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Practice Paper 3 Question 11

Let \(I_n=\int_{-\pi/2}^{\pi/2} \cos^n\!x \,dx\) for any non-negative integer \(n.\) \(\;(i)\) Find a recursive expression for \(I_n.\) \((ii)\) Find the simplest non-recursive expression for \(I_{2n}\) that contains \(\binom{2n}{n},\) where \(\binom{n}{k}=\frac{n!}{k!(n-k)!}.\)

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Warm-up Questions

  1. Evaluate \(\int_{-\pi}^{\pi}x^2\,dx.\)
  2. Find \(\int\ln x \,dx.\)
  3. A geometric sequence starts with \(\frac{1}{5}, \frac{2}{15}, \frac{4}{45} \ldots.\) Find a recursive formula for the sequence and solve it to get the closed form.


  • Hint 1
    Try to split \(\cos^n x\) into some terms, one of which you can integrate easily.
  • Hint 2
    Have you tried integrating by parts?
  • Hint 3
    Integrate and simplify \(I_n\) as much as you can.
  • Hint 4
    Try to rewrite \(I_n\) as a recurrence relation.
  • Hint 5
    ... you may find a trigonometric identity helpful.
  • Hint 6
    Which initial term can you find to help solve the recurrence for \(I_{2n}?\)
  • Hint 7
    Try writing out and unwinding \(I_{2n}.\)
  • Hint 8
    ... remembering that your expression should contain \(\frac{(2n)!}{(n!)^2}.\)


Writing \(\cos^n x = \cos x \cdot \cos^{n-1} x\) allows us to use integration by parts to get \(I_n = [\sin x \cdot \cos^{n-1} x]_{-\pi/2}^{\pi/2} + \int_{-\pi/2}^{\pi/2} (n-1) \sin^2\!x \cos^{n-2}\!x \,dx.\) Simplifying using a trigonometric identity yields \(I_n = (n-1) \int_{-\pi/2}^{\pi/2} (\cos^{n-2} x-\cos^{n} x) \,dx,\) so \(I_n = (n-1)(I_{n-2}-I_{n}).\) Rearranging gives \(I_{n} = \frac{n-1}{n} I_{n-2}.\)

To solve this recurrence relation for \(I_{2n},\) we first find \(I_0=\int_{-\pi/2}^{\pi/2}dx=\pi.\) Unwinding the recursion we get \(I_{2n} = \frac{2n-1}{2n}\cdot\frac{2n-3}{2n-2}\cdot\frac{2n-5}{2n-4}\cdots\frac{1}{2}\cdot\pi\) which has the product of all odd terms in the numerator and the product of all even terms in the denominator. To get \((2n)!\) in the numerator, we need to fill in the even terms, so we multiply both the numerator and denominator with the denominator, which yields \(I_{2n} = \pi \cdot \frac{(2n)!}{2^2 4^2 6^2 \cdots (2n)^2}.\) Factorising \(2^2\) from each term in the denominator gives \(I_{2n}=\pi \cdot \frac{(2n)!}{2^{2n} (n!)^2}=\binom{2n}{n}\frac{\pi}{4^n}.\)

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