2022TrialShort answer

Q13 (a)

3 marks

Question

Find $\displaystyle\int\cos^{-1}x\,dx$ .

Try the question before revealing the answer.

Solution

I=\int \cos^{-1}x\,dx

u=\cos^{-1}x\qquad v'=1

u'=-\frac{1}{\sqrt{1-x^2}}\qquad v=x

I=x\cos^{-1}x-\int x\left(\frac{-1}{\sqrt{1-x^2}}\right)dx

=x\cos^{-1}x-\frac12\int\left(\frac{-2x}{\sqrt{1-x^2}}\right)dx

=x\cos^{-1}x-\frac12\int\left(-2x(1-x^2)^{-1/2}\right)dx

=x\cos^{-1}x-\frac12\left(\frac{(1-x^2)^{1/2}}{1/2}\right)+C

=x\cos^{-1}x-\sqrt{1-x^2}+C

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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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