By using the substitution \(x = \sin t\) , find \(\int {\frac{{{x^3}}}{{\sqrt {1 - {x^2}} }}{\text{d}}x} \) .

\(x = \sin t,{\text{ d}}x = \cos t\,{\text{d}}t\)

\(\int {\frac{{{x^3}}}{{\sqrt {1 - {x^2}} }}{\text{d}}x}  = \int {\frac{{{{\sin }^3}t}}{{\sqrt {1 - {{\sin }^2}t} }}\cos t\,{\text{d}}t} \)     M1

\( = \int {{{\sin }^3}t\,{\text{d}}t} \)     (A1)

\( = \int {{{\sin }^2}t\sin t\,{\text{d}}t} \)

\( = \int {(1 - {{\cos }^2}t)\sin t\,{\text{d}}t} \)     M1A1

\( = \int {\sin t\,{\text{d}}t - \int {{{\cos }^2}t\sin t\,{\text{d}}t} } \)

\( = - \cos t + \frac{{{{\cos }^3}t}}{3} + C\)     A1A1

\( = - \sqrt {1 - {x^2}}  + \frac{1}{3}{\left( {\sqrt {1 - {x^2}} } \right)^3} + C\)     A1

\(\left( { = - \sqrt {1 - {x^2}} \left( {1 - \frac{1}{3}(1 - {x^2})} \right) + C} \right)\)

\(\left( { = - \frac{1}{3}\sqrt {1 - {x^2}} (2 + {x^2}) + C} \right)\)

[7 marks]

Just a few candidates got full marks in this question. Substitution was usually incorrectly done and lead to wrong results. A cosine term in the denominator was a popular error. Candidates often chose unhelpful trigonometric identities and attempted integration by parts. Results such as \(\int {{{\sin }^3}t\,{\text{d}}t = \frac{{{{\sin }^4}t}}{4} + C} \) were often seen along with other misconceptions concerning the manipulation/simplification of integrals were also noticed. Some candidates unsatisfactorily attempted to use \(\arcsin x\) . However, there were some good solutions involving an expression for the cube of \(\sin t\) in terms of \(\sin t\) and \(\sin 3t\) . Very few candidates re-expressed their final result in terms of x.