Using the substitution \(x = 2\sin \theta \) , show that\[\int {\sqrt {4 - {x^2}} } {\text{d}}x = Ax\sqrt {4 - {x^2}}  + B\arcsin \frac{x}{2} + {\text{constant ,}}\]where \(A\) and \(B\) are constants whose values you are required to find.

\(\int {\sqrt {4 - {x^2}} } {\text{d}}x\)

\(x = 2\sin \theta \)

\({\text{d}}x = 2\cos \theta {\text{d}}\theta \)     A1

\( = \int {\sqrt {4 - 4{{\sin }^2}\theta } }  \times 2\cos \theta {\text{d}}\theta \)     M1A1

\( = \int {2\cos \theta  \times } 2\cos \theta {\text{d}}\theta \)

\( = 4\int {{{\cos }^2}\theta {\text{d}}\theta } \)

now \(\int {{{\cos }^2}\theta {\text{d}}\theta } \)

\( = \int {\left( {\frac{1}{2}\cos 2\theta  + \frac{1}{2}} \right)} {\text{d}}\theta \)     M1A1

\( = \left( {\frac{{\sin 2\theta }}{4} + \frac{1}{2}\theta } \right)\)    A1

so original integral

\( = 2\sin 2\theta  + 2\theta \)

\( = 2\sin \theta \cos \theta  + 2\theta \)

\( = \left( {2 \times \frac{x}{2} \times \frac{{\sqrt {4 - {x^2}} }}{2}} \right) + 2\arcsin \left( {\frac{x}{2}} \right)\)

\( = \frac{{x\sqrt {4 - {x^2}} }}{2} + 2\arcsin \left( {\frac{x}{2}} \right) + C\)     A1A1

Note: Do not penalise omission of \(C\).

\(\left( {A = \frac{1}{2},{\text{ }}B = 2} \right)\)

[8 marks]

For many candidates this was an all or nothing question. Examiners were surprised at the number of candidates who were unable to change the variable in the integral using the given substitution. Another stumbling block, for some candidates, was a lack of care with the application of the trigonometric version of Pythagoras' Theorem to reduce the integrand to a multiple of \({\cos ^2}\theta \) . However, candidates who obtained the latter were generally successful in completing the question.