Find \(\int {\frac{{{{\rm{e}}^x}}}{{1 + {{\rm{e}}^x}}}} {\rm{d}}x\) .

Find \(\int {\sin 3x\cos 3x{\rm{d}}x} \) .

attempt to use substitution or inspection     M1

e.g. \(u = 1 + {{\rm{e}}^x}\) so \(\frac{{{\rm{d}}u}}{{{\rm{d}}x}} = {{\rm{e}}^x}\)

correct working     A1

e.g. \(\int {\frac{{{\rm{d}}u}}{u}}  = \ln u\)

\(\ln (1 + {{\rm{e}}^x}) + C\)     A1     N3

[3 marks]

METHOD 1

attempt to use substitution or inspection     M1

e.g. let \(u = \sin 3x\)

\(\frac{{{\rm{d}}u}}{{{\rm{d}}x}} = 3\cos 3x\)     A1

\(\frac{1}{3}\int {u{\rm{d}}u = } \frac{1}{3} \times \frac{{{u^2}}}{2} + C\)    A1

\(\int {\sin 3x\cos 3x{\rm{d}}x = } \frac{{{{\sin }^2}3x}}{6} + C\)     A1     N2

METHOD 2

attempt to use substitution or inspection     M1

e.g. let \(u = \cos 3x\)

\(\frac{{{\rm{d}}u}}{{{\rm{d}}x}} = - 3\sin 3x\)     A1

\( - \frac{1}{3}\int {u{\rm{d}}u = } - \frac{1}{3} \times \frac{{{u^2}}}{2} + C\)     A1

\(\int {\sin 3x\cos 3x{\rm{d}}x = } \frac{{{{\cos }^2}3x}}{6} + C\)     A1     N2

METHOD 3

recognizing double angle     M1

correct working     A1

e.g. \(\frac{1}{2}\sin 6x\)

\(\int {\sin 6x{\rm{d}}x = } \frac{{ - \cos 6x}}{6} + C\)     A1

\(\int {\frac{1}{2}\sin 6x{\rm{d}}x = - \frac{{\cos 6x}}{{12}}}  + C\)     A1     N2

[4 marks]