2020HSCShort answer

Q16 (b) (ii)

3 marks

Question

Using the result of part (i), deduce that
$I_n=\frac{2^{2n}(n!)^2}{(2n+1)!}.$

Try the question before revealing the answer.

Solution

Criteria	Marks
Provides correct solution	3
Writes $I_n$ in terms of $I_0$ and evaluates $I_0$ , or equivalent merit	2
Calculates $I_0$ <br>OR<br>Writes $I_n$ in terms of $I_{n-2}$ <br>OR<br>Equivalent merit	1

Sample answer:

Applying the reduction formula,

I_n=\frac{2n}{2n+1}I_{n-1}

=\frac{2n(2n-2)}{(2n+1)(2n-1)}I_{n-2}

=\frac{2n(2n-2)(2n-4)\cdots2}{(2n+1)(2n-1)(2n-3)\cdots3}\times\frac23I_0

=\frac{[2n(2n-2)\cdots2][2n(2n-2)\cdots2]\times I_0}{(2n+1)(2n)(2n-1)(2n-2)\cdots3\times2}

=\frac{2^{2n}(n!)^2\times I_0}{(2n+1)!}

Now

I_0=\int_0^{\pi/2}\sin(2\theta)\,d\theta

=-\frac12\cos(2\theta)\Big|_0^{\pi/2}=1

\therefore I_n=\frac{2^{2n}(n!)^2}{(2n+1)!}

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Syllabus alignment

CalculusWorking Mathematically

Further IntegrationApply substitution, partial fractions, integration by parts and recurrence relationships to theoretical and practical problems.

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