By considering integration as the reverse of differentiation, show that for

\(0 \leqslant x < \frac{\pi }{2}\)

\[\int {\sec x{\text{d}}x = \ln (\sec x + \tan x) + C.} \]

Hence, using integration by parts, show that

\[\int {{{\sec }^3}x{\text{d}}x = \frac{1}{2}\left( {\sec x\tan x + \ln (\sec x + \tan x)} \right) + C.} \]

Find an integrating factor and hence solve the differential equation, giving your answer in the form \(y = f(x)\).

Starting with the differential equation, show that

\[\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + y = 2{\sec ^2}x\tan x.\]

Hence, by using your calculator to draw two appropriate graphs or otherwise, find the \(x\)-coordinate of the point of inflection on the graph of \(y = f(x)\).

\(\frac{{\text{d}}}{{{\text{d}}x}}\left( {\ln (\sec x + \tan x)} \right) = \frac{{\sec x\tan x + {{\sec }^2}x}}{{\sec x + \tan x}}\)     M1

\( = \sec x\)     A1

therefore \(\int {\sec x{\text{d}}x = \ln (\sec x + \tan x) + C} \)     AG

[4 marks]

\(\int {{{\sec }^3}x{\text{d}}x = \int {\sec x \times {{\sec }^2}x{\text{d}}x} } \)     M1

\( = \sec x\tan x - \int {\sec x{{\tan }^2}x{\text{d}}x} \)     A1A1

\( = \sec x\tan x - \int {\sec x({{\sec }^2}x - 1){\text{d}}x} \)     A1

\( = \sec x\tan x - \int {{{\sec }^3}x{\text{d}}x + \int {\sec x{\text{d}}x} } \)

\( = \sec x\tan x - \int {{{\sec }^3}x{\text{d}}x + \ln (\sec x + \tan x)} \)     A1

\(2\int {{{\sec }^3}x{\text{d}}x = \left( {\sec x\tan x + \ln (\sec x + \tan x)} \right)} \)     A1

therefore

\(\int {{{\sec }^3}x{\text{d}}x = \frac{1}{2}\left( {\sec x\tan x + \ln (\sec x + \tan x)} \right) + C} \)     AG

[4 marks]

\({\text{int factor}} = {{\text{e}}^{\int {\tan x{\text{d}}x} }}\)     (M1)

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

\( = \sec x\)     A1

the differential equation can be written as

\(\frac{{\text{d}}}{{{\text{d}}x}}(y\sec x) = 2{\sec ^3}x\)     M1A1

integrating,

\(y\sec x = \sec x\tan x + \ln (\sec x + \tan x) + C\)     A1

putting \(x = 0,{\text{ }}y = 1,\)     M1

\(C = 1\)     A1

the solution is \(y = \cos x\left( {\sec x\tan x + \ln (\sec x + \tan x) + 1} \right)\)     A1

[??? marks]

differentiating the differential equation,

\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + \frac{{{\text{d}}y}}{{{\text{d}}x}}\tan x + y{\sec ^2}x = 4{\sec ^2}x\tan x\)     A1A1

\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + (2{\sec ^2}x - y\tan x)\tan x + y{\sec ^2}x = 4{\sec ^2}x\tan x\)     A1

\(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + y = 2{\sec ^2}x\tan x\)     AG

[??? marks]

at a point of inflection, \(\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = 0\) so \(y = 2{\sec ^2}x\tan x\)     (M1)

therefore the point of inflection can be found as the point of intersection of the graphs of \(y = \cos x\left( {\sec x\tan x + \ln (\sec x + \tan x) + 1} \right)\)

and \(y = 2{\sec ^2}x\tan x\)     (M1)

drawing these graphs on the calculator, \(x = 0.605\)     A2

[??? marks]