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.