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2016 Paper 2 Q7
D: 1600.0 B: 1516.0
integration substitution symmetry trigonometric logarithm definite integral integral identity

Show that \[ \int_0^a \f(x) \d x= \int _0^a \f(a-x) \d x\,, \tag{\(*\)} \] where f is any function for which the integrals exist.

  1. Use (\(*\)) to evaluate \[ \int_0^{\frac12\pi} \frac{\sin x}{\cos x + \sin x} \, \d x \,. \]
  2. Evaluate \[ \int_0^{\frac14\pi} \frac{\sin x}{\cos x + \sin x} \, \d x \,. \]
  3. Evaluate \[ \int_0^{\frac14\pi} \ln (1+\tan x) \, \d x \,. \]
  4. Evaluate \[ \int_0^{\frac14 \pi} \frac x {\cos x \, (\cos x + \sin x)}\, \d x \,. \]

Solution: \begin{align*} u = a-x, \d u = - \d x: && \int_0^a f(x) \d x &= \int_{u=a}^{u=0} f(a-u) (-1) \d u \\ &&&= \int_0^a f(a-u) \d u \\ &&&= \int_0^a f(a-x) \d x \end{align*}

  1. \begin{align*} && I &= \int_0^{\frac12 \pi} \frac{\sin x}{\cos x + \sin x } \d x\\ &&&= \int_0^{\frac12 \pi} \frac{\sin (\frac12 \pi - x)}{\cos (\frac12 \pi-x) + \sin (\frac12 \pi-x) } \d x\\ &&&= \int_0^{\frac12 \pi} \frac{\cos x}{\sin x + \cos x } \d x\\ \Rightarrow && 2I &= \int_0^{\frac12 \pi} 1 \d x \\ \Rightarrow && I &= \frac{\pi}{4} \end{align*}
  2. \begin{align*} && I &= \int_0^{\frac14 \pi} \frac{\sin x}{\cos x + \sin x } \d x\\ &&&= \int_0^{\frac14 \pi} \frac{\sin (\frac14 \pi - x)}{\cos (\frac14 \pi-x) + \sin (\frac14 \pi-x) } \d x\\ &&&= \int_0^{\frac14 \pi} \frac{\frac1{\sqrt{2}} \cos x - \frac{1}{\sqrt{2}} \sin x}{\frac1{\sqrt{2}} \cos x + \frac{1}{\sqrt{2}} \sin x + \frac1{\sqrt{2}} \cos x - \frac{1}{\sqrt{2}} \sin x} \d x \\ &&&= \int_0^{\frac14 \pi} \frac{\cos x - \sin x}{2 \cos x} \d x \\ &&&= \left [\frac12 x + \ln(\cos x) \right]_0^{\pi/4} \\ &&&= \frac{\pi}{8} -\frac12\ln2 - 1 \end{align*}
  3. \begin{align*} && I &= \int_0^{\frac14\pi} \ln (1+\tan x) \, \d x \\ &&&= \int_0^{\frac14 \pi} \ln \left (1 + \tan \left(\frac{\pi}{4} - x\right) \right) \, \d x\\ &&&= \int_0^{\frac14 \pi} \ln \left (1 +\frac{1 - \tan x}{1+ \tan x} \right) \, \d x\\ &&&= \int_0^{\frac14 \pi} \ln \left (\frac{2}{1+ \tan x} \right) \, \d x\\ &&&= \frac{\pi}{4} \ln 2 - I \\ \Rightarrow && I &= \frac{\pi}{8} \ln 2 \end{align*}
  4. \begin{align*} && I &= \int_0^{\frac14 \pi} \frac x {\cos x \, (\cos x + \sin x)}\, \d x \\ &&&= \int_0^{\frac14 \pi} \frac {\frac14 \pi - x} {(\frac1{\sqrt{2}} \cos x + \frac{1}{\sqrt{2}} \sin x) \, (\frac{2}{\sqrt{2}}\cos x)}\, \d x \\ &&&= \int_0^{\frac14 \pi} \frac {\frac14 \pi - x} {\cos x \, (\cos x + \sin x)}\, \d x \\ \\ \Rightarrow && I &= \frac{\pi}{8} \int_0^{\pi/4} \frac{\sec^2 x}{1 + \tan x} \d x\\ &&&= \frac{\pi}{8} \left [\ln (1 + \tan x) \right]_0^{\pi/4} \\ &&&= \frac{\pi}{8} \ln 2 \end{align*}

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2010 Paper 2 Q4
D: 1600.0 B: 1516.0
integration substitution symmetry definite integral logarithmic trigonometric clever substitution integral identity

  1. Let \[ I=\int_0^a \frac {\f(x)}{\f(x)+\f(a-x)} \, \d x\,. \] Use a substitution to show that \[ I = \int_0^a \frac {\f(a-x)}{\f(x)+\f(a-x)} \, \d x\, \] and hence evaluate \(I\) in terms of \(a\). Use this result to evaluate the integrals \[ \int_0^1 \frac{\ln (x+1)}{\ln (2+x-x^2)}\, \d x \ \ \ \ \ \ \text{ and }\ \ \ \ \ \int_0^{\frac\pi 2} \frac{\sin x } {\sin(x+\frac \pi 4 )} \, \d x \,. \]
  2. Evaluate \[ \int_{\frac12}^2 \frac {\sin x}{x \big(\sin x + \sin \frac 1 x\big)} \, \d x\,. \]

Solution:

  1. \(\,\) \begin{align*} && I &= \int_0^a \frac{f(x)}{f(x)+f(a-x)} \d x \\ u =a-x, \d u = - \d x: &&& \int_{u=a}^{u=0} \frac{f(a-u)}{f(a-u)+f(u)} (-1) \d u \\ &&&= \int_0^a \frac{f(a-u)}{f(u)+f(a-u)} \d u \\ &&&= \int_0^a \frac{f(a-x)}{f(x)+f(a-x)} \d x \\ \Rightarrow && 2 I &= \int_0^a \left ( \frac{f(x)}{f(x)+f(a-x)} + \frac{f(a-x)}{f(x)+f(a-x)} \right) \d x \\ &&&= \int_0^a 1 \d x \\ &&&= a \\ \Rightarrow && I &= \frac{a}{2} \end{align*} \begin{align*} && J &= \int_0^1 \frac{\ln (x+1)}{\ln (2+x-x^2)}\, \d x \\ &&&= \int_0^1 \frac{\ln (x+1)}{\ln((x+1)(2-x))} \d x \\ &&&= \int_0^1 \frac{\ln (x+1)}{\ln(x+1) + \ln ((1-x)+1)} \d x \\ &&&= \frac{1}{2} \tag{\(f(x) = \ln (x+1)\)} \\ \\ && K &= \int_0^{\frac\pi 2} \frac{\sin x } {\sin(x+\frac \pi 4 )} \, \d x \\ &&&= \int_0^{\frac{\pi}{2}} \frac{\sin x }{\sin x \cos \frac{\pi}{4} + \cos x \sin \frac{\pi}{4}} \\ &&&= \sqrt{2} \int_0^{\frac{\pi}{2}} \frac{\sin x }{\sin x + \sin (\frac{\pi}{2}-x)} \d x\\ &&&= \frac{\pi}{2\sqrt{2}} \end{align*}
  2. \(\,\) \begin{align*} &&I &= \int_{\frac12}^2 \frac{\sin x }{x(\sin x + \sin \frac1x)} \d x \\ u = 1/x, \d u = -1/x^2 \d x : &&&= \int_{u = 2}^{u=\frac12} \frac{\sin \frac1u}{\frac{1}{u}(\sin \frac1u + \sin u)} (-\frac{1}{u^2} ) \d u \\ &&&= \int_{\frac12}^2 \frac{\sin \frac1u}{u (\sin u + \sin \frac1u)} \d u \\ \Rightarrow && 2I &= \int_{\frac12}^2 \left ( \frac{\sin x }{x(\sin x + \sin \frac1x)} + \frac{\sin \frac1x }{x(\sin x + \sin \frac1x)}\right) \d x \\ &&&= \int_{\frac12}^2 \frac{1}{x} \d x\\ &&&= 2\ln2 \\ \Rightarrow && I &= \ln 2 \end{align*}

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