METHOD 1

\({\text{area}} = \mathop \smallint \limits_0^{\sqrt 3 } \arctan x{\text{d}}x\)     A1

attempting to integrate by parts     M1

\( = \left[ {x\arctan x} \right]_0^{\sqrt 3 } - \mathop \smallint \limits_0^{\sqrt 3 } x\frac{1}{{1 + {x^2}}}{\text{d}}x\)     A1A1

\( = \left[ {x\arctan x} \right]_0^{\sqrt 3 } - \left[ {\frac{1}{2}\ln \left( {1 + {x^2}} \right)} \right]_0^{\sqrt 3 }\)     A1

Note: Award A1 even if limits are absent.

\( = \frac{\pi }{{\sqrt 3 }} - \frac{1}{2}\ln 4\)     A1

\(\left( { = \frac{{\pi \sqrt 3 }}{3} - \ln 2} \right)\)

METHOD 2

\({\text{area}} = \frac{{\pi \sqrt 3 }}{3} - \mathop \smallint \limits_0^{\frac{\pi }{3}} \tan y{\text{d}}y\)     M1A1A1

\(\ { = \frac{{\pi \sqrt 3 }}{3} + \left[ {\ln \left| {\cos y} \right|} \right]_0^{\frac{\pi }{3}}} \)     M1A1

\( = \frac{{\pi \sqrt 3 }}{3} + \ln \frac{1}{2}\)     \(\left( { = \frac{{\pi \sqrt 3 }}{3} - \ln 2} \right)\)     A1

[6 marks]

Many candidates were able to write down the correct expression for the required area, although in some cases with incorrect integration limits. However, very few managed to achieve any further marks due to a number of misconceptions, in particular \(\arctan x = \cot x = \frac{{\cos x}}{{\sin x}}\). Candidates who realised they should use integration by parts were in general very successful in answering this question. It was pleasing to see a few alternative correct approaches to this question.