(a)     Integrate $\int {\frac{{\sin \theta }}{{1 - \cos \theta }}} {\text{d}}\theta$ .

(b)     Given that $\int_{\frac{\pi }{2}}^a {\frac{{\sin \theta }}{{1 - \cos \theta }}} {\text{d}}\theta  = \frac{1}{2}$ and $\frac{\pi }{2} < a < \pi$, find the value of $a$ .

(a)     $\int {\frac{{\sin \theta }}{{1 - \cos \theta }}} {\text{d}}\theta  = \int {\frac{{\left( {1 - \cos \theta } \right)'}}{{1 - \cos \theta }}} {\text{d}}\theta  = \ln \left( {1 - \cos \theta } \right) + C$     (M1)A1A1

Note: Award A1 for $\ln \left( {1 - \cos \theta } \right)$ and A1 for C.

(b)     $\int_{\frac{\pi }{2}}^a {\frac{{\sin \theta }}{{1 - \cos \theta }}} {\text{d}}\theta  = \frac{1}{2} \Rightarrow \left[ {\ln \left( {1 - \cos \theta } \right)} \right]_{\frac{\pi }{2}}^a = \frac{1}{2}$     M1

$1 - \cos a = {{\text{e}}^{\frac{1}{2}}} \Rightarrow a = \arccos \left( {1 - \sqrt {\text{e}} } \right)$) or $2.28$     A1     N2

[5 marks]

Generally well answered, although many students did not include the constant of integration.