Year: 2007
Paper: 1
Question Number: 3
Course: LFM Pure
Section: Integration
There were significantly more candidates attempting this paper this year (an increase of nearly 50%), but many found it to be very difficult and only achieved low scores. In particular, the level of algebraic skill required by the questions was often lacking. The examiners' express their concern that this was the case despite a conscious effort to make the paper more accessible than last year's. At this level, the fluent, confident and correct handling of mathematical symbols (and numbers) is necessary and is expected; many good starts to questions soon became unstuck after a simple slip. Graph sketching was usually poor: if future candidates wanted to improve one particular skill, they would be well advised to develop this. There were of course some excellent scripts, full of logical clarity and perceptive insight. It was pleasing to note that the applied questions were more popular this year, and many candidates scored well on at least one of these. It was however surprising how rarely answers to questions such as 5, 9, 10, 11 and 12 began with a diagram. However, the examiners were left with the overall feeling that some candidates had not prepared themselves well for the examination. The use of past papers to ensure adequate preparation is strongly recommended. A student's first exposure to STEP questions can be a daunting, demanding experience; it is a shame if that takes place during a public examination on which so much rides. Further, and fuller, discussion of the solutions to these questions can be found in the Hints and Answers document.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
Prove the identities
$\cos^4\theta -\sin^4\theta \equiv \cos 2\theta$ and
$\cos^4 \theta + \sin^4 \theta \equiv 1 - {\frac12}
\sin^2 2 \theta$. Hence or otherwise evaluate
\[
\int_0^{\frac{1}{2}\pi} \cos^4 \theta \; \d \theta \;\;\;\;
\mbox{and}\;\;\;\; \int_0^{\frac{1}{2}\pi} \sin^4 \theta \; \d \theta \,.
\]
Evaluate also
\[
\int_0^{\frac{1}{2}\pi} \cos^6 \theta \; \d \theta \;\;\;\;
\mbox{and}\;\;\;\; \int_0^{\frac{1}{2}\pi} \sin^6 \theta \; \d \theta \,.
\]\begin{align*}
&& \cos^4 \theta - \sin^4 \theta &= (\cos^2 \theta - \sin^2 \theta)(\cos^2 \theta + \sin^2 \theta) \\
&&&= \cos^2 \theta - \sin^2 \theta \\
&&&= \cos 2 \theta \\
\\
&& 1&= (\cos^2 \theta + \sin^2 \theta)^2 \\
&&&= \cos^4 \theta + \sin^4 \theta + 2 \sin^2 \theta \cos^2 \theta \\
&&&= \cos^4 \theta + \sin^4 \theta + \frac12 ( \sin^2 2 \theta) \\
\Rightarrow && \cos^4 \theta + \sin^4 \theta &= 1 - \tfrac12 \sin^2 2 \theta
\end{align*}
\begin{align*}
&& I &= \int_{0}^{\pi/2} \cos^4 \theta \d \theta \\
&& J &= \int_0^{\pi/2} \sin^4 \theta \d \theta \\
&& I-J &= \int_0^{\pi/2} \cos 2 \theta \d \theta = 0 \\
&& I+J &= \int_0^{\pi/2} (1- \frac12 \sin^2 2 \theta) \d \theta \\
&&&= \frac{\pi}{2} - \frac14 \int_0^{\pi} \sin^2 \theta \d \theta \\
&&&= \frac{\pi}{2} - \frac{\pi}{8} \\
&&&= \frac{3\pi}{8} \\
\Rightarrow && I=J &= \frac{3\pi}{16}
\end{align*}
\begin{align*}
&& \cos^6 \theta + \sin^6 \theta &= (\cos^2 \theta + \sin^2 \theta)(\cos^4 \theta - \cos^2 \theta \sin^2 \theta + \sin^4 \theta) \\
&&&= 1-\tfrac12 \sin^2 2\theta - \tfrac14 \sin^2 2 \theta \\
&&&= 1 - \tfrac34 \sin^2 2 \theta \\
%&& \cos^6 \theta - \sin^6 \theta &= (\cos^2 \theta - \sin^2 \theta)(\cos^4 \theta + \cos^2 \theta \sin^2 \theta + \sin^4 \theta) \\
%&&&= \cos 2 \theta (1 - \tfrac12 \sin^2 2 \theta + \tfrac14 \sin^2 2 \theta) \\
%&&&= \cos 2 \theta (1 - \tfrac14 \sin^2 2 \theta) \\
\end{align*}
\begin{align*}
&& I &= \int_{0}^{\pi/2} \cos^6 \theta \d \theta \\
&& J &= \int_0^{\pi/2} \sin^6 \theta \d \theta \\
&& I-J &= 0 \\
&& I+J &= \int_0^{\pi/2} (1 - \tfrac34 \sin^2 2 \theta) \d \theta \\
&&&= \frac{\pi}{2} - \frac{3\pi}{16} = \frac{5\pi}{16} \\
\Rightarrow && I = J &= \frac{5\pi}{32}
\end{align*}
This was the most popular question on the paper, and many different methods were seen. The intended method was to use the identities cos4θ – sin4 θ ≡ cos 2 θ and cos4 θ + sin4 θ ≡ 1 – ½ sin2 2 θ to evaluate the integrals of cos4 θ – sin4 θ and cos4 θ + sin4 θ, and hence be able to write down separately the values of the integrals of cos4 θ and sin4 θ. A similar approach works well for cos6 θ – sin6 θ and cos6 θ + sin6 θ. Other methods were, of course, acceptable, and many candidates received high marks for this question.