Find y in terms of x, given that \((1 + {x^3})\frac{{{\text{d}}y}}{{{\text{d}}x}} = 2{x^2}\tan y\) and \(y = \frac{\pi }{2}\) when x = 0.

\((1 + {x^3})\frac{{{\text{d}}y}}{{{\text{d}}x}} = 2{x^2}\tan y \Rightarrow \int {\frac{{{\text{d}}y}}{{\tan y}} = \int {\frac{{2{x^2}}}{{1 + {x^3}}}{\text{d}}x} } \)     M1

\(\int {\frac{{\cos y}}{{\sin y}}{\text{d}}y = \frac{2}{3}\int {\frac{{3{x^2}}}{{1 + {x^3}}}{\text{d}}x} } \)     (A1)(A1)

\(\ln \left| {\sin y} \right| = \frac{2}{3}\ln \left| {1 + {x^3}} \right| + C\)     A1A1

Notes: Do not penalize omission of modulus signs.

Do not penalize omission of constant at this stage.

EITHER

\(\ln \left| {\sin \frac{\pi }{2}} \right| = \frac{2}{3}\ln \left| 1 \right| + C \Rightarrow C = 0\)     M1

OR

\(\left| {\sin y} \right| = A{\left| {1 + {x^3}} \right|^{\frac{2}{3}}},{\text{ }}A = {{\text{e}}^C}\)

\(\left| {\sin \frac{\pi }{2}} \right| = A{\left| {1 + {0^3}} \right|^{\frac{2}{3}}} \Rightarrow A = 1\)     M1

THEN

\(y = \arcsin \left( {{{(1 + {x^3})}^{\frac{2}{3}}}} \right)\)     A1

Note: Award M0A0 if constant omitted earlier.

[7 marks]

Many candidates separated the variables correctly but were then unable to perform the integrations.