Given that \(\frac{{{\rm{d}}x}}{{{\rm{d}}y}} + 2y\tan x = \sin x\) , and \(y = 0\) when \(x = \frac{\pi }{3}\) , find the maximum value of y.

integrating factor \( = {{\rm{e}}^{\int {2\tan xdx} }}\)    M1

\( = {{\rm{e}}^{2\ln \sec x}}\)     A1 

\( = {\sec ^2}x\)     A1

it follows that

\(y{\sec ^2}x = \int {\sin x{{\sec }^2}x{\rm{d}}x} \)     M1

\( = \int {\sec x\tan x{\rm{d}}x} \)     (A1)

\( = \sec x + C\)     A1

substituting,

\(0 = 2 + C\) so \(C = - 2\)     M1A1

the solution is

\(y = \cos x - 2{\cos ^2}x\)    A1

EITHER

using a GDC

maximum value of \(y\) is \(0.125\)     A2

OR

\(y' = - \sin x + 4\sin x\cos x = 0\)     M1

\( \Rightarrow \cos x = \frac{1}{4}\) (or \(\sin x = 0\) which leads to a minimum)

\( \Rightarrow y = \frac{1}{8}\)     A1

[11 marks]

Solutions were often disappointing with some candidates even unable to find the integrating factor because of an inability to integrate \(2\tan x\) . Some candidates who found the integrating factor correctly were then unable to integrate \(\sin x{\sec ^2}x\) and others omitted the constant of integration. Some of the candidates who obtained the correct expression for y failed to realise that the quickest way to find the maximum value was to plot the graph of \(y|) on their calculator.