2020HSCShort answer

Q16 (b) (i)

3 marks

Question

Let
$I_n=\int_0^{\pi/2}\sin^{2n+1}(2\theta)\,d\theta,\qquad n=0,1,\ldots.$

(i) Prove that
$I_n=\frac{2n}{2n+1}I_{n-1},\qquad n\geq1.$

Try the question before revealing the answer.

Solution

Criteria	Marks
Provides correct proof	3
Correctly applies integration by parts and attempts to reduce to integrals involving only $\sin(2\theta)$ , or equivalent merit	2
Attempts to use integration by parts, or equivalent merit	1

Sample answer:

I_n=\int_0^{\pi/2}\sin^{2n+1}(2\theta)\,d\theta

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

Using integration by parts with

u=\sin^{2n}(2\theta),\quad\frac{dv}{d\theta}=\sin2\theta,

\frac{du}{d\theta}=4n\sin^{2n-1}(2\theta)\cos2\theta,\quad v=-\frac{\cos(2\theta)}2

I_n=2n\left[\int_0^{\pi/2}\sin^{2n-1}(2\theta)\,d\theta-\int_0^{\pi/2}\sin^{2n+1}(2\theta)\,d\theta\right]

I_n(2n+1)=2nI_{n-1}

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