Find \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\).

Find \(\int_0^1 {{\text{arccos}}\left( {\frac{x}{2}} \right){\text{d}}x} \).

\(y = {\text{arccos}}\left( {\frac{x}{2}} \right) \Rightarrow \frac{{{\text{d}}y}}{{{\text{d}}x}} =  - \frac{1}{{2\sqrt {1 - {{\left( {\frac{x}{2}} \right)}^2}} }}\left( { =  - \frac{1}{{\sqrt {4 - {x^2}} }}} \right)\)    M1A1

Note: M1 is for use of the chain rule.

[2 marks]

attempt at integration by parts     M1

\(u = {\text{arccos}}\left( {\frac{x}{2}} \right) \Rightarrow \frac{{{\text{d}}u}}{{{\text{d}}x}} =  - \frac{1}{{\sqrt {4 - {x^2}} }}\)

\(\frac{{{\text{d}}v}}{{{\text{d}}x}} = 1 \Rightarrow v = x\)     (A1)

\(\int_0^1 {{\text{arccos}}\left( {\frac{x}{2}} \right){\text{d}}x}  = \left[ {x\,\,{\text{arccos}}\left( {\frac{x}{2}} \right)} \right]_0^1 + \int_0^1 {\frac{1}{{\sqrt {4 - {x^2}} }}} dx\)      A1

using integration by substitution or inspection      (M1)

\(\left[ {x\,\,{\text{arccos}}\left( {\frac{x}{2}} \right)} \right]_0^1 + \left[ { - {{\left( {4 - {x^2}} \right)}^{\frac{1}{2}}}} \right]_0^1\)      A1

Note: Award A1 for \({ - {{\left( {4 - {x^2}} \right)}^{\frac{1}{2}}}}\) or equivalent.

Note: Condone lack of limits to this point.

attempt to substitute limits into their integral     M1

\( = \frac{\pi }{3} - \sqrt 3  + 2\)     A1

[7 marks]