A-Level Mathematics May/June 2024 Q8(b): Hence find the exact value of ∫(from 0 to π/12) 3/((3 cos2x - √3sin 2x)²) dx, simplifyi…

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  From part (a), we have the identity: $3 \cos(2x) - \sqrt{3}\sin(2x) = 2\sqrt{3} \cos(2x + \frac{\pi}{6})$

  Using this result, the integral becomes: $I = \int_{0}^{\frac{\pi}{12}} \frac{3}{(3 \cos(2x) - \sqrt{3}\sin(2x))^2} \,dx = \int_{0}^{\frac{\pi}{12}} \frac{3}{(2\sqrt{3} \cos(2x + \frac{\pi}{6}))^2} \,dx$

  Simplifying the denominator and the constant factor: $I = \int_{0}^{\frac{\pi}{12}} \frac{3}{(2\sqrt{3})^2 \cos^2(2x + \frac{\pi}{6})} \,dx = \int_{0}^{\frac{\pi}{12}} \frac{3}{12 \cos^2(2x + \frac{\pi}{6})} \,dx$

  $I = \int_{0}^{\frac{\pi}{12}} \frac{1}{4} \sec^2(2x + \frac{\pi}{6}) \,dx$

  Now, we perform the integration: $I = \frac{1}{4} \left[ \frac{1}{2} \tan(2x + \frac{\pi}{6}) \right]_{0}^{\frac{\pi}{12}}$

  $I = \left[ \frac{1}{8} \tan(2x + \frac{\pi}{6}) \right]_{0}^{\frac{\pi}{12}}$

  Substitute the limits of integration: $I = \frac{1}{8} \tan(2(\frac{\pi}{12}) + \frac{\pi}{6}) - \frac{1}{8} \tan(2(0) + \frac{\pi}{6})$

  $I = \frac{1}{8} \tan(\frac{\pi}{6} + \frac{\pi}{6}) - \frac{1}{8} \tan(\frac{\pi}{6})$

  $I = \frac{1}{8} \tan(\frac{\pi}{3}) - \frac{1}{8} \tan(\frac{\pi}{6})$

  Using exact values, $\tan(\frac{\pi}{3}) = \sqrt{3}$ and $\tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}$: $I = \frac{1}{8} (\sqrt{3} - \frac{1}{\sqrt{3}})$

  To simplify, find a common denominator inside the bracket: $I = \frac{1}{8} \left( \frac{3 - 1}{\sqrt{3}} \right)$ $I = \frac{1}{8} \left( \frac{2}{\sqrt{3}} \right)$ $I = \frac{2}{8\sqrt{3}} = \frac{1}{4\sqrt{3}}$

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  How the marks are awarded

  • B1FT — Correctly substituting the R-form from part (a) and simplifying the integrand to the form A ∫ sec²(2x + α) dx, specifically ∫ (1/4) sec²(2x + π/6) dx.
  • B1FT — Correctly integrating sec²(2x + π/6) to get (1/2)tan(2x + π/6), remembering the factor of 1/2 from the chain rule.
  • B1FT — Combining the constant factors to obtain the correct integrated expression (1/8)tan(2x + π/6).
  • M1 — Correctly substituting the limits x = π/12 and $x = 0$ into the integrated expression and subtracting, leading to (1/8)tan(π/3) - (1/8)tan(π/6).
  • A1 — Evaluating the expression using exact trigonometric values and simplifying fully to the final answer 1/(4√3) or an equivalent single term.

  Common mistakes

  • Forgetting the factor of 1/2 from the chain rule when integrating sec²(2x + π/6), leading to a coefficient of 1/4 instead of 1/8.
  • Making errors with exact trigonometric values, such as swapping tan(π/3) and tan(π/6), or using a calculator in degrees mode.
  • Incorrectly simplifying the final expression, for example (√3 - 1/√3) becoming (√3 - √3)/√3 = 0.
  • Leaving the answer in an unsimplified form like (1/8)(√3 - 1/√3), which the question requires to be simplified.

  Examiner tip: Recognise that 'Hence' questions often involve using a harmonic form R cos(θ ± α) from part (a) to transform a complex integral into a standard form like ∫ sec²(u) du.

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