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]