Year: 2012
Paper: 1
Question Number: 5
Course: LFM Pure
Section: Integration
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1485.6
Banger Comparisons: 1
Show that
\[
\int_0^{\frac14\pi} \sin (2x) \ln(\cos x)\, \d x = \frac14(\ln 2 -1)\,,
\]
and that
\[
\int_0^{\frac14\pi} \cos (2x) \ln(\cos x)\, \d x = \frac18(\pi -\ln 4-2)\,.
\]
Hence evaluate
\[
\int_{\frac14\pi}^{\frac12\pi}
\big ( \cos(2x) + \sin (2x)\big) \, \ln \big( \cos x + \sin x\big)\, \d x\,.
\]\begin{align*}
&&\int_0^{\frac14\pi} \sin (2x) \ln(\cos x)\, \d x &= \int_0^{\frac14 \pi} 2 \sin x \cos x \ln (\cos x) \d x \\
u = \cos \theta :&&&= \int_{u=1}^{u=\frac1{\sqrt2}} -2u \ln u \d u \\
&&&= \int_{\frac1{\sqrt{2}}}^1 2u \ln u \d u \\
&&&= \left [u^2 \ln u \right]_{\frac1{\sqrt{2}}}^1-\int_{\frac1{\sqrt{2}}}^1 u \d u \\
&&&= -\frac12 \ln \frac{1}{\sqrt{2}} - \l\frac12 - \frac14 \r \\
&&&= \frac14 (\ln 2 - 1)
\end{align*}
\begin{align*}
&& \int_0^{\frac14\pi} \cos (2x) \ln(\cos x)\, \d x &= \left [ \frac12 \sin 2x \ln (\cos x) \right]_0^{\frac14\pi}- \int_0^{\frac14\pi} \frac12 \sin 2x \frac{-\sin x}{\cos x} \d x \\
&&&=\frac12 \ln \frac{1}{\sqrt{2}}+\int_0^{\frac14\pi} \sin^2 x \d x \\
&&&= -\frac14 \ln 2 + \int_0^{\frac14\pi} \frac{1-\cos 2x }{2} \d x \\
&&&= -\frac14 \ln 2 +\frac{\pi}{8} -\frac{1}{4} \\
&&&= \frac18 (\pi - 2\ln 2 - 2) \\
&&&= \frac18 (\pi - \ln 4 - 2) \\
\end{align*}
Notice that $\cos x + \sin x = \sqrt{2} \cos (x -\frac{\pi}{4})$, so:
\begin{align*}
&&\int_{\frac14\pi}^{\frac12\pi}
\big ( \cos(2x) + \sin (2x)\big) \ln \big( \cos x + \sin x\big)\d x &= \int_{\frac14\pi}^{\frac12\pi}
\big ( \cos(2x) + \sin (2x)\big) \ln (\sqrt{2} \cos ( x - \frac{\pi}{4}) ) \d x \\
&&&= \int_{u=0}^{u=\frac{\pi}{4}} \l \cos(2u+\frac{\pi}{2})+\sin(2u+\frac{\pi}{2}) \r \cdot \l \frac12 \ln 2 +\ln (\cos u) \r \d x \\
&&&= \int_{0}^{\frac{\pi}{4}} \l -\sin 2u+\cos 2u \r \cdot \l \frac12 \ln 2 +\ln (\cos u) \r \d x \\
&&&= \frac14 \ln 2\left [ \cos 2u + \sin 2u \ \right]_{0}^{\frac{\pi}{4}} - \frac14(\ln2 - 1) + \frac18\pi - \frac14(\ln 2 +1) \\
&&&= \frac{\pi}{8}-\frac12 \ln 2
\end{align*}
This was another popular question, with attempts by a large number of candidates. Integration by parts was required for both of the main methods for the first integral, either preceded or followed by the application of some trigonometric identities. A number of candidates managed to obtain the correct answers for each of the first two integrals, but then struggled to relate them to the final part of the question, in some cases ignoring the different limits and in others incorrectly manipulating the logarithms.