Integration

Showing 76-100 of 112 problems
2000 Paper 3 Q2
D: 1700.0 B: 1484.2

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 \(q={(a+b)/2}\).

1999 Paper 1 Q8
D: 1500.0 B: 1500.0

The function \(\f\) satisfies \(0\leqslant\f(t)\leqslant K\) when \(0\leqslant t\leqslant x\). Explain by means of a sketch, or otherwise, why \[0\leqslant\int_{0}^{x} \f (t)\,{\mathrm d}t \leqslant Kx.\] By considering \(\displaystyle \int_{0}^{1}\frac{t}{n(n-t)}\,{\mathrm d}t\), or otherwise, show that, if \(n>1\), \[ 0\le \ln \left( \frac n{n-1}\right) -\frac 1n \le \frac 1 {n-1} - \frac 1n \] and deduce that \[ 0\le \ln N -\sum_{n=2}^N \frac1n \le 1. \] Deduce that as \(N\to \infty\) \[ \sum_{n=1}^N \frac1n \to\infty. \] Noting that \(2^{10}=1024\), show also that if \(N<10^{30}\) then \[ \sum_{n=1}^N \frac1n <101. \]

1999 Paper 2 Q6
D: 1600.0 B: 1484.0

Find \(\displaystyle \ \frac{\d y}{\d x} \ \) if $$ y = \frac{ax+b}{cx+d}. \tag{*} $$ By using changes of variable of the form \((*)\), or otherwise, show that \[ \int_0^1 \frac{1}{(x+3)^2} \; \ln \left(\frac{x+1}{x+3}\right)\d x = {\frac16} \ln3 - {\frac14}\ln 2 - \frac 1{12}, \] and evaluate the integrals \[ \int_0^1 \frac{1}{(x+3)^2} \; \ln \left(\frac{x^2+3x+2}{(x+3)^2}\right)\d x \mbox{ and } \int_0^1 \frac{1}{(x+3)^2} \; \ln\left(\frac{x+1}{x+2}\right)\d x . \] [Not on original paper:] By changing to the variable \(y\) defined by $$ y=\frac{2x-3}{x+1},$$ evaluate the integral $$ \int_2^4 \frac{2x-3}{(x+1)^3}\; \ln\!\left(\frac{2x-3}{x+1}\right)\d x.$$ Evaluate the integral $$ \int_9^{25} {\big({2z^{-3/2} -5z^{-2}}\big)}\ln{\big(2-5z^{-1/2}\big)}\; \d z.$$

Show Solution
\begin{align*} && y &= \frac{ax+b}{cx+d} \\ &&&= \frac{\frac{a}{c}(cx+d) - \frac{da}{c} + b}{cx+d} \\ \Rightarrow && y' &= \left (b - \frac{da}{c} \right)(-1)(cx+d)^{-2} \cdot c \\ &&&= (ad-bc)(cx+d)^{-2} \end{align*} \begin{align*} && y &= \frac{x+1}{x+3} \\ && \frac{\d y}{\d x} &= \frac{2}{(x+3)^2} \\ \Rightarrow && I &= \int_0^1 \frac{1}{(x+3)^2} \; \ln \left(\frac{x+1}{x+3}\right)\d x \\ &&&= \int_{y=1/3}^{y=1/2} \frac12 \ln y \, \d y \\ &&&= \frac12 \left [ y \ln y - y \right]_{1/3}^{1/2} \\ &&&= \frac12 \left ( \frac12\ln \frac12 - \frac12 - \frac13 \ln\frac13 + \frac13 \right) \\ &&&= \frac16 \ln 3 -\frac14 \ln 2 -\frac1{12} \end{align*} \begin{align*} && J &= \int_0^1 \frac1{(x+3)^2} \ln \left ( \frac{x^2+3x+2}{(x+3)^2} \right) \d x \\ &&&= \int_0^1 \frac1{(x+3)^2} \left ( \ln \frac{x+1}{x+3} + \ln \frac{x+2}{x+3} \right) \d x \\ &&&= I + \int_0^1 \frac1{(x+3)^2} \ln \left ( \frac{x+2}{x+3} \right) \d x \\ &&&= I + \int_{y=2/3}^{y=3/4} \ln y\, \d y \\ &&&= I + \left [ y \ln y- y\right]_{2/3}^{3/4} \\ &&&= I + \left ( \frac34 \ln \frac34 - \frac34 - \frac23 \ln \frac23 + \frac23 \right) \\ &&&= I + \left ( \frac34 \ln 3 - \frac32 \ln 2- \frac1{12} - \frac23 \ln 2 + \frac23 \ln 3\right) \\ &&&= I + \left ( \frac{17}{12} \ln 3 - \frac{13}6 \ln 2- \frac1{12} \right) \\ &&&= \frac16 \ln 3 -\frac14 \ln 2 -\frac1{12} + \left ( \frac{17}{12} \ln 3 - \frac{13}6 \ln 2- \frac1{12} \right) \\ &&&= \frac{19}{12} \ln 3 -\frac{29}{12}\ln 2 - \frac16 \end{align*} \begin{align*} && K &= \int_0^1 \frac{1}{(x+3)^2} \; \ln\left(\frac{x+1}{x+2}\right)\d x \\ &&&= \int_0^1 \frac{1}{(x+3)^2} \; \left ( \ln\left(\frac{x+1}{x+3}\right) - \ln \left ( \frac{x+3}{x+2} \right) \right)\d x \\ &&&= \frac16 \ln 3 -\frac14 \ln 2 -\frac1{12} - \left ( \frac{17}{12} \ln 3 - \frac{13}6 \ln 2- \frac1{12} \right) \\ &&&= -\frac54 \ln 3 +\frac{23}{12} \ln 2 \end{align*}
1998 Paper 1 Q2
D: 1500.0 B: 1516.0

Show, by means of a suitable change of variable, or otherwise, that \[ \int_{0}^{\infty}\mathrm{f}((x^{2}+1)^{1/2}+x)\,{\mathrm d}x =\frac{1}{2} \int_{1}^{\infty}(1+t^{-2})\mathrm{f}(t)\,{\mathrm d}t. \] Hence, or otherwise, show that \[ \int_{0}^{\infty}((x^{2}+1)^{1/2}+x)^{-3}\,{\mathrm d}x =\frac{3}{8}. \]

Show Solution
\begin{align*} && t &= (x^2+1)^{1/2}+x \\ && 1&=t^2-2tx \\ && x &= \frac{t^2-1}{2t} = \frac12 \left (t - \frac1t\right) \\ && \frac{\d x}{\d t} &= \frac12 \left ( 1+ \frac{1}{t^2} \right) \\ \Rightarrow && \int_0^{\infty} f((x^2+1)^{1/2}+x) \d x &= \int_{t=1}^{t = \infty}f(t) \frac12(1 + t^{-2}) \d t\\ &&&= \frac12 \int_1^{\infty}(1+t^{-2})f(t) \d t \end{align*} \begin{align*} \int_{0}^{\infty}((x^{2}+1)^{1/2}+x)^{-3}\,{\mathrm d}x &= \frac12 \int_1^{\infty}(1+t^{-2})t^{-3} \d t \\ &= \frac12 \left [\frac{-1}{2}t^{-2}-\frac{1}{4}t^{-4} \right]_{1}^{\infty} \\ &= \frac12 \left ( \frac12 + \frac14\right) = \frac38 \end{align*}
1998 Paper 2 Q4
D: 1600.0 B: 1470.2

The integral \(I_n\) is defined by $$I_n=\int_0^\pi(\pi/2-x)\sin(nx+x/2)\,{\rm cosec}\,(x/2)\,\d x,$$ where \(n\) is a positive integer. Evaluate \(I_n-I_{n-1}\), and hence evaluate \(I_n\) leaving your answer in the form of a sum.

Show Solution
\begin{align*} && I_n - I_{n-1} &= \int_0^\pi \left ( \frac{\pi}{2} - x \right) \left ( \sin\left(nx + \frac{x}{2}\right) - \sin \left ((n-1)x + \frac{x}{2} \right)\right) \cosec \frac{x}{2} \d x \\ &&&= \int_0^\pi \left ( \frac{\pi}{2} - x \right) \left (2 \sin \left ( \frac{nx + \frac{x}{2} - (n-1)x - \frac{x}{2} }{2}\right)\cos nx \right) \cosec \frac{x}{2} \d x \\ &&&= \int_0^\pi \left ( \frac{\pi}{2} - x \right) \left (2 \sin \left ( \frac{x}{2}\right)\cos nx \right) \cosec \frac{x}{2} \d x \\ &&&=2 \int_0^\pi \left ( \frac{\pi}{2} - x \right) \cos nx \d x \\ &&&=\pi \left [ \frac{\sin nx}{n}\right]_0^{\pi} - 2\int_0^\pi x \cos n x \d x \\ &&&= 0 - 2\left[ \frac{x \sin nx}{n} \right]_0^{\pi} + 2\int_0^\pi \frac{\sin nx}{n} \d x \\ &&&= 2\left[ -\frac{\cos nx}{n^2} \right]_0^{\pi} \\ &&&=2 \frac{1-(-1)^{n}}{n^2} \\ \\ && I_0 &= \int_0^\pi (\pi/2 - x) \d x =0 \\ \Rightarrow && I_{2k+2} = I_{2k+1} &= 4 \left (\frac{1}{1^2} + \frac{1}{3^2} + \cdots + \frac{1}{(2k+1)^2} \right) \end{align*}
1998 Paper 3 Q2
D: 1700.0 B: 1500.0

Let $$ {\rm I}(a,b) = \int_0^1 t^{a}(1-t)^{b} \, \d t \; \qquad (a\ge0,\ b\ge0) .$$

  1. Show that \({\rm I}(a,b)={\rm I}(b,a)\),
  2. Show that \({\rm I}(a,b)={\rm I}(a+1,b)+{\rm I}(a,b+1)\).
  3. Show that \((a+1){\rm I}(a,b)=b{\rm I}(a+1,b-1)\) when \(a\) and \(b\) are positive and hence calculate \({\rm I}(a,b)\) when \(a\) and \(b\) are positive integers.

Show Solution
  1. Let \(u = 1-t, \d u = -\d t\), then: \begin{align*} \mathrm{I}(a,b) &= \int_0^1 t^a(1-t)^b \d t \\ &= \int_{u=1}^{u=0} -(1-u)^a u^b \d u \\ &= \int_0^1(1-u)^a u^b \d u \\ &= \mathrm{I}(b, a) \end{align*}
  2. \begin{align*} \mathrm{I}(a+1,b)+\mathrm{I}(a,b+1) &= \int_0^1 t^{a+1}(1-t)^b + t^a(1-t)^{b+1} \d t \\ &= \int_0^1 (t+(1-t))t^a(1-t)^b \d t \\ &= \int_0^1 t^a(1-t)^b \d t \\ &= \mathrm{I}(a,b) \end{align*}
  3. Integrating by parts with \(\frac{du}{dt} = t^a, v = (1-t)^{b}\)\begin{align*} \mathrm{I}(a,b) &= \int_0^1 t^a (1-t)^b \d t \\ &= \left [ \frac{t^{a+1}}{a+1} (1-t)^b \right ]_0^1 + \int_0^1 \frac{t^{a+1}}{a+1} b(1-t)^{b-1} \\ &= \frac{b}{a+1} \int_0^1 t^{a+1}(1-t)^{b-1} \d t \\ &= \frac{b}{a+1} \mathrm{I}(a+1,b-1) \end{align*} Claim: \(\mathrm{I}(a,b) = \frac{a!b!}{(a+b+1)!}\) Proof: Note that \(I(a,0) = \frac{1}{a+1}\) so the formula holds for this case. We will induct on \(b\). The base case is done. Suppose that for \(b = k\) our formula is true, ie: \(\mathrm{I}(a,k) = \frac{a!k!}{(a+k+1)!}\) for all \(a\) (and fixed \(k\)) \begin{align*} \mathrm{I}(a,k+1) &= \frac{k+1}{a+1} \mathrm{I}(a+1,k) \\ &= \frac{k+1}{a+1} \frac{(a+1)!k!}{(a+1+k+1)!} \\ &= \frac{a!(k+1)!}{(a+(k+1)+1)!} \end{align*} So the formula is true for \(b=k+1\). Therefore, since it is true if \(b=0\) and if \(b=k\) is true then \(b=k+1\) is true, it is true for all values of \(b\).
1997 Paper 1 Q6
D: 1516.0 B: 1500.0

Find constants \(a_{0}\), \(a_{1}\), \(a_{2}\), \(a_{3}\), \(a_{4}\), \(a_{5}\), \(a_{6}\) and \(b\) such that \[x^{4}(1-x)^{4}=(a_{6}x^{6}+a_{5}x^{5}+a_{4}x^{4}+a_{3}x^{3}+ a_{2}x^{2}+a_{1}x+a_{0})(x^{2}+1)+b.\] Hence, or otherwise, prove that \[\int_{0}^{1}\frac{x^{4}(1-x)^{4}}{1+x^{2}}{\rm d}x =\frac{22}{7}-\pi.\] Evaluate \(\displaystyle{\int_{0}^{1}x^{4}(1-x)^{4}{\rm d}x}\) and deduce that \[\frac{22}{7}>\pi>\frac{22}{7}-\frac{1}{630}.\]

Show Solution
Plugging in \(x = i\) we obtain \((1-i)^4 = (-2i)^2 = -4 \Rightarrow b = -4\). \begin{align*} x^4(1-x)^4 &= x^4(1-4x+6x^2-4x^3+x^4) \\ &= x^8-4x^7+6x^6-4x^5+x^4 \\ &= x^6(x^2+1) - x^6 -4x^7+6x^6-4x^5+x^4 \\ &= x^6(x^2+1) -4x^5(x^2+1)+4x^5 +5x^6-4x^5+x^4 \\ &= (x^6-4x^5)(x^2+1) +5x^4(x^2+1)-5x^4+x^4 \\ &= (x^6-4x^5+5x^4)(x^2+1) -4x^2(x^2+1)+4x^2 \\ &= (x^6-4x^5+5x^4-4x^2)(x^2+1) +4(x^2+1)-4 \\ &= (x^6-4x^5+5x^4-4x^2+4)(x^2+1) -4 \\ \end{align*} So \begin{align*} \int_0^1 \frac{x^4(1-x)^4}{1+x^2} \d x &= \int_0^1 (x^6-4x^5+5x^4-4x^2+4) - \frac{4}{1+x^2} \d x \\ &= \frac17 - \frac46+1-\frac43+4 - \pi \\ &= \frac{22}7 - \pi \end{align*} \begin{align*} \int_0^1 x^4(1-x)^4 \d x &= B(5,5) \\ &= \frac{4!4!}{9!} \\ &= \frac1{630} \end{align*} Therefore since \(0 < \frac{x^4(1-x)^4}{1+x^2} < x^4(1-x)^4\) we must have that \begin{align*} && 0 &< \frac{22}7 - \pi \\ \Rightarrow && \pi & < \frac{22}{7} \\ && \frac{22}{7} - \pi &< \frac1{630} \\ \Rightarrow && \frac{22}{7} - \frac1{630} &< \pi \end{align*} which is what we wanted.
1997 Paper 2 Q3
D: 1600.0 B: 1500.0

Find constants \(a,\,b,\,c\) and \(d\) such that $$\frac{ax+b}{ x^2+2x+2}+\frac{cx+d}{ x^2-2x+2}= \frac{1}{ x^4+4}.$$ Show that $$\int_0^1\frac {\d x}{ x^4+4}\;= \frac{1}{16} \ln 5 +\frac{1}{8} \tan^{-1}2 .$$

Show Solution
First notice that \((x^2+2+2x)(x^2+2-2x) = (x^2+2)^2-4x^2 = x^2+4\) and so \begin{align*} && \frac{1}{x^4+4} &= \frac{ax+b}{ x^2+2x+2}+\frac{cx+d}{ x^2-2x+2} \\ \Rightarrow && 1 &= (ax+b)(x^2-2x+2) + (cx+d)(x^2+2x+2) \\ &&&= (a+c)x^3+(b-2a+d+2c)x^2+(2a-2b+2c+2d)x+2b+2d \\ \Rightarrow && 0 &= a+c \\ && 2a &= b+d+2c \\ && b &= a+c+d \\ && \frac12 &= b+d \\ \Rightarrow && c &= -a \\ && 2a &= \frac12+2(-a) \\ \Rightarrow && a &= \frac18, c = -\frac18 \\ && b &= d \\ \Rightarrow && b &= \frac14, d = \frac14 \end{align*} Therefore \begin{align*} \int_0^1\frac {\d x}{ x^4+4} &= \int_0^1 \frac18\left ( \frac{x+2}{x^2+2x+2} -\frac{x-2}{x^2-2x+2}\right) \d x \\ &= \frac1{16}\int_0^1 \left ( \frac{2x+2+2}{x^2+2x+2} -\frac{2x-2-2}{x^2-2x+2}\right) \d x \\ &= \frac1{16} \left [ \ln(x^2+2x+2) -\ln(x^2-2x+2)\right]_0^1 + \frac1{16} \int_0^1 \left ( \frac{2}{(x+1)^2+1} + \frac{2}{(x-1)^2+1} \right) \d x \\ &= \frac1{16} \left (\ln 5 - \ln 1 -(\ln 2-\ln 2) \right) + \frac18 \left [ \tan^{-1} (x+1)+\tan^{-1}(x-1) \right]_0^1 \\ &= \frac1{16} \ln 5 + \frac18 \left (\tan^{-1} 2+\tan^{-1} 0 - \tan^{-1}1-\tan^{-1} (-1) \right) \\ &= \frac1{16} \ln 5+ \frac18 \tan^{-1} 2 \end{align*}
1996 Paper 1 Q2
D: 1484.0 B: 1500.0

  1. Show that \[ \int_{0}^{1}\left(1+(\alpha-1)x\right)^{n}\,\mathrm{d}x=\frac{\alpha^{n+1}-1}{(n+1)(\alpha-1)} \] when \(\alpha\neq1\) and \(n\) is a positive integer.
  2. Show that if \(0\leqslant k\leqslant n\) then the coefficient of \(\alpha^{k}\) in the polynomial \[ \int_{0}^{1}\left(\alpha x+(1-x)\right)^{n}\,\mathrm{d}x \] is \[ \binom{n}{k}\int_{0}^{1}x^{k}(1-x)^{n-k}\,\mathrm{d}x\,. \]
  3. Hence, or otherwise, show that \[ \int_{0}^{1}x^{k}(1-x)^{n-k}\,\mathrm{d}x=\frac{k!(n-k)!}{(n+1)!}\,. \]

Show Solution
  1. \begin{align*} u = 1+(\alpha-1)x: && \int_0^1 (1 + (\alpha - 1)x)^n \d x &= \int_{u=1}^{u=\alpha} u^n \frac{1}{\alpha - 1} \d u \\ &&&= \left [\frac{u^{n+1}}{(n+1)(\alpha-1)} \right]_1^\alpha \\ &&&= \frac{\alpha^{n+1}-1}{(n+1)(\alpha-1)} \end{align*}
  2. \begin{align*} && \int_0^1 (\alpha x + (1-x))^n \d x &= \int_0^1 \sum_{k=0}^n \binom{n}{k} \alpha^k x^k (1-x)^{n-k} \d x \\ &&&= \sum_{k=0}^n \alpha^k \int_0^1 \binom{n}{k} x^k (1-x)^{n-k} \d x \end{align*} Therefore the coefficient of \(\alpha^k\) is \(\displaystyle \int_0^1 \binom{n}{k} x^k (1-x)^{n-k} \d x\)
  3. The coefficient of \(\alpha^{k}\) in \(\displaystyle \frac{\alpha^{n+1}-1}{(n+1)(\alpha-1)}\) is \(\displaystyle \frac1{n+1}\). Therefore \begin{align*} && \frac1{n+1} &= \binom{n}{k} \int_0^1 x^k(1-x)^{n-k} \d x \\ \Rightarrow && \int_0^1 x^k (1-x)^{n-k} \d x &= \frac{k!(n-k)!}{(n+1)n!} \\ &&&= \frac{k!(n-k)!}{(n+1)!} \end{align*}
1996 Paper 1 Q4
D: 1484.0 B: 1500.0

Show that \[ \int_{0}^{1}\frac{1}{x^{2}+2ax+1}\,\mathrm{d}x=\begin{cases} \dfrac{1}{\sqrt{1-a^{2}}}\tan^{-1}\sqrt{\dfrac{1-a}{1+a}} & \text{ if }\left|a\right|<1,\\ \dfrac{1}{2\sqrt{a^{2}-1}}\ln |a+\sqrt{a^{2}-1}| & \text{ if }\left|a\right|>1. \end{cases} \]

Show Solution
First suppose \(|a| < 1\), then \begin{align*} && I &= \int_0^1 \frac{1}{x^2+2ax+1} \d x \\ &&&= \int_0^1 \frac{1}{(x+a)^2 +1-a^2} \d x \\ &&&= \int_0^1 \frac{1}{(x+a)^2 +(\sqrt{1-a^2})^2} \d x \tag{\(1-a^2 > 0\)}\\ &&&= \left [\frac{1}{\sqrt{1-a^2}} \tan^{-1} \frac{x+a}{\sqrt{1-a^2}} \right]_0^1 \\ &&&= \frac{1}{\sqrt{1-a^2}} \left ( \tan^{-1} \frac{a+1}{\sqrt{1-a^2}} - \tan^{-1} \frac{a}{\sqrt{1-a^2}} \right) \\ &&&= \frac{1}{\sqrt{1-a^2}} \tan^{-1} \left ( \frac{\frac{a+1}{\sqrt{1-a^2}}-\frac{a}{\sqrt{1-a^2}}}{1+\frac{(a+1)a}{1-a^2}} \right) \\ &&&= \frac{1}{\sqrt{1-a^2}} \tan^{-1} \left ( \frac{\frac{1}{\sqrt{1-a^2}}}{\frac{1+a}{1-a^2}} \right) \\ &&&= \frac{1}{\sqrt{1-a^2}} \tan^{-1} \left ( \frac{1-a}{\sqrt{1-a^2}}\right) \\ &&&= \frac{1}{\sqrt{1-a^2}} \tan^{-1} \sqrt { \frac{1-a}{1+a}} \\ \end{align*} Second, suppose \(|a| > 1\), then \begin{align*} && I &= \int_0^1 \frac{1}{x^2+2ax+1} \d x \\ &&&= \int_0^1 \frac{1}{(x+a)^2-(a^2-1)} \d x \\ &&&= \int_0^1 \frac{1}{(x+a-\sqrt{a^2-1})(x+a+\sqrt{a^2-1})} \d x \tag{\(a^2-1 > 0\)} \\ &&&= \frac{1}{2\sqrt{a^2-1}}\int_0^1 \left ( \frac{1}{x+a-\sqrt{a^2-1}} - \frac{1}{x+a+\sqrt{a^2-1}} \right) \d x \\ &&&= \frac{1}{2\sqrt{a^2-1}} \left [ \ln |x+a-\sqrt{a^2-1}|- \ln |x+a+\sqrt{a^2-1}| \right]_0^1 \\ &&&= \frac{1}{2\sqrt{a^2-1}} \left ( \ln |1+a-\sqrt{a^2-1}| - \ln|1+a+\sqrt{a^2-1}| - \ln|a-\sqrt{a^2-1}| +\ln|a + \sqrt{a^2-1}| \right) \\ &&&= \frac{1}{2\sqrt{a^2-1}} \ln | \frac{(1+a-\sqrt{a^2-1})(a+\sqrt{a^2-1})}{(1+a+\sqrt{a^2-1})(a-\sqrt{a^2-1})}|\\ &&&= \frac{1}{2\sqrt{a^2-1}} \ln |\frac{a+a^2-(a^2-1) +\sqrt{a^2-1}}{1+a-\sqrt{a^2-1}}| \\ &&&= \frac{1}{2\sqrt{a^2-1}} \ln |\frac{(1+a +\sqrt{a^2-1})^2}{(1+a)^2-(a^2-1)}| \\ &&&= \frac{1}{2\sqrt{a^2-1}} \ln |\frac{1+2a+a^2+a^2-1+2(1+a)\sqrt{a^2-1}}{2+2a}| \\ &&&= \frac{1}{2\sqrt{a^2-1}} \ln |a+\sqrt{a^2-1}| \\ \end{align*}
1996 Paper 1 Q6
D: 1500.0 B: 1500.0

Let \(\mathrm{f}(x)=\dfrac{\sin(n+\frac{1}{2})x}{\sin\frac{1}{2}x}\) for \(0 < x\leqslant\pi.\)

  1. Using the formula \[ 2\sin\tfrac{1}{2}x\cos kx=\sin(k+\tfrac{1}{2})x-\sin(k-\tfrac{1}{2})x \] (which you may assume), or otherwise, show that \[ \mathrm{f}(x)=1+2\sum_{k=1}^{n}\cos kx\,. \]
  2. Find \({\displaystyle \int_{0}^{\pi}\mathrm{f}(x)\,\mathrm{d}x}\) and \({\displaystyle \int_{0}^{\pi}\mathrm{f}(x)\cos x\,\mathrm{d}x}.\)

Show Solution
  1. \(\,\) \begin{align*} && 2\sin \tfrac12 x \sum_{k=1}^n \cos kx &= \sum_{k=1}^n 2\sin \tfrac12 x \cos kx \\ &&&= \sum_{k=1}^n \left ( \sin(k+\tfrac12)x - \sin(k - \tfrac12)x \right) \\ &&&= \left ( \sin\tfrac32x - \sin\tfrac12x \right) + \\ &&&\quad \quad \left ( \sin\tfrac52x - \sin \tfrac32 x \right) + \\ &&&\quad \quad \quad +\cdots + \\ &&&\quad \quad \quad \quad +\left ( \sin(n+\tfrac12)x - \sin(n - \tfrac12)x \right) \\ &&&= \sin(n+\tfrac12)x - \sin \tfrac12 x \\ \Rightarrow && \sin(n+\tfrac12)x &= \sin \tfrac12 x + 2\sin \tfrac12 x \sum_{k=1}^n \cos kx \\ \Rightarrow && f(x) &= 1 + 2 \sum_{k=1}^n \cos kx \end{align*}
  2. \(\,\) \begin{align*} && \int_0^{\pi} f(x) \d x &= \int_0^{\pi} \left (1 + 2 \sum_{k=1}^n \cos kx \right) \d x \\ &&&= \pi + 2 \left [ \sum_{k=1}^n \frac{1}{k} \sin k x\right]_0^\pi \\ &&&= \pi \\ \\ && \int_0^{\pi} f(x) \cos x \d x &= \int_0^{\pi} \left (\cos x + 2 \sum_{k=1}^n \cos kx \cos x \right) \d x \\ &&&= 0 + \sum_{k=1}^n \left ( \int_0^{\pi} 2 \cos k x \cos x \d x \right) \\ &&&= \sum_{k=1}^n \left ( \int_0^{\pi} (\cos (k+1)x - \cos (k-1) x)\d x\right) \\ &&&= -\pi \end{align*}
1996 Paper 2 Q4
D: 1600.0 B: 1470.2

Show that \(\cos 4u=8\cos^{4}u-8\cos^{2}u+1\). If \[ I=\int_{-1}^{1} \frac{1}{\vphantom{{\big(}^2}\; \surd(1+x)+\surd(1-x)+2\; }\;{\rm d}x ,\] show, by using the change of variable \(x=\cos t\), that \[ I= \int_0^\pi \frac{\sin t}{4\cos^{2}\left(\frac{t}{4}-\frac{\pi}{8}\right)}\,{\rm d}t.\] By using the further change of variable \(u=\frac{t}{4}-\frac{\pi}{8}\), or otherwise, show that \[I=4\surd{2}-\pi-2.\] \noindent[You may assume that \(\tan\frac{\pi}{8}=\surd{2}-1\).]

Show Solution
\begin{align*} && \cos 4u &= 2\cos^2 2u - 1 \\ &&&= 2 (2\cos^2 u - 1)^2 - 1 \\ &&&= 2(4\cos^4u - 4\cos^2 u + 1) - 1\\ &&&= 8\cos^4u - 8\cos^2 u + 1 \end{align*} \begin{align*} && I &= \int_{-1}^1 \frac{1}{\sqrt{1+x}+\sqrt{1-x}+2} \d x \\ x = \cos t, \d x = - \sin t \d t: &&&= \int_{t = \pi}^{t=0} \frac{1}{\sqrt{1+\cos t} + \sqrt{1-\cos t} + 2} (- \sin t ) \d t \\ &&&= \int_0^\pi \frac{\sin t}{\sqrt{2 \cos^2 \frac{t}{2}}+\sqrt{2 \sin^2 \frac{t}{2}}+2} \d t \\ &&&= \int_0^\pi \frac{\sin t}{\sqrt{2}(\cos \frac{t}{2} + \sin \frac{t}{2}) + 2} \d t \\ &&&= \int_0^\pi \frac{\sin t}{\sqrt{2}(\sqrt{2} \cos (\frac{t}{2}-\frac{\pi}{4})) + 2} \d t \\ &&&= \int_0^\pi \frac{\sin t}{2(1+\cos (\frac{t}{2}-\frac{\pi}{4}))} \d t \\ &&&= \int_0^\pi \frac{\sin t}{4\cos^2(\frac{t}{4}-\frac{\pi}{8})} \d t \\ \\ u = \tfrac{t}{4} -\tfrac{\pi}{8}, \d u = \tfrac14 \d t:&&&=\int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} \frac{\sin (4u+\frac{\pi}{2})}{4 \cos^2 u} 4 \d u \\ &&&= \int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} \frac{\cos4u}{\cos^2 u} \d u \\ &&&= \int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} 4 (2 \cos^2 u-1)-4 + \sec^2 u \d u \\ &&&= \int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} 4 \cos 2u-4 + \sec^2 u \d u \\ &&&= \left [2\sin 2u - 4u + \tan u \right]_{-\pi/8}^{\pi/8} \\ &&&= 4 \sin \frac{\pi}{4} - \pi+ 2\tan \frac{\pi}{8} \\ &&&= \frac{4}{\sqrt{2}} - \pi + 2\sqrt{2}-2 \\ &&&= 4\sqrt{2} - \pi - 2 \end{align*}
1996 Paper 3 Q3
D: 1700.0 B: 1500.0

Find \[ \int_{0}^{\theta}\frac{1}{1-a\cos x}\,\mathrm{d}x\,, \] where \(0 < \theta < \pi\) and \(-1 < a < 1.\) Hence show that \[ \int_{0}^{\frac{1}{2}\pi}\frac{1}{2-a\cos x}\,\mathrm{d}x=\frac{2}{\sqrt{4-a^{2}}}\tan^{-1}\sqrt{\frac{2+a}{2-a}}\,, \] and also that \[ \int_{0}^{\frac{3}{4}\pi}\frac{1}{\sqrt{2}+\cos x}\,\mathrm{d}x=\frac{\pi}{2}\,. \]

Show Solution
Let \(t = \tan \tfrac{x}{2}\), then \(\cos x = \frac{1-t^2}{1+t^2}, \frac{d t}{d x} =\tfrac12 (1+t^2)\) so the integral is: \begin{align*} \int_0^{\theta} \frac{1}{1-a \cos x} \d x &= \int_{0}^{\tan \frac{\theta}{2}} \frac{1}{1-a \left (\frac{1-t^2}{1+t^2} \right)} \frac{2}{1+t^2} \d t \\ &= \int_0^{\tan \frac{\theta}{2}} \frac{2}{1+t^2-a+at^2} \d t \\ &= \int_0^{\tan \frac{\theta}{2}} \frac{2}{1-a+(1+a) t^2} \d t \\ &= \frac{2}{1+a}\int_0^{\tan \tfrac{\theta}{2}} \frac{1}{\left (\frac{1-a}{1+a} \right)+t^2} \d t \\ &= \frac{2}{1+a} \sqrt{\frac{1+a}{1-a}} \tan^{-1} \left ( \sqrt{\frac{1+a}{1-a}} \tan \frac{\theta}{2} \right) + C \\ &= \frac{2}{\sqrt{1-a^2}}\tan^{-1} \left ( \sqrt{\frac{1+a}{1-a}} \tan \frac{\theta}{2} \right) + C \end{align*} Therefore \begin{align*} \int_{0}^{\frac{1}{2}\pi}\frac{1}{2-a\cos x}\,\mathrm{d}x &= \frac12 \int_0^{\frac12 \pi} \frac{1}{1-\tfrac{a}{2} \cos x} \d x \\ &= \left [\frac12 \frac{2}{\sqrt{1-\frac{a^2}{4}}} \tan^{-1} \left ( \sqrt{\frac{1+\frac{a}{2}}{1-\frac{a}{2}} } \tan\frac{\theta}{2} \right) \right]_0^{\pi/2} \\ &= \frac12 \frac{2}{\sqrt{1-\frac{a^2}{4}}} \tan^{-1} \left ( \sqrt{\frac{1+\frac{a}{2}}{1-\frac{a}{2}} } \tan\frac{\pi}{4} \right) \\ &= \frac{2}{\sqrt{4-a^2}} \tan^{-1} \left ( \sqrt{\frac{2+a}{2-a} } \right) \\ \end{align*} \begin{align*} \int_{0}^{\frac{3}{4}\pi}\frac{1}{\sqrt{2}+\cos x}\,\mathrm{d}x &= \frac{1}{\sqrt{2}} \int_0^{\frac34 \pi} \frac{1}{1 -\left(- \frac{1}{\sqrt{2}} \right)\cos x} \d x \\ &= \frac{1}{\sqrt{2}} \left [ \frac{2}{\sqrt{1-\tfrac12}} \tan^{-1} \left ( \sqrt{\frac{1-\frac{1}{\sqrt{2}}}{1+\frac{1}{\sqrt{2}}} } \tan\frac{\theta}{2} \right) \right]_0^{3\pi/4} \\ &= \frac{1}{\sqrt{2}} \frac{2}{\sqrt{1/2}} \tan^{-1} \left ( \sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1} } \tan\frac{3\pi}{8} \right) \\ &= 2 \tan^{-1} \left ( \sqrt{\frac{(\sqrt{2}-1)^2}{2-1} } \tan\frac{3\pi}{8} \right)\\ &= 2 \tan^{-1} \left ( (\sqrt{2}-1) \tan\frac{3\pi}{8} \right) \end{align*} If \(t = \tan \tfrac{3\pi}{8}\), then \(-1 = \tan \tfrac{3\pi}{4} = \frac{2t}{1-t^2} \Rightarrow t^2-2t-1 = 0 \Rightarrow t = 1\pm \sqrt{2}\), since \( t > 0\), we must have \(t = 1 + \sqrt{2}\), so \begin{align*} \int_{0}^{\frac{3}{4}\pi}\frac{1}{\sqrt{2}+\cos x}\,\mathrm{d}x &= 2 \tan^{-1} \left ((\sqrt{2}-1)(\sqrt{2}+1) \right) \\ &= 2 \tan^{-1} 1 \\ &= 2 \frac{\pi}{4} \\ &= \frac{\pi}{2} \end{align*}
1995 Paper 1 Q2
D: 1500.0 B: 1516.0

  1. Suppose that \[ S=\int\frac{\cos x}{\cos x+\sin x}\,\mathrm{d}x\quad\mbox{ and }\quad T=\int\frac{\sin x}{\cos x+\sin x}\,\mathrm{d}x. \] By considering \(S+T\) and \(S-T\) determine \(S\) and \(T\).
  2. Evaluate \({\displaystyle \int_{\frac{1}{4}}^{\frac{1}{2}}(1-4x)\sqrt{\frac{1}{x}-1}\,\mathrm{d}x}\) by using the substitution \(x=\sin^{2}t.\)

Show Solution
  1. \begin{align*} && S + T &= \int \frac{\cos x + \sin x }{\cos x + \sin x} \d x \\ &&&= \int \d x \\ &&&= x + C \\ && S - T &= \int \frac{\cos x - \sin x}{\cos x + \sin x} \d x \\ &&&= \ln( \cos x + \sin x) + C \\ \Rightarrow && 2S &= x + \ln(\cos x + \sin x) + C \\ \Rightarrow && S &= \frac12 \left ( x + \ln(\cos x + \sin x) \right) + C \\ \Rightarrow && 2T &= x - \ln(\cos x + \sin x) + C \\ \Rightarrow && T &= \frac12 \left ( x - \ln(\cos x + \sin x) \right) + C \end{align*}
  2. \begin{align*} && I &= \int_{1/4}^{1/2} (1-4x)\sqrt{\frac1x-1} \d x \\ x = \sin^2 t, \d x = 2 \sin t \cos t \d t: &&&= \int_{\pi/6}^{\pi/4} (1-4\sin^2 t) \sqrt{\frac{1-\sin^2 t}{\sin^2 t}} 2 \sin t \cos t \d t\\ &&&=\int_{\pi/6}^{\pi/4} (1-4\sin^2 t)\frac{\cos t}{\sin t} 2 \sin t \cos t \d t \\ &&&= \int_{\pi/6}^{\pi/4} (1-4\sin^2 t) 2 \cos^2 t \d t \\ &&&= \int_{\pi/6}^{\pi/4} \left ( 2\cos^2t - 8 \sin^2t \cos^2 t \right) \d t \\ &&&= \int_{\pi/6}^{\pi/4} \left ( 1+\cos2t - 2 \sin^2 2t \right) \d t \\ &&&= \int_{\pi/6}^{\pi/4} \left ( 1+\cos2t +(\cos 4t-1)\right) \d t \\ &&&= \left[\frac12 \sin 2t + \frac14 \sin 4t \right]_{\pi/6}^{\pi/4} \\ &&&= \left ( \frac12 \right) - \left (\frac12 \frac{\sqrt{3}}{2} + \frac14 \frac{\sqrt{3}}{2} \right) \\ &&&= \frac{4-3\sqrt{3}}{8} \end{align*}
1995 Paper 1 Q5
D: 1500.0 B: 1500.0

If \[ \mathrm{f}(x)=nx-\binom{n}{2}\frac{x^{2}}{2}+\binom{n}{3}\frac{x^{3}}{3}-\cdots+(-1)^{r+1}\binom{n}{r}\frac{x^{r}}{r}+\cdots+(-1)^{n+1}\frac{x^{n}}{n}\,, \] show that \[ \mathrm{f}'(x)=\frac{1-(1-x)^{n}}{x}\,. \] Deduce that \[ \mathrm{f}(x)=\int_{1-x}^{1}\frac{1-y^{n}}{1-y}\,\mathrm{d}y. \] Hence show that \[ \mathrm{f}(1)=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\,. \]

Show Solution
\begin{align*} f(x) & =nx-\binom{n}{2}\frac{x^{2}}{2}+\binom{n}{3}\frac{x^{3}}{3}-\cdots+(-1)^{r+1}\binom{n}{r}\frac{x^{r}}{r}+\cdots+(-1)^{n+1}\frac{x^{n}}{n} \\ f'(x) &= n - \binom{n}{2} x + \binom{n}{3}x^2 - \cdots (-1)^{r+1} \binom{n}{r} + \cdots + (-1)^{n+1} x^{n-1} \\ &= \frac{1-(1-x)^n}{x} \end{align*} Therefore, since \(\displaystyle f(x) = \int_0^xf'(t)\,dt\) \begin{align*} f(x) &= \int_0^x \frac{1 - (1-t)^n}{t} \, dt \\ &= \int_{1}^{1-x} \frac{1-y^n}{1-y} (-1)\, dy \tag{Let \(y = 1-t, \frac{dy}{dt} = -1\)} \\ &= \boxed{\int_{1-x}^1 \frac{1-y^n}{1-y} dy} \\ &= \int_{1-x}^1 \l 1 + y + y^2 + \cdots + y^{n-1} \r \, dy \\ &= \left [ y + \frac{y^2}{2} + \frac{y^3}{3} + \cdots + \frac{y^n}{n} \right]_{1-x}^1 \\ \end{align*} So when \(x = 1, 1-x = 0\) so we exactly have the sum required.
1994 Paper 1 Q4
D: 1484.0 B: 1628.6

Show that

  1. \(\dfrac{1-\cos\alpha}{\sin\alpha}=\tan\frac{1}{2}\alpha,\)
  2. if \(\left|k\right|<1\) then \({\displaystyle \int\frac{\mathrm{d}x}{1-2kx+x^{2}}=\frac{1}{\sqrt{1-k^{2}}}\tan^{-1}\left(\frac{x-k}{\sqrt{1-k^{2}}}\right)+C,}\) where \(C\) is a constant of integration.
Hence, or otherwise, show that if \(0<\alpha<\pi\) then \[ \int_{0}^{1}\frac{\sin\alpha}{1-2x\cos\alpha+x^{2}}\,\mathrm{d}x=\frac{\pi-\alpha}{2}. \]

Show Solution
  1. \begin{align*} \frac{1-\cos \alpha}{\sin \alpha} &= \frac{1-(1-2\sin^2 \frac{\alpha}{2})}{2 \sin \frac \alpha2 \cos \frac\alpha2} \\ &= \frac{2 \sin^2 \frac \alpha2}{2 \sin \frac \alpha2 \cos \frac\alpha2} \\ &= \frac{\sin \frac \alpha2}{ \cos \frac\alpha2} \\ &= \tan \tfrac{\alpha}{2} \end{align*}
  2. \begin{align*} \int\frac{\mathrm{d}x}{1-2kx+x^{2}} &= \int \frac{\d x}{(x-k)^2+1-k^2} \\ &= \frac{1}{1-k^2}\int \frac{\d x}{\left (\frac{x-k}{\sqrt{1-k^2}} \right)^2+1} \\ &= \frac{1}{\sqrt{1-k^2}} \tan^{-1} \left (\frac{x-k}{\sqrt{1-k^2}} \right)+C \end{align*}
\begin{align*} \int_{0}^{1}\frac{\sin\alpha}{1-2x\cos\alpha+x^{2}}\,\mathrm{d}x &= \sin \alpha \left [\frac{1}{\sqrt{1-\cos ^2\alpha}} \tan^{-1} \left ( \frac{x - \cos \alpha}{\sqrt{1-\cos^2\alpha}} \right) \right]_0^1 \\ &= \tan^{-1} \left ( \frac{1 - \cos \alpha}{\sin \alpha} \right) -\tan^{-1} \left ( \frac{- \cos \alpha}{\sin \alpha} \right) \\ &= \tan^{-1} \tan \tfrac{\alpha}{2} + \tan^{-1} \cot \alpha \\ &= \frac{\alpha}{2} + \frac{\pi}{2} - \alpha \\ &= \frac{\pi-\alpha}{2} \end{align*}
1994 Paper 1 Q8
D: 1516.0 B: 1500.8

By means of the change of variable \(\theta=\frac{1}{4}\pi-\phi,\) or otherwise, show that \[ \int_{0}^{\frac{1}{4}\pi}\ln(1+\tan\theta)\,\mathrm{d}\theta=\tfrac{1}{8}\pi\ln2. \] Evaluate \[ {\displaystyle \int_{0}^{1}\frac{\ln(1+x)}{1+x^{2}}\,\mathrm{d}x}\qquad\mbox{ and }\qquad{\displaystyle \int_{0}^{\frac{1}{2}\pi}\ln\left(\frac{1+\sin x}{1+\cos x}\right)\,\mathrm{d}x}. \]

Show Solution
\begin{align*} && I &= \int_0^{\frac14\pi} \ln (1 + \tan \theta) \d \theta \\ \theta = \tfrac14\pi - \phi, \d \theta = -\d\phi: &&&= \int_0^{\frac14 \pi} \ln ( 1 + \tan (\tfrac14\pi - \phi)) \d \phi \\ &&&= \int_0^{\frac14 \pi} \ln \left ( 1 + \frac{1 - \tan \phi}{1+\tan \phi} \right) \d \phi \\ &&&= \int_0^{\frac14 \pi} \ln \left ( \frac{2}{1+\tan \phi} \right) \d \phi \\ &&&= \tfrac14 \pi \ln 2 - I \\ \Rightarrow && I &= \tfrac18\pi \ln 2 \end{align*} \begin{align*} && J &= \int_0^1 \frac{\ln(1+x)}{1+x^2} \d x \\ x= \tan \theta \d \theta, \d \theta = \frac{\d x}{1+x^2} &&&= \int_0^{\frac14 \pi} \ln(1 + \tan \theta) \d \theta \\ &&&= \tfrac18 \pi \ln 2 \end{align*} \begin{align*} && K &= \int_0^{\frac12 \pi} \ln \left ( \frac{1 + \sin x}{1 + \cos x} \right) \d x \\ y = \tfrac12\pi - x, \d y = -\d x: &&&= \int_0^{\frac12\pi} \ln \left ( \frac{1+\cos y}{1+\sin y}\right) \d y \\ &&&= -K \\ \Rightarrow && K &= 0 \end{align*}
1994 Paper 2 Q2
D: 1600.0 B: 1516.0

If \(\mathrm{Q}\) is a polynomial, \(m\) is an integer, \(m\geqslant1\) and \(\mathrm{P}(x)=(x-a)^{m}\mathrm{Q}(x),\) show that \[ \mathrm{P}'(x)=(x-a)^{m-1}\mathrm{R}(x) \] where \(\mathrm{R}\) is a polynomial. Explain why \(\mathrm{P}^{(r)}(a)=0\) whenever \(1\leqslant r\leqslant m-1\). (\(\mathrm{P}^{(r)}\) is the \(r\)th derivative of \(\mathrm{P}.\)) If \[ \mathrm{P}_{n}(x)=\frac{\mathrm{d}^{n}}{\mathrm{d}x^{n}}(x^{2}-1)^{n} \] for \(n\geqslant1\) show that \(\mathrm{P}_{n}\) is a polynomial of degree \(n\). By repeated integration by parts, or otherwise, show that, if \(n-1\geqslant m\geqslant0,\) \[ \int_{-1}^{1}x^{m}\mathrm{P}_{n}(x)\,\mathrm{d}x=0 \] and find the value of \[ \int_{-1}^{1}x^{n}\mathrm{P}_{n}(x)\,\mathrm{d}x. \] {[}Hint. \textit{You may use the formula \[ \int_{0}^{\frac{\pi}{2}}\cos^{2n+1}t\,\mathrm{d}t=\frac{(2^{2n})(n!)^{2}}{(2n+1)!} \] without proof if you need it. However some ways of doing this question do not use this formula.}{]}

Show Solution
\begin{align*} && P(x) &= (x-a)^mQ(x) \\ \Rightarrow && P'(x) &= m(x-a)^{m-1}Q(x) + (x-a)^mQ'(x) \\ &&&= (x-a)^{m-1}(\underbrace{mQ(x) + (x-a)Q'(x)}_{\text{a polynomial}}) \\ &&&= (x-a)^{m-1}R(x) \end{align*} Therefore \(P^{(r)}(a) = 0\) for \(1 \leq r \leq m-1\) since each time we differentiate we will have a factor of \((x-a)^{m-r}\) which is zero when we evaluate at \(x = a\). If \(P_n(x) = \frac{\d^n}{\d x^n}(x^2-1)^n\) then we are differentiating a degree \(2n\) polynomial \(n\) times. Each time we differentiate we reduce the degree by \(1\), therefore the degree of \(P_n\) is \(n\). \begin{align*} && \int_{-1}^1 x^mP_n(x) \d x &= \left [x^m \underbrace{\frac{\d^{n-1}}{\d x^{n-1}}\left ( (x-1)^{n} (x+1)^{n} \right)}_{\text{has a factor of }x-1\text{ and }x+1}\right]_{-1}^1 - \int_{-1}^1 mx^{m-1}\frac{\d^{n-1}}{\d x^{n-1}}\left ( (x-1)^{n} (x+1)^{n} \right) \d x\\ &&&= 0 - \int_{-1}^1 mx^{m-1}\frac{\d^{n-1}}{\d x^{n-1}}\left ( (x-1)^{n} (x+1)^{n} \right) \d x\\ &&&= -\left [mx^{m-1} \underbrace{\frac{\d^{n-2}}{\d x^{n-2}}\left ( (x-1)^{n} (x+1)^{n} \right)}_{\text{has a factor of }x-1\text{ and }x+1}\right]_{-1}^1+ \int_{-1}^1 m(m-1)x^{m-2}\frac{\d^{n-2}}{\d x^{n-2}}\left ( (x-1)^{n} (x+1)^{n} \right) \d x\\ &&&= m(m-1)\int_{-1}^1 x^{m-2}\frac{\d^{n-2}}{\d x^{n-2}}\left ( (x-1)^{n} (x+1)^{n} \right) \d x\\ &&& \cdots \\ &&&= (-1)^m m!\int_{-1}^1 \frac{\d^{n-m}}{\d x^{n-m}} \left ( (x-1)^{n} (x+1)^{n} \right) \d x\\ &&&= 0 \end{align*} If \(n = m\), we have \begin{align*} && \int_{-1}^1 x^n P_n(x) \d x&= (-1)^nn! \int_{-1}^1 (x^2-1)^n \d x \\ && &= (-1)^{2n}n! \cdot 2\int_{0}^1 (1-x^2)^n \d x \\ x = \sin \theta, \d x = \cos \theta \d \theta: &&&= 2 \cdot n!\int_{0}^{\pi/2} \cos^{2n} \theta \cdot \cos \theta \d \theta \\ &&&= 2 \cdot n!\int_{0}^{\pi/2} \cos^{2n+1} \theta \d \theta \\ &&&= 2 \cdot n!\frac{(2^{2n})(n!)^{2}}{(2n+1)!} \\ &&&= \frac{(2^{2n+1})(n!)^{3}}{(2n+1)!} \\ \end{align*}
1994 Paper 2 Q4
D: 1600.0 B: 1500.0

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 \[ 4ab\sin^{-1}\left(\frac{b}{\sqrt{a^{2}+b^{2}}}\right). \]

1993 Paper 1 Q4
D: 1484.0 B: 1516.0

By making the change of variable \(t=\pi-x\) in the integral \[ \int_{0}^{\pi}x\mathrm{f}(\sin x)\,\mathrm{d}x, \] or otherwise, show that, for any function \(\mathrm{f},\) \[ \int_{0}^{\pi}x\mathrm{f}(\sin x)\,\mathrm{d}x=\frac{\pi}{2}\int_{0}^{\pi}\mathrm{f}(\sin x)\,\mathrm{d}x\,. \] Evaluate \[ \int_{0}^{\pi}\frac{x\sin x}{1+\cos^{2}x}\,\mathrm{d}x\quad\mbox{ and }\quad\int_{0}^{2\pi}\frac{x\sin x}{1+\cos^{2}x}\,\mathrm{d}x\,. \]

Show Solution
\begin{align*} && I &= \int_0^{\pi} x f(\sin x) \d x \\ t = \pi - x, \d t = -\d t : &&&= \int_{t = \pi}^{t = 0} (\pi - t) f(\sin (\pi - t)) -\d t \\ &&&= \int_0^{\pi} (\pi - t) f(\sin t) \d t \\ \Rightarrow && 2 I &= \pi \int_0^\pi f(\sin t) \d t \\ \Rightarrow && I &= \frac{\pi}{2} \int_0^{\pi} f(\sin x) \d x \end{align*} \begin{align*} && I &= \int_{0}^{\pi}\frac{x\sin x}{1+\cos^{2}x}\,\mathrm{d}x \\ &&&= \frac{\pi}{2}\int_0^\pi \frac{\sin x}{1 + \cos^2 x} \d x \\ &&&= \frac{\pi}{2}\left [ -\tan^{-1} \cos x\right]_0^{\pi} \\ &&&= \tan 1 - \tan (-1) = \frac{\pi^2}{4} \\ \\ && I &= \int_{0}^{2\pi}\frac{x\sin x}{1+\cos^{2}x}\,\mathrm{d}x \\ &&&= \int_{0}^{\pi}\frac{x\sin x}{1+\cos^{2}x}\,\mathrm{d}x + \int_{\pi}^{2\pi} \frac{x\sin x}{1+\cos^{2}x}\,\mathrm{d}x \\ u = x - \pi, \d u = \d x: &&&= \frac{\pi^2}{4} + \int_{0}^{\pi} \frac{(u+\pi)(-\sin u)}{1 + \cos^2 u}\d u \\ &&&= \frac{\pi^2}{4} -\frac{3\pi}{2} \int_0^{\pi} \frac{\sin u}{1+\cos^2 u} \d u \\ &&&= - \frac{\pi^2}2 \end{align*}
1993 Paper 2 Q2
D: 1600.0 B: 1531.5

  1. Evaluate \[ \int_{0}^{2\pi}\cos(mx)\cos(nx)\,\mathrm{d}x, \] where \(m,n\) are integers, taking into account any special cases that arise.
  2. Find \({\displaystyle \int\sqrt{1+\frac{1}{x}}\,\mathrm{d}x}.\)

1992 Paper 1 Q3
D: 1500.0 B: 1486.1

Evaluate

  1. \({\displaystyle \int_{-\pi}^{\pi}\left|\sin x\right|\,\mathrm{d}x,}\)
  2. \({\displaystyle \int_{-\pi}^{\pi}\sin\left|x\right|\,\mathrm{d}x},\)
  3. \({\displaystyle \int_{-\pi}^{\pi}x\sin x\,\mathrm{d}x},\)
  4. \({\displaystyle \int_{-\pi}^{\pi}x^{10}\sin x\,\mathrm{d}x.}\)

Show Solution
  1. \begin{align*} \int_{-\pi}^\pi |\sin x | \d x &= \int_{-\pi}^{0} - \sin x \d x + \int_0^\pi \sin x \d x \\ &= \left [\cos x \right]_{-\pi}^{0} +[-\cos x]_0^{\pi} \\ &= 1-(-1)+(1)-(-1) \\ &= 4 \end{align*}
  2. \begin{align*} \int_{-\pi}^\pi \sin | x | \d x &= \int_{-\pi}^0 - \sin x \d x + \int_0^\pi \sin x \d x \\ &= 4 \end{align*}
  3. \begin{align*} \int_{-\pi}^\pi x \sin x \d x &= \left [ -x \cos x \right]_{-\pi}^\pi + \int_{-\pi}^{\pi} \cos x \d x \\ &= \pi -(-\pi) + \left [\sin x \right]_{-\pi}^\pi \\ &= 2\pi \end{align*}
  4. \begin{align*} \int_{-\pi}^{\pi} x^{10} \sin x \d x &\underbrace{=}_{x^{10}\sin x \text{ is odd}} 0 \end{align*}
1992 Paper 2 Q3
D: 1600.0 B: 1485.7

In the figure, the large circle with centre \(O\) has radius \(4\) and the small circle with centre \(P\) has radius \(1\). The small circle rolls around the inside of the larger one. When \(P\) was on the line \(OA\) (before the small circle began to roll), the point \(B\) was in contact with the point \(A\) on the large circle.

TikZ diagram
Sketch the curve \(C\) traced by \(B\) as the circle rolls. Show that if we take \(O\) to be the origin of cartesian coordinates and the line \(OA\) to be the \(x\)-axis (so that \(A\) is the point \((4,0)\)) then \(B\) is the point \[ (3\cos\phi+\cos3\phi,3\sin\phi-\sin3\phi). \] It is given that the area of the region enclosed by the curve \(C\) is \[ \int_{0}^{2\pi}x\frac{\mathrm{d}y}{\mathrm{d}\phi}\,\mathrm{d}\phi, \] where \(B\) is the point \((x,y).\) Calculate this area.

1991 Paper 1 Q8
D: 1500.0 B: 1516.9

  1. By a substitution of the form \(y=k-x\) for suitable \(k\), prove that, for any function \(\mathrm{f}\), \[ \int_{0}^{\pi}x\mathrm{f}(\sin x)\,\mathrm{d}x=\pi\int_{0}^{\frac{1}{2}\pi}\mathrm{f}(\sin x)\,\mathrm{d}x. \] Hence or otherwise evaluate \[ \int_{0}^{\pi}\frac{x}{2+\sin x}\,\mathrm{d}x. \]
  2. Evaluate \[ \int_{0}^{1}\frac{(\sin^{-1}t)\cos\left[(\sin^{-1}t)^{2}\right]}{\sqrt{1-t^{2}}}\,\mathrm{d}t. \] {[}No credit will be given for numerical answers obtained by use of a calculator.{]}

Show Solution
  1. \begin{align*} y = \pi - x, \d y = -\d x: && \int_0^{\pi} x f(\sin x) &= \int_{y = \pi}^{y = 0}(\pi - y) f(\sin(\pi-y))- \d y \\ &&&= \int_0^{\pi} (\pi -y) f(\sin y) \d y \\ \Rightarrow && 2 \int_0^{\pi} x f(\sin x)\d x &= \pi \int_0^{\pi} f(\sin x) \d x \\ &&&= \pi \int_0^{\pi/2} f(\sin x ) \d x + \pi \int_{\pi/2}^{\pi} f(\sin x ) \d x \\ &&&= \pi \int_0^{\pi/2} f(\sin x ) \d x +\pi \int_{y=\pi/2}^{y=0} f(\sin (\pi-y) ) (-\d y) \\ &&&= 2 \pi \int_0^{\pi/2} f(\sin x) \d x \\ \Rightarrow && \int_0^{\pi} x f(\sin x)\d x &= \pi \int_0^{\pi/2} f(\sin x) \d x \end{align*} Therefore if \(f(x) = \frac1{2+\sin x}\), letting \(t = \tan \frac{x}{2}\) we have \(\sin x = \frac{2 t}{1+t^2}, \frac{dt}{\d x} = \frac12 (1+t^2)\) \begin{align*} && \int_0^{\pi} \frac{x}{2 + \sin x } \d x &= \pi \int_0^{\pi/2} \frac{1}{2 + \sin x} \d x \\ &&&= \pi \int_{t = 0}^{t = 1} \frac{1}{2+\frac{2t}{1+t^2}} \frac{2}{1+t^2} \d t \\ &&&=\pi \int_0^1 \frac{2}{2t^2+2t+2} \d t\\ &&&=\pi \int_0^1 \frac{1}{(t+\tfrac12)^2 + \tfrac34} \d t\\ &&&= \pi \left [\frac{1}{\sqrt{3/4}} \tan^{-1} \frac{u}{\sqrt{3/4}} \right ]_{u=1/2}^{3/2} \\ &&&= \frac{2 \pi}{\sqrt{3}} \left ( \tan^{-1} \sqrt{3} - \tan^{-1} \frac1{\sqrt{3}} \right) \\ &&&= \frac{2 \pi}{\sqrt{3}} \left ( \frac{\pi}{3} - \frac{\pi}{6} \right) \\ &&&= \frac{\pi^2}{3\sqrt{3}} \end{align*}
  2. Let \(u = (\sin^{-1} t)^2, \frac{\d u}{\d t} = 2(\sin^{-1} t) \frac{1}{\sqrt{1-t^2}}\) \begin{align*} \int_{0}^{1}\frac{(\sin^{-1}t)\cos\left[(\sin^{-1}t)^{2}\right]}{\sqrt{1-t^{2}}}\,\mathrm{d}t &= \int_{u=0}^{\pi^2/4} \frac12 \cos u \d u \\ &= \frac12 \sin \frac{\pi^2}{4} \end{align*}
1991 Paper 2 Q6
D: 1600.0 B: 1485.5

Show by means of a sketch, or otherwise, that if \(0\leqslant\mathrm{f}(y)\leqslant\mathrm{g}(y)\) for \(0\leqslant y\leqslant x\) then \[ 0\leqslant\int_{0}^{x}\mathrm{f}(y)\,\mathrm{d}y\leqslant\int_{0}^{x}\mathrm{g}(y)\,\mathrm{d}y. \] Starting from the inequality \(0\leqslant\cos y\leqslant1,\) or otherwise, prove that if \(0\leqslant x\leqslant\frac{1}{2}\pi\) then \(0\leqslant\sin x\leqslant x\) and \(\cos x\geqslant1-\frac{1}{2}x^{2}.\) Deduce that \[ \frac{1}{1800}\leqslant\int_{0}^{\frac{1}{10}}\frac{x}{(2+\cos x)^{2}}\,\mathrm{d}x\leqslant\frac{1}{1797}. \] Show further that if \(0\leqslant x\leqslant\frac{1}{2}\pi\) then \(\sin x\geqslant x-\frac{1}{6}x^{3}.\) Hence prove that \[ \frac{1}{3000}\leqslant\int_{0}^{\frac{1}{10}}\frac{x^{2}}{(1-x+\sin x)^{2}}\,\mathrm{d}x\leqslant\frac{2}{5999}. \]