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2015 Paper 2 Q6
D: 1600.0 B: 1504.2
integration trigonometric identity substitution symmetry definite integral sec squared Weierstrass substitution integration by parts

  1. Show that \[ \mathrm{sec}^2\left(\tfrac14\pi-\tfrac12 x\right)=\frac{2}{1+\sin x} \,. \] Hence integrate \(\dfrac{1}{1+\sin x}\) with respect to \(x\).
  2. By means of the substitution \(y=\pi -x\), show that \[ \int_0^\pi x \f (\sin x)\, \d x = \frac \pi 2 \int_0^\pi \f(\sin x) \, \d x ,\] where \(\mathrm{f}\) is any function for which these integrals exist. Hence evaluate \[ \int_0^\pi \frac x {1+\sin x} \, \d x \,. \]
  3. Evaluate \[ \int_0^\pi\frac{ 2x^3 -3\pi x^2}{(1+\sin x)^2}\, \d x .\]

Solution:

  1. \(\,\) \begin{align*} && \sec^2\left(\tfrac14\pi-\tfrac12 x\right) &= \frac{1}{\cos^2 \left(\tfrac14\pi-\tfrac12 x\right)} \\ &&&= \frac{1}{\frac{1+\cos 2\left(\tfrac14\pi-\tfrac12 x\right)}{2}} \\ &&&= \frac{2}{1 + \cos \left(\tfrac12\pi- x\right)} \\ &&&= \frac{2}{1+\sin x} \\ \\ && \int \frac{1}{1+\sin x} \d x &= \int \tfrac12\sec^2\left(\tfrac14\pi-\tfrac12 x\right) \d x\\ &&&= - \tan\left(\tfrac14\pi-\tfrac12 x\right) + C \end{align*}
  2. \(\,\) \begin{align*} && I &= \int_0^{\pi} x f(\sin x) \d x \\ y = \pi - x, \d y = - \d x: &&&= \int_{y=\pi}^{y = 0} (\pi - y) f(\sin(\pi - y))(-1) \d y \\ &&&= \int_0^\pi (\pi - y) f(\sin y) \d y \\ &&&= \pi \int_0^\pi f(\sin y) \d y - I \\ \Rightarrow && I &= \frac{\pi}{2} \int_0^\pi f(\sin x) \d x \\ \\ \Rightarrow && \int_0^{\pi} \frac{x}{1 + \sin x} \d x &= \frac{\pi}{2} \int_0^{\pi} \frac{1}{1 + \sin x} \d x\\ &&&=\frac{\pi}{2} \left [- \tan\left(\tfrac14\pi-\tfrac12 x\right) \right]_0^{\pi} \\ &&&= \frac{\pi}{2} \left (-\tan (-\tfrac{\pi}{4}) + \tan \tfrac{\pi}{4} \right) \\ &&&= \pi \end{align*}
  3. \(\,\) \begin{align*} && J &= \int_0^{\pi} \frac{2x^3-3\pi x^2}{(1+\sin x)^2} \d x \\ y = \pi - x: &&&= \int_0^{\pi} \frac{2(\pi-y)^3-3\pi (\pi - y)^2}{(1+\sin x)^2 } \d y \\ &&&= \int_0^{\pi} \frac{-2 y^3 + 3 \pi y^2 - \pi^3}{(1+ \sin x)^2}\\ &&&= -\pi^3 \int_0^{\pi} \frac{1}{(1 + \sin x)^2} \d x -J \\ \Rightarrow && J &= -\frac{\pi^3}{2} \int_0^{\pi} \frac{1}{(1 + \sin x)^2} \d x\\ &&&= -\frac{\pi^3}{2} \int_0^\pi \tfrac14 \sec^4\left(\tfrac14\pi-\tfrac12 x\right) \d x \\ &&&= -\frac{\pi^3}{8} \int_0^\pi \sec^2\left(\tfrac14\pi-\tfrac12 x\right)\left (1 + \tan^2\left(\tfrac14\pi-\tfrac12 x\right) \right) \d x \\ &&&= -\frac{\pi^3}{8} \left [-\frac23 \tan^3\left(\tfrac14\pi-\tfrac12 x\right) - 2 \tan\left(\tfrac14\pi-\tfrac12 x\right) \right]_0^{\pi} \\ &&&= -\frac{\pi^3}{8} \left (\frac43+4 \right) \\ &&&= -\frac{2\pi^3}{3} \end{align*}

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2013 Paper 3 Q1
D: 1700.0 B: 1484.0
integration Weierstrass substitution t-substitution trigonometric reduction formula arctan recurrence relation

Given that \(t= \tan \frac12 x\), show that \(\dfrac {\d t}{\d x} = \frac12(1+t^2)\) and \( \sin x = \dfrac {2t}{1+t^2}\,\). Hence show that \[ \int_0^{\frac12\pi} \frac 1{1+a \sin x}\, \d x = \frac2 {\sqrt{1-a^2}} \arctan \frac{\sqrt{1-a}}{\sqrt{1+a}}\, \qquad \quad (0 < a < 1). \] Let \[ I_n = \int_0^{\frac12\pi} \frac{ \sin ^nx}{2+\sin x} \, \d x \qquad \quad (n\ge0). \] By considering \(I_{n+1}+2I_{n}\,\), or otherwise, evaluate \(I_3\).

Solution: Let \(t = \tan \frac12 x\), then \begin{align*} \frac{\d t}{\d x} &= \tfrac12 \sec^2 \tfrac12 t \\ &= \tfrac12 (1 + \tan^2 \tfrac12 ) \\ &= \tfrac12 (1 + t^2) \\ \\ \sin x &= 2 \sin \tfrac12 x \cos \tfrac12 \\ &= \frac{2 \frac{\sin \tfrac12 x}{ \cos \tfrac12x}}{\frac{1}{\cos^2 \tfrac12 x}} \\ &= \frac{2 \tan \tfrac12 x}{\sec^2 \tfrac12 } \\ &= \frac{2t }{1+t^2} \end{align*} Now consider \begin{align*} t = \tan \tfrac12 x: && \int_0^{\frac12\pi} \frac 1{1+a \sin x}\, \d x &= \int_{t=0}^{t = 1} \frac{1}{1 + a \frac{2t}{1+t^2}} \frac{2}{1+t^2} \d t \\ &&&= \int_0^1 \frac{2}{1+2at+t^2} \d t \\ &&&= \int_0^1 \frac{2}{(t+a)^2 + 1-a^2} \d t \\ (1-a^2) > 0: &&&= \left [ \frac{2}{\sqrt{1-a^2}} \arctan \frac{t+a}{\sqrt{1-a^2}} \right]_0^1 \\ &&&= \frac{2}{\sqrt{1-a^2}} \left ( \arctan \frac{1+a}{\sqrt{1-a^2}} - \arctan \frac{a}{\sqrt{1-a^2}} \right) \\ &&&= \frac{2}{\sqrt{1-a^2}} \arctan \left ( \frac{\frac{1+a}{\sqrt{1-a^2}}-\frac{a}{\sqrt{1-a^2}}}{1+\frac{1+a}{\sqrt{1-a^2}}\frac{a}{\sqrt{1-a^2}}} \right) \\ &&&= \frac{2}{\sqrt{1-a^2}} \arctan \left ( \frac{\frac{1}{\sqrt{1-a^2}}}{\frac{1+a}{1-a^2}} \right) \\ &&&= \frac{2}{\sqrt{1-a^2}} \arctan \left ( \frac{\sqrt{1-a}}{\sqrt{1+a}} \right) \end{align*} as required. Let \[ I_n = \int_0^{\frac12\pi} \frac{ \sin ^nx}{2+\sin x} \, \d x \qquad \quad (n\ge0). \] and consider \begin{align*} I_{n+1} + 2I_n &= \int_0^{\frac12\pi} \frac{ \sin ^{n+1}x+2\sin^{n} x}{2+\sin x} \, \d x \\ &= \int_0^{\frac12\pi} \frac{ \sin^n x (2 + \sin x)}{2+\sin x} \, \d x \\ &= \int_0^{\frac12\pi} \sin^n x \d x \end{align*} Therefore we can compute \begin{align*} I_0 &= \int_0^{\pi/2} \frac{1}{2 + \sin x} \d x \\ &= \frac12 \int_0^{\pi/2} \frac{1}{1 + \frac12 \sin x} \d x \\ &= \frac{1}{\sqrt{3/4}} \arctan \frac{\sqrt{1/2}}{\sqrt{3/2}} \\ &= \frac{2}{\sqrt{3}} \arctan \frac{1}{\sqrt{3}} \\ &= \frac{\pi}{3\sqrt{3}} \\ \\ I_1 &= \int_0^{\pi/2} 1 \d x - 2 I_0 \\ &= \frac{\pi}{2} - \frac{2\pi}{3\sqrt{3}} \\ I_2 &= \int_0^{\pi/2} \sin x \d x - 2I_1 \\ &= 1 - \pi + \frac{4\pi}{3\sqrt{3}} \\ I_3 &= \int_0^{\pi/2} \sin^2 x \d x - 2I_2 \\ &= \frac12 \int_0^{\pi/2} \sin^2 + \cos^2 x \d x - 2I_2 \\ &= \frac{\pi}{4} - 2 + 2\pi - \frac{8\pi}{3\sqrt{3}} \\ &= -2 + \frac{9\pi}{4} - \frac{8\pi}{3\sqrt{3}} \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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2000 Paper 2 Q6
D: 1600.0 B: 1585.4
integration Weierstrass substitution t-substitution trigonometric trigonometric identities definite integral half-angle substitution

Show that \[ \sin\theta = \frac {2t}{1+t^2}, \ \ \ \cos\theta = \frac{1-t^2}{1+t^2}, \ \ \ \frac{1+\cos\theta}{\sin\theta} = \tan (\tfrac{1}{2}\pi-\tfrac{1}{2}\theta), \] where \(t =\tan\frac{1}{2}\theta\). Use the substitution \(t =\tan\frac{1}{2}\theta\) to show that, for \(0<\alpha<\frac{1}{2}\pi\), \[ \int_0^{\frac{1}{2}\pi} {1 \over {1 + \cos\alpha \sin \theta}} \,\d\theta =\frac{\alpha}{\sin\alpha}\,, \] and deduce a similar result for \[ \int_0^{\frac{1}{2}\pi} {1 \over {1 + \sin\alpha \cos \theta}} \,\d\theta \,. \]

Solution: \begin{align*} && \frac{2t}{1+t^2} &= \frac{2 \sin \tfrac12 \theta \cos\tfrac12 \theta }{\cos^2 \tfrac12 \theta + \sin^2 \tfrac12 \theta} \\ &&&= \frac{\sin \theta}{1} = \sin \theta \\ \\ && \frac{1-t^2}{1+t^2} &= \frac{\cos^2 \tfrac12 \theta - \sin^2 \tfrac12 \theta}{\cos^2 \tfrac12 \theta + \sin^2 \tfrac12 \theta} \\ &&&= \frac{\cos \theta }{1} = \cos \theta \\ \\ && \tan(\tfrac12 \pi - \tfrac12 \theta) &= \frac{1}{t} \\ && \frac{1+\cos \theta}{\sin \theta} &= \frac{1 + \frac{1-t^2}{1+t^2}}{\frac{2t}{1+t^2}} \\ &&&= \frac{2}{2t} = \frac1t = \tan(\tfrac12\pi - \tfrac12 \theta) \end{align*} Notice also that \(\frac{\d t}{\d \theta} = \tfrac12 \sec^2 \tfrac12 \theta = \tfrac12(1 + t^2)\) so \begin{align*} && I &= \int_0^{\frac12 \pi} \frac{1}{1 + \cos \alpha \sin \theta} \d \theta \\ t = \tan \tfrac12 \theta, \d \theta = \frac{2}{1+t^2} \d t: &&&= \int_{0}^{1} \frac{1}{1 + \cos \alpha \frac{2t}{1+t^2}}\frac{2}{1+t^2} \d t \\ &&&= \int_0^1 \frac{2}{1+t^2 + 2\cos \alpha t} \d t \\ &&&= \int_0^1 \frac{2}{(t + \cos \alpha)^2+\sin^2 \alpha} \d t \\ &&&= \left [ \frac{2}{\sin \alpha} \tan^{-1} \left ( \frac{t+ \cos \alpha}{\sin \alpha} \right) \right]_0^1 \\ &&&= \frac{2}{\sin \alpha} \left ( \tan^{-1} \left ( \frac{1+ \cos \alpha}{\sin \alpha} \right) - \tan^{-1} \left ( \frac{ \cos \alpha}{\sin \alpha} \right) \right) \\ &&&= \frac{2}{\sin \alpha} \left ( \tan^{-1} \left (\tan (\tfrac12 \pi - \tfrac12 \alpha \right) - \tan^{-1} \left (\tan(\tfrac12\pi - \alpha )\right) \right) \\ &&&= \frac{2}{\sin \alpha} \left ( \tfrac12 \pi - \tfrac12 \alpha - \tfrac12 \pi + \alpha \right) \\ &&&= \frac{\alpha}{\sin \alpha} \end{align*} \begin{align*} && J &= \int_0^{\tfrac12 \pi} \frac{1}{1 + \sin \alpha \cos \theta} \d \theta \\ &&&= \int_0^{\tfrac12 \pi} \frac{1}{1 + \cos (\tfrac12 \pi - \alpha) \sin \theta} \d \theta \\ &&&= \frac{\tfrac12 \pi - \alpha}{\cos \alpha} \end{align*}

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1999 Paper 2 Q5
D: 1600.0 B: 1500.0
trigonometry R-formula t-substitution Weierstrass substitution quadratic equation inequality cosine proof

Show that if \(\alpha\) is a solution of the equation $$ 5{\cos x} + 12{\sin x} = 7, $$ then either $$ {\cos }{\alpha} = \frac{35 -12\sqrt{120}}{169} $$ or \(\cos \alpha\) has one other value which you should find. Prove carefully that if \(\frac{1}{2}\pi< \alpha < \pi\), then \(\alpha < \frac{3}{4}\pi\).

Solution: \begin{align*} && 5 \cos x + 12\sin x &= 7 \\ \Rightarrow && 5 \cos x - 7 &= -12 \sin x \\ \Rightarrow && 25 \cos^2 x - 70\cos x + 49 &= 144 \sin^2 x \\ \Rightarrow && 25 \cos^2 x - 70\cos x + 49 &= 144 (1-\cos^2 x) \\ \Rightarrow && 169 \cos^2 x - 70 \cos x -95 &= 0 \\ \Rightarrow && \cos \alpha &= \frac{70 \pm \sqrt{70^2 - 4 \cdot 169 \cdot (-95)}}{2 \cdot 169} \\ &&&= \frac{35 \pm \sqrt{35^2 + 169 \cdot 95} }{169} \\ &&&= \frac{35 \pm 12\sqrt{120}}{169} \end{align*} If \(\frac12 \pi < \alpha < \pi\) then \(\cos \alpha\) is negative, in particular \(\cos \alpha = \frac{35 -12\sqrt{120}}{169}\). Since \(\cos\) is decreasing over this range, if \(\cos \alpha > \cos \frac34 \pi = -\frac{\sqrt{2}}2\), then we will have shown \(\alpha < \frac34 \pi\) \begin{align*} && \cos \alpha &= \frac{35 - 12 \sqrt{120}}{169} \\ &&&> \frac{35 - 12 \cdot \sqrt{121}}{169} \\ &&&= \frac{35 - 12 \cdot 11}{169} \\ &&&= \frac{35 - 132}{169} \\ &&&= -\frac{97}{169} \\ &&&> -\frac{8}{13} \end{align*} but \(\left ( \frac{8}{13} \right)^2 = \frac{64}{169} < \frac12\), so we are done.

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1996 Paper 2 Q4
D: 1600.0 B: 1470.2
integration trigonometric substitution cos nθ Weierstrass substitution double angle formula definite integral algebraic manipulation

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 \(\tan\frac{\pi}{8}=\surd{2}-1\).]

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*}

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1996 Paper 3 Q3
D: 1700.0 B: 1500.0
integration Weierstrass substitution t-substitution trigonometric definite integral inverse trigonometric functions

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 \[ \int_{0}^{\frac{3}{4}\pi}\frac{1}{\sqrt{2}+\cos x}\,\mathrm{d}x=\frac{\pi}{2}\,. \]

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*}

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1994 Paper 1 Q8
D: 1516.0 B: 1500.8
integration substitution logarithmic trigonometric definite integral symmetry argument Weierstrass substitution tan addition formula

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 \[ {\displaystyle \int_{0}^{1}\frac{\ln(1+x)}{1+x^{2}}\,\mathrm{d}x}\qquad\mbox{ and }\qquad{\displaystyle \int_{0}^{\frac{1}{2}\pi}\ln\left(\frac{1+\sin x}{1+\cos x}\right)\,\mathrm{d}x}. \]

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*}

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1992 Paper 2 Q8
D: 1600.0 B: 1484.0
integration partial fractions Weierstrass substitution t-substitution hyperbolic functions trigonometric logarithmic integration

Calculate the following integrals

  1. \({\displaystyle \int\frac{x}{(x-1)(x^{2}-1)}\,\mathrm{d}x}\);
  2. \({\displaystyle \int\frac{1}{3\cos x+4\sin x}\,\mathrm{d}x}\);
  3. \({\displaystyle \int\frac{1}{\sinh x}\,\mathrm{d}x}.\)

Solution:

  1. \begin{align*} \int\frac{x}{(x-1)(x^{2}-1)}\,\mathrm{d}x &= \int \frac{x}{(x-1)^2 (x+1)} \d x \\ &= \int \frac{1}{2(x-1)^2} + \frac{1}{4(x-1)} - \frac{1}{4(x+1)} \d x \\ &= -\frac12 (x-1)^{-1} + \frac14 \ln(x-1) - \frac14 \ln (x+1) + C \end{align*}
  2. \begin{align*} \int \frac{1}{3 \cos x + 4 \sin x } \d x &= \int \frac{1}{5 \cos (x - \cos^{-1}(3/5))} \d x \\ &= \frac15 \int \sec (x - \cos^{-1}(3/5)) \d x\\ &= \frac15 \left (\ln | \sec (x - \cos^{-1}(3/5)) + \tan (x - \cos^{-1}(3/5)) | \right) + C \end{align*}
  3. \begin{align*} \int \frac{1}{\sinh x} \d x &= \int \frac{2}{e^x - e^{-x}} \\ &= \int \frac{2e^x}{e^{2x}-1} \d x \\ &=\int \frac{e^x}{e^x-1} - \frac{e^x}{e^x+1} \d x \\ &= \ln (e^x - 1) + \ln (e^x+1) + C \end{align*}

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1991 Paper 1 Q8
D: 1500.0 B: 1516.9
integration substitution symmetry argument Weierstrass substitution t-substitution trigonometric inverse trigonometric functions definite integral

  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.{]}

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*}

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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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