2022TrialShort answer

Q15 (b)

4 marks

Question

Consider
$I_n=\int_0^1\frac{x^n}{\sqrt{1-x}}\,dx.$
Use integration by parts to show that
$I_n=\frac{2n}{2n+1}I_{n-1}.$

Try the question before revealing the answer.

Solution

I_n=\int_0^1 \frac{x^n}{\sqrt{1-x}}\,dx

u=x^n \qquad v'=\frac{1}{\sqrt{1-x}}=(1-x)^{-\frac12}

u'=nx^{n-1} \qquad v=-2(1-x)^{\frac12}=-2\sqrt{1-x}

I_n= \left[-2x^n\sqrt{1-x}\right]_0^1 -\int_0^1 -2nx^{n-1}\sqrt{1-x}\,dx

I_n= 0+2n\int_0^1 \frac{x^{n-1}(1-x)}{\sqrt{1-x}}\,dx

I_n= 2n\int_0^1 \frac{x^{n-1}-x^n}{\sqrt{1-x}}\,dx

I_n=2n(I_{n-1}-I_n)

I_n=2nI_{n-1}-2nI_n

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

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

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Syllabus alignment

Calculus

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

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