2022TrialShort answer

Q13 (b)

3 marks

Question

Use the substitution $t=\tan\frac x2$ , or otherwise, to evaluate
$\int_0^{\pi/2}\frac{1}{3\sin x-4\cos x+5}\,dx.$

Try the question before revealing the answer.

Solution

\int_0^{\pi/2}\frac{1}{3\sin x-4\cos x+5}\,dx =\int_0^1\frac{2}{3(2t/(1+t^2))-4((1-t^2)/(1+t^2))+5(1+t^2)/(1+t^2)}\frac{1}{1+t^2}\,dt

=\int_0^1\frac{2}{6t-4(1-t^2)+5(1+t^2)}\,dt

=\int_0^1\frac{2}{6t+1+9t^2}\,dt

=\frac{2}{9}\int_0^1\frac{1}{t^2+\frac23t+\frac19}\,dt

=\frac{2}{9}\int_0^1\frac{1}{\left(t+\frac13\right)^2}\,dt

=\frac{2}{9}\int_0^1\left(t+\frac13\right)^{-2}dt

=\frac{2}{9}\left[-\left(t+\frac13\right)^{-1}\right]_0^1

=-\frac{2}{9}\left[(1+\tfrac13)^{-1}-(0+\tfrac13)^{-1}\right]

=-\frac{2}{9}\left[\frac34-3\right]

=-\frac{2}{9}\left[-\frac94\right]

=\frac12

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Calculus

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

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