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2002 Paper 1 Q7
D: 1484.0 B: 1500.0
integration trigonometric symmetry addition of integrals logarithmic result partial fractions definite integral

Let \[ I= \int_0^a \frac {\cos x}{\sin x + \cos x} \; \d x \, \quad \mbox{ and } \quad J= \int_0^a \frac {\sin x}{\sin x + \cos x} \; \d x \;, \] where \(0\le a < \frac{3}{4}\pi\,\). By considering \(I+J\) and \(I-J\), show that $ 2I= a + \ln (\sin a +\cos a)\;. $ Find also:

  1. \(\displaystyle \int_0^{\frac{1}{2}\pi} \frac {\cos x}{p\sin x + q\cos x} \; \d x \,\), where \(p\) and \(q\) are positive numbers; %
  2. [(ii)] %\(\displaystyle \int_0^{\frac{1}{2}\pi/2} \frac {\cos x}{\sin (x+k)} \; \d x \,\), where \(0 < k < \pi/2\,\);
  3. \(\displaystyle \int_0^{\frac{1}{2}\pi} \frac {\cos x+4}{3\sin x + 4\cos x+ 25} \; \d x \,\).

Solution: \begin{align*} && I + J &= \int_0^a \frac{\sin x + \cos x}{\sin x + \cos x } \d x = a \\ && I - J &= \int_0^a \frac{\cos x - \sin x}{\sin x + \cos x} \d x \\ &&&= \left [\ln ( \sin x + \cos x) \right]_0^a = \ln (\sin a + \cos a) - \ln 1 = \ln(\sin a + \cos a) \\ \\ \Rightarrow && 2I &= a + \ln(\sin a + \cos a) \end{align*}

  1. Let \(\displaystyle I = \int_0^{\frac12 \pi} \frac{\cos x}{p \sin x + q \cos x} \d x, J = \int_0^{\frac12 \pi} \frac{\sin x}{p \sin x + q \cos x} \d x\) so \begin{align*} && qI + pJ &= \frac{\pi}{2} \\ && pI - qJ &= \int_0^{\frac12 \pi} \frac{p \cos x - q \sin x}{p \sin x + q \cos x } \d x \\ &&&= \left [\ln (p \sin x + q \cos x) \right]_0^{\pi/2} \\ &&&= \ln(p) - \ln(q) = \ln \frac{p}{q} \end{align*}
  2. \(\,\) \begin{align*} && \int_0^{\frac{\pi}{2}} \frac{\cos x}{\sin(x + k)} \d x &= \int_0^{\frac{\pi}{2}} \frac{\cos x}{\sin(x) \cos(k) + \cos(x) \sin (k)} \d x \\ &&&= \ln \tan k \end{align*}
  3. Let \(\displaystyle I = \int_0^{\pi/2} \frac{\cos x + 4}{3 \sin x + 4 \cos x + 25} \d x, J = \int_0^{\pi/2} \frac{\sin x + 3}{3 \sin x + 4 \cos x + 25} \d x\), so \begin{align*} && 4I + 3J &= \int_0^{\pi/2} \frac{3 \sin x + 4 \cos x + 25}{3 \sin x + 4 \cos x + 25} \d x \\ &&&= \frac{\pi}{2} \\ && 3I - 4J &= \int_0^{\pi/2} \frac{3\cos x - 4 \sin x}{3 \sin x + 4 \cos x + 25} \d x \\ &&&= \left [\ln(3 \sin x + 4 \cos x + 25) \right]_0^{\pi/2} \\ &&&= \ln (28) - \ln (29) = \ln \frac{28}{29} \\ \Rightarrow && 25I &= 2\pi + 3 \ln \frac{28}{29} \\ \Rightarrow && I &= \frac{2}{25} \pi + \frac{3}{25} \ln \frac{28}{29} \end{align*}

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