(b-a)(\pi +3{\surd3} -6)}{12}, $$ where \(p= {(3a+b)/4}\) and \(q={(a+b)/2}\).

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1998 Paper 3 Q11
D: 1700.0 B: 1500.0
SHM simple pendulum large amplitude energy conservation elliptic integral trigonometric substitution binomial expansion approximation period

Consider a simple pendulum of length \(l\) and angular displacement \(\theta\), which is {\bf not} assumed to be small. Show that $$ {1\over 2}l \left({\d\theta\over \d t}\right)^2 = g(\cos\theta -\cos\gamma)\,, $$ where \(\gamma\) is the maximum value of \(\theta\). Show also that the period \(P\) is given by $$ P= 2 \sqrt{l\over g} \int_0^\gamma \left( \sin^2(\gamma/2)-\sin^2(\theta/2) \right)^{-{1\over 2}} \,\d\theta \,. $$ By using the substitution \(\sin(\theta/2)=\sin(\gamma/2) \sin\phi\), and then finding an approximate expression for the integrand using the binomial expansion, show that for small values of \(\gamma\) the period is approximately $$ 2\pi \sqrt{l\over g} \left(1+{\gamma^2\over 16}\right) \,. $$

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1995 Paper 3 Q2
D: 1700.0 B: 1586.3
integration reduction formula definite integral trigonometric substitution recurrence relation Wallis integral beta function

If \[ \mathrm{I}_{n}=\int_{0}^{a}x^{n+\frac{1}{2}}(a-x)^{\frac{1}{2}}\,\mathrm{d}x, \] show that \(\mathrm{I}_{0}=\pi a^{2}/8.\) Show that \((2n+4)\mathrm{I}_{n}=(2n+1)a\mathrm{I}_{n-1}\) and hence evaluate \(\mathrm{I}_{n}\).

Solution: \begin{align*} && I_n &= \int_{0}^{a}x^{n+\frac{1}{2}}(a-x)^{\frac{1}{2}}\,\mathrm{d}x\\ && I_0 &= \int_0^a x^{\frac12}(a-x)^{\frac12} \d x \\ x = a \sin^2 \theta, \d x = 2a \sin \theta \cos \theta \d \theta &&&= \int_{\theta =0}^{\theta = \pi/2} \sqrt{a}\sin \theta\sqrt{a} \cos \theta 2a \sin \theta \cos \theta \d \theta \\ &&&= \frac{a^2}{2} \int_0^{\pi/2} \sin^2 2 \theta \d \theta \\ &&&= \frac{a^2}{4} \int_0^{\pi/2}(1- \underbrace{\cos 4\theta}_{\text{runs round the whole unit circle}}) \d \theta \\ &&&= \frac{\pi a^2}{8} \\ \\ && I_n &= \int_0^a x^{n+\frac12}(a-x)^{\frac12} \d x \\ &&&=\underbrace{\left [-\frac23x^{n+\frac12}(a-x)^\frac32 \right]_0^a}_{=0} + \frac23 \left(n+\frac12\right) \int_0^ax^{n-1+\frac12}(a-x)^\frac32 \d x \\ &&&= \frac23 \left(n+\frac12\right) \int_0^ax^{n-1+\frac12}(a-x)(a-x)^\frac12 \d x \\ &&&= \frac23 \left(n+\frac12\right)aI_{n-1}-\frac23 \left(n+\frac12\right)I_{n} \\ \Rightarrow && \left(n+\frac12+\frac32\right)I_{n} &= \left(n+\frac12\right)aI_{n-1}\\ \Rightarrow && (2n+4)I_n &= (2n+1)aI_{n-1} \\ \\ \Rightarrow && I_n &= \frac{2n+1}{2n+4}a I_{n-1} \\ &&&=\frac{2n+1}{2n+4}\frac{2n-1}{2n+2}a^2 I_{n-2} \\ &&&= \frac{(2n+1)!!}{(2n+4)!!} \pi a^{n+2} \end{align*}

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1994 Paper 2 Q4
D: 1600.0 B: 1500.0
integration area under curve circle ellipse trigonometric substitution geometric proof intersection of regions inverse trigonometric functions

By considering the area of the region defined in terms of Cartesian coordinates \((x,y)\) by \[ \{(x,y):\ x^{2}+y^{2}=1,\ 0\leqslant y,\ 0\leqslant x\leqslant c\}, \] show that \[ \int_{0}^{c}(1-x^{2})^{\frac{1}{2}}\,\mathrm{d}x=\tfrac{1}{2}[c(1-c^{2})^{\frac{1}{2}}+\sin^{-1}c], \] if \(0 < c\leqslant1.\) Show that the area of the region defined by \[ \left\{ (x,y):\ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,\ 0\leqslant y,\ 0\leqslant x\leqslant c\right\} , \] is \[ \frac{ab}{2}\left[\frac{c}{a}\left(1-\frac{c^{2}}{a^{2}}\right)^{\frac{1}{2}}+\sin^{-1}\left(\frac{c}{a}\right)\right], \] if \(0 < c\leqslant a\) and \(0 < b.\) Suppose that \(0 < b\leqslant a.\) Show that the area of intersection \(E\cap F\) of the two regions defined by \[ E=\left\{ (x,y):\ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\leqslant1\right\} \qquad\mbox{ and }\qquad F=\left\{ (x,y):\ \frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}\leqslant1\right\} \] is \[ 4ab\sin^{-1}\left(\frac{b}{\sqrt{a^{2}+b^{2}}}\right). \]

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1988 Paper 1 Q16
D: 1500.0 B: 1498.6
probability continuous probability distribution probability density function conditional probability Bayes' theorem integration trigonometric substitution rust/decay model

Wondergoo is applied to all new cars. It protects them completely against rust for three years, but thereafter the probability density of the time of onset of rust is proportional to \(t^{2}/(1+t^{2})^{2}\) for a car of age \(3+t\) years \((t\geqslant0)\). Find the probability that a car becomes rusty before it is \(3+t\) years old. Every car is tested for rust annually on the anniversary of its manufacture. If a car is not rusty, it will certainly pass; if it is rusty, it will pass with probability \(\frac{1}{2}.\) Cars which do not pass are immediately taken off the road and destroyed. What is the probability that a randomly selected new car subsequently fails a test taken on the fifth anniversary of its manufacture? Find also the probability that a car which was destroyed immediately after its fifth anniversary test was rusty when it passed its fourth anniversary test.

Solution: Given the probability density after \(3\) years is proportional to \(\frac{t^2}{(1+t^2)^2}\) then we must have that: \begin{align*} && 1 &= A \int_0^{\infty} \frac{t^2}{(1+t^2)^2} \, \d t \\ &&&= A \left [ -\frac12 \frac{t}{1+t^2} \right]_0^{\infty} + \frac{A}2 \int_0^{\infty} \frac{1}{1+t^2} \d t \\ &&&= \frac{A}{2} \frac{\pi}{2} \\ \Rightarrow && A &= \frac{4}{\pi} \end{align*} In order to fail a test on the fifth anniversary, there are two possibilities for when we went faulty. We could have gone faulty before \(4\) years, got lucky once and then failed the second test, or gone faulty in the next year and then failed the first test. \begin{align*} \P(\text{rusty before } 4 \text{ years}) &=\frac{4}{\pi} \int_0^1 \frac{t^2}{(1+t^2)^2} \d t \\ &= \frac{4}{\pi} \left [ -\frac12 \frac{t}{1+t^2} \right]_0^{1} + \frac{2}{\pi} \int_0^{1} \frac{1}{1+t^2} \d t \\ &= -\frac{1}{\pi} + \frac{2}{\pi} \frac{\pi}{4} \\ &= \frac12 - \frac{1}{\pi} \\ &\approx 0.181690\cdots \\ \\ \P(\text{rusty before } 5 \text{ years}) &=\frac{4}{\pi} \int_0^1 \frac{t^2}{(1+t^2)^2} \d t \\ &= \frac{4}{\pi} \left [ -\frac12 \frac{t}{1+t^2} \right]_0^{2} + \frac{2}{\pi} \int_0^{2} \frac{1}{1+t^2} \d t \\ &= -\frac{4}{5\pi} + \frac{2}{\pi} \tan^{-1} 2 \\ &\approx 0.450184\cdots \\ \end{align*} Therefore: \begin{align*} \P(\text{fails 5th anniversary}) &= \P(\text{rusty before } 4 \text{ years}) \P(\text{pass one, fail other}) + \\ & \quad \quad + \P(\text{rusty between 4 and 5 years}) \P(\text{fail}) \\ &= 0.181690\cdots \cdot \frac{1}{4} + \frac{1}{2} ( 0.450184\cdots- 0.181690\cdots) \\ &= \frac{1}{2} 0.450184\cdots - \frac{1}{4} 0.181690\cdots \\ &= 0.1796688\cdots \\ &= 18.0\%\,\, (3\text{ s.f.}) \end{align*} We also must have that: \begin{align*} \P(\text{rusty at 4 years}|\text{destroyed at 5}) &= \frac{\P(\text{rusty at 4 years and destroyed at 5})}{\P(\text{destroyed at 5})} \\ &= \frac{0.181690\cdots \cdot \frac{1}{4}}{\frac{1}{2} 0.450184\cdots - \frac{1}{4} 0.181690\cdots} \\ &= 0.252811\cdots \\ &= 25.3\%\,\,(3\text{ s.f.}) \end{align*}

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1988 Paper 2 Q7
D: 1600.0 B: 1500.0
integration substitution trigonometric substitution t-substitution symmetry rational function definite integral change of limits

The integral \(I\) is defined by \[ I=\int_{1}^{2}\frac{(2-2x+x^{2})^{k}}{x^{k+1}}\,\mathrm{d}x \] where \(k\) is a constant. Show that \[ I=\int_{0}^{1}\frac{(1+x^{2})^{k}}{(1+x)^{k+1}}\,\mathrm{d}x=\int_{0}^{\frac{1}{4}\pi}\frac{\mathrm{d}\theta}{\left[\sqrt{2}\cos\theta\cos\left(\frac{1}{4}\pi-\theta\right)\right]^{k+1}}=2\int_{0}^{\frac{1}{8}\pi}\frac{\mathrm{d}\theta}{\left[\sqrt{2}\cos\theta\cos\left(\frac{1}{4}\pi-\theta\right)\right]^{k+1}}. \] Hence show that \[ I=2\int_{0}^{\sqrt{2}-1}\frac{(1+x^{2})^{k}}{(1+x)^{k+1}}\,\mathrm{d}x \] Deduce that \[ \int_{1}^{\sqrt{2}}\left(\frac{2-2x^{2}+x^{4}}{x^{2}}\right)^{k}\frac{1}{x}\,\mathrm{d}x=\int_{1}^{\sqrt{2}}\left(\frac{2-2x+x^{2}}{x}\right)^{k}\frac{1}{x}\,\mathrm{d}x \]

Solution: \begin{align*} I &=\int_{1}^{2}\frac{(2-2x+x^{2})^{k}}{x^{k+1}}\,\mathrm{d}x \\ u = x-1 &, \quad \d u = \d x \\ &= \int_{u = 0}^{u=1} \frac{(u^2+1)^k}{(u+1)^{k+1}} \d u \\ &= \boxed{\int_0^1 \frac{(1+x^2)^k}{(1+x)^{k+1}} \d x} \\ x = \tan \theta &, \quad \d x = \sec^2 \theta \d \theta \\ &= \int_{\theta = 0}^{\theta = \pi/4} \frac{\sec^{2k+2} \theta }{(1 + \tan \theta)^{k+1}} \d \theta \\ &= \int_0^{\pi/4} \frac{\d \theta}{\cos^{2k+2} \theta (\frac{\sin \theta + \cos \theta}{\cos \theta})^{k+1}} \\ &= \int_0^{\pi/4} \frac{\d \theta}{\cos^{k+1} \theta ({\sin \theta + \cos \theta})^{k+1}} \\ &= \int_0^{\pi/4} \frac{\d \theta}{\cos^{k+1} \theta (\sqrt{2} \cos (\frac{\pi}{4} - \theta))^{k+1}} \\ I &= \boxed{ \int_0^{\pi/4} \frac{\d \theta}{(\sqrt{2}\cos \theta \cos (\frac{\pi}{4} - \theta))^{k+1}}} \\ \end{align*} Since \(f(\theta) = \cos \theta \cos (\frac{\pi}{4} - \theta)\) is symmetric about \(\frac{\pi}{8}\) this integral is twice the integral to \(\frac{\pi}{8}\). \(\tan 2 \theta = \frac{2\tan \theta}{1 - \tan^2 \theta} \Rightarrow 1 = \frac{2 \tan \frac{\pi}{8}}{1 - \tan^2 \frac{\pi}{8}} \Rightarrow \tan \frac{\pi}{8} = \sqrt{2}-1\). Therefore, using the same substitution we must have: \[ I=2\int_{0}^{\sqrt{2}-1}\frac{(1+x^{2})^{k}}{(1+x)^{k+1}}\,\mathrm{d}x \] Let \(u = x^2\), then \(\d u = 2 x\d x\) \begin{align*} \int_{1}^{\sqrt{2}}\left(\frac{2-2x^{2}+x^{4}}{x^{2}}\right)^{k}\frac{1}{x}\,\mathrm{d}x &= \int_{u = 1}^{u = 2} \l \frac{2-2u+u^2}{u}\r^k \frac{1}{2u} \d u \\ &= \frac12 I \\ &= \int_{0}^{\sqrt{2}-1}\frac{(1+x^{2})^{k}}{(1+x)^{k+1}}\,\mathrm{d}x \\ u = 1+x & \quad \d u = \d x \\ &= \int_1^{\sqrt{2}} \frac{(1+(u-1)^2)^k}{u^{k+1}} \d u \\ &= \int_{1}^{\sqrt{2}}\left(\frac{2-2u+u^{2}}{u}\right)^{k}\frac{1}{u}\,\mathrm{d}x \\ &= \int_{1}^{\sqrt{2}}\left(\frac{2-2x+x^{2}}{x}\right)^{k}\frac{1}{x}\,\mathrm{d}x \end{align*}

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1987 Paper 1 Q5
D: 1500.0 B: 1500.0
integration trigonometric substitution definite integral square root algebraic manipulation Weierstrass substitution

Using the substitution \(x=\alpha\cos^{2}\theta+\beta\sin^{2}\theta,\) show that, if \(\alpha<\beta\), \[ \int_{\alpha}^{\beta}\frac{1}{\sqrt{(x-\alpha)(\beta-x)}}\,\mathrm{d}x=\pi. \] What is the value of the above integral if \(\alpha>\beta\)? Show also that, if \(0<\alpha<\beta\), \[ \int_{\alpha}^{\beta}\frac{1}{x\sqrt{(x-\alpha)(\beta-x)}}\,\mathrm{d}x=\frac{\pi}{\sqrt{\alpha\beta}}. \]

Solution: Using the suggested substitution, we can find. \begin{align*} && x &=\alpha\cos^{2}\theta+\beta\sin^{2}\theta \\ && x-\alpha &=\alpha(\cos^{2}\theta-1)+\beta\sin^{2}\theta \\ &&& = (\beta - \alpha) \sin^2 \theta \\ && \beta - x &= -\alpha\cos^{2}\theta+\beta(1-\sin^{2}\theta) \\ &&&= (\beta-\alpha)\cos^2 \theta \\ && x &=\alpha\cos^{2}\theta+\beta\sin^{2}\theta \\ \Rightarrow && \frac{dx}{d\theta} &= (\beta - \alpha) 2 \cos \theta \sin\theta \\ \\ &&\int_{\alpha}^{\beta}\frac{1}{\sqrt{(x-\alpha)(\beta-x)}}\,\mathrm{d}x &= \int_0^{\pi/2} \frac{1}{(\beta - \alpha)\sin\theta \cos \theta} (\beta - \alpha) 2 \cos \theta \sin \theta \, d \theta \\ &&&= \int_0^{\pi/2} \frac{1}{\bcancel{(\beta - \alpha)}\bcancel{\sin\theta \cos \theta}} \bcancel{(\beta - \alpha)} 2 \bcancel{\cos \theta \sin \theta} \, d \theta \\ &&&= \int_0^{\pi/2} 2 d \theta \\ && &= 2 \frac{\pi}{2} = \boxed{\pi} \end{align*} If \(\alpha > \beta\) we can rewrite the integral as: \begin{align*} \int_{\alpha}^{\beta}\frac{1}{\sqrt{(x-\alpha)(\beta-x)}}\,\mathrm{d}x &= \int_{\alpha}^{\beta}\frac{1}{\sqrt{(x-\beta)(\alpha-x)}}\,\mathrm{d}x \\ &= -\int_{\beta}^{\alpha}\frac{1}{\sqrt{(x-\beta)(\alpha-x)}}\,\mathrm{d}x \\ &= -\pi \end{align*} Where the last step we are directly using the first integral with the use of \(\alpha\) and \(\beta\) reversed. Finally, using the substitution \(xt = 1\), we fortunately lose the \(\frac1{x}\) term: \begin{align*} && x &= \frac{1}{t} \\ && \frac{dx}{dt} &= -\frac1{t^2} \\ \\ && \int_{\alpha}^{\beta}\frac{1}{x\sqrt{(x-\alpha)(\beta-x)}}\,\mathrm{d}x &= \int_{\alpha}^{\beta}\frac{t}{\sqrt{(\frac{1}{t}-\alpha)(\beta-\frac{1}{t})}} \frac{-1}{t^2}\,\mathrm{d}t \\ && &= \int_{\frac1{\alpha}}^{\frac1\beta}\frac{-1}{\sqrt{(1-t\alpha)(t\beta-1)}}\,\mathrm{d}t \\ && &= \int_{\frac1{\alpha}}^{\frac1\beta}\frac{-1}{\sqrt{\alpha\beta}\sqrt{(\frac1{\alpha}-t)(t-\frac1{\beta})}}\,\mathrm{d}t \\ && &= \frac1{\sqrt{\alpha\beta}}\int_{\frac1{\alpha}}^{\frac1\beta}\frac{-1}{\sqrt{(\frac1{\alpha}-t)(t-\frac1{\beta})}}\,\mathrm{d}t \\ &&&= \boxed{\frac{\pi}{\sqrt{\alpha\beta}}} \end{align*} Where again the last step we are using the intermediate integral, with the roles of \(\alpha\) and \(\beta\) replaced with \(\frac{1}{\beta}\) and \(\frac1{\alpha}\)

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